151 linear mathematics spring, 2018 crn 11979...

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151 Linear Mathematics Spring, 2018 CRN 11979 SECTION 100 Wednesday 6:30 p.m. – 9:35 p.m. Room 210 Contents: Syllabus. Recommended homework. Supplement: linear regression and matrices. Reviews for tests 1 – 3 with solutions.

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151 Linear Mathematics

Spring 2018 CRN 11979

SECTION 100 Wednesday

630 pm ndash 935 pm Room 210

Contents

Syllabus

Recommended homework

Supplement linear regression and matrices

Reviews for tests 1 ndash 3 with solutions

SYLLABUS Course MATH 151 Linear Mathematics Catalog description

Cartesian coordinates linear equations in two variables graphing lines systems of linear equations and inequalities Gauss-Jordan elimination matrices matrix addition and multiplication matrix inversion geometric solution of linear programming problems the Simplex method duality Prerequisite MATH 118 with a grade of ldquoCrdquo or better or MATH 161 (or higher) placement

Learning outcomes

Upon successful completion of this course students will be able to

1 Graph lines linear inequalities and systems of linear inequalities in the plane

2 Determine whether a system of linear equations is independent dependent or inconsistent and solve systems of linear equations using matrices

3 Solve linear programming problems graphically and using the simplex method

Instructor Dr Arkady Kitover Office 327 Email (the best way to contact me) akitoverccpedu or akitoverhotmailcom Web Page httpfacultyccpeduFACULTYakitover The web page contains the syllabus and the reviews with complete solutions Phone (215) 751-8723 Office hours MWR 500 ndash 630 pm

Book Community College of Philadelphia Linear Math (Math 151)

Course outline Straight Lines and Linear Functions Sections 11 ndash 12 Systems of linear equations Section 21 Review 1 Test 1 Matrices and their applications Sections 23 ndash 26 Supplement linear regression and matrices Review 2 Test 2 Linear programming Chapters 3 and 4 Review 3 Test 3 Quizzes for extra credit will be given in class at the discretion of the instructor Grading I will add the results of three tests (100 points each) plus

what you earn on extra credit quizzes A 270 ndash 300 points B 240 ndash 269 points C 210 ndash 239 points D 180 ndash 209 points F 0 ndash 179 points Class Rules No food is allowed in the classroom Put your cell phones in the vibration mode before the class starts You may not use cell phones during a test You may not use electronic devices in the classroom for texting surfing the web et cetera Attendance will be taken every time the class meets people missing three classes or more without a valid reason will be dropped from the course

151 Linear Mathematics Review 1 1 The straight line goes through the points (-3 2) and (4 -1) Find the slope of the line

According to the formula 2

2 1

y ym 1

x xminus

=minus

we have 1 2 34 ( 3) 7

m minus minus minus= =

minus minus

2 Write a point-slope equation for the line from Problem 1 If we know the slope m and a point on the straight line we can write an equation in the point-slope form as If we use eg the point (-32) then we can write

( )a b(y b m x aminus = minus )

3 32 [ ( 3)] or 2 (7 7

y x y xminus = minus minus minus minus = minus + 3)

3 Write the slope-intercept equation for the line from Problem 1 We will solve the equation we got in the previous problem for y

3 3 9 14( 3) 27 7 7 7

y x x x= minus + + = minus minus + = minus +3 57 7

4 Write an equation of the line parallel to the line from Problem 1 and going through the point (13) Parallel lines have equal slopes and therefore we can write an equation of this parallel

line in the point-slope form 33 (7

y xminus = minus minus1) Solving for y we get an equation in the

slope-intercept form 3 3 3 237 7 7 7

y x x= minus + + = minus +4

5 Write an equation of the line perpendicular to the line from Problem 1 and going through the point (13) The slopes of perpendicular lines are negative reciprocals ( 1 2 1m m = minus ) and therefore the

slope of the perpendicular line is 73

and its equation in the point-slope form

is 73 ( 13

y xminus = minus ) Solving for y we get an equation in the slope-intercept form

7 7 733 3 3

y x x= minus + = +23

6 Graph the lines from Problems 1 and 5 in the same coordinate system

7 Find the point of intersection of the lines from Problems 1 and 5 We have to solve the system of equations

3 57 7

7 23 3

y x

y x

= minus +

= +

Because the left parts are equal so are the right parts 3 5 77 7 3 3

x xminus + = +2 Multiplying both parts by the common denominator 21 we get

9 15 49 14x xminus + = + or and58 1xminus = minus 158

x = After we plug this value of x into the firs

equation we get 3 1 5 287 77 58 7 406

y = minus times + = asymp

8 The straight line has the x-intercept -2 and the y-intercept 3 Write an equation of the

line We use the formula 1x ya b+ = where a and b are the x-intercept and the y-

intercept respectively In our case 12 3x y+ =

minus or 2 3y x 6minus =

9 If the price of a CD-player is $40 the demand is 4000 If the price of the same CD- player is $60 the demand is 3000 Assuming that the demand is a linear function of the price find the demand if the price is $55 First we need the slope-intercept equation of the line through the points (40 4000) and

(60 3000) The slope of the line is 3000 4000 1000 5060 40 20

m minus minus= = = minus

minus Using for example

the first point we can write an equation of the line in the point-slope form From here4000 50( 40)d minus = minus minusp 50 6000d p= minus + If

then 55p = 50 55 6000 3250d = minus times + =10 A small company produces dolls The permanent monthly expenses are $6000 and the Cost of producing a doll is $3 If the dolls sell for $10 each what is the breakeven point The cost of producing x dolls is ( ) 3 6000C x x= + The corresponding revenue is ( ) 10R x = x To find the break-even point we have to solve the equation ( ) ( )R x C x= or10 We have 3 6000x x= + 7 600x 0= or 6000 7 857x = asymp

151 Linear Mathematics Review 2 1 Solve the system

2 3 94 3 13 2 4

x y zx y zx y z

+ + =minus + = minus+ minus = minus

Solution The augmented matrix of the system is2 3 1 94 1 3 13 1 2 4

minus minus minus minus

To get rid of x in the

second equation we perform the operation 2 12R Rminus The result

is2 3 1 9 2 3 1 9

4 2 2 1 2 3 3 2 1 1 2 9 0 7 1 193 1 2 4 3 1 2 4

minus sdot minus minus sdot minus sdot minus minus sdot = minus minus minus minus minus minus

Next we get rid of x in

the third equation with the help of the operation 3 12 3R Rminus The result

is2 3 1 9 2 3 1 90 7 1 19 0 7 1 19

2 3 3 2 2 1 3 3 2( 2) 3 1 2( 4) 3 9 0 7 7 35

minus minus = minus minus sdot minus sdot sdot minus sdot minus minus sdot minus minus sdot minus minus minus

Finally to

bring the matrix to the triangular form we perform the operation 3 2R Rminus with the

result2 3 1 9 2 3 1 90 7 1 19 0 7 1 190 7 ( 7) 7 1 35 ( 19) 0 0 8 16

minus minus = minus minus minus minus minus minus minus minus minus minus minus minus

The corresponding

system is 2 3 9

7 198 16

x y zy z

z

+ + =minus + = minus

minus = minus

From the last equation we get 2z = Then we plug in this value of z into the second equation 7 2 19 7 21 3y y yminus + = minus minus = minus = Finally we plug in these values of y and z into the first equation 2 3 3 2 9 2 2 1x x x+ sdot + = = minus = minus The system has the unique solution 1 3 2x y z= minus = = Plugging in these numbers into the original system we check that the solution is correct 2 Show that the following system has no solutions

5 4 2 45 3 1715 5 3 24

x y zx y zx y z

minus + =+ minus =minus + =

Solution

2 1 3 1

3 2

5 4 2 4 5 4 2 4 5 4 2 45 3 1 17 0 7 3 13 3 0 7 3 13

15 5 3 24 15 5 3 24 0 7 3 12

5 4 2 40 7 3 130 0 0 1

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a contradiction 0 = 1 Therefore the system has no solutions 3 Show that the following system has infinitely many solutions and express its solutions in parametric form

3

2 3 5 58 7 13 21

x y zx y zx y z

minus + =+ minus =+ minus =

Solution

2 1 3 1

3 2

1 1 1 3 1 1 1 3 1 1 1 32 3 5 5 2 0 5 7 1 8 0 5 7 18 7 13 21 8 7 13 21 0 15 21 3

1 1 1 33 0 5 7 1

0 0 0 0

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a tautology 0 = 0 and we are left with two equations

35 7 1

x y zy z

minus + =minus = minus

Let us make z the parameter z t= where t can be any real number Then 7 15 7 1 5 7 1

5ty t y t y minus

minus = minus = minus =

and 7 1 7 1 15 5 7 1 2 143 35 5 5 5

t t t t tx t x tminus minus minus + minus +minus + = = minus + = = The solutions in

parametric form are 2 14

57 1

5

tx

ty

z t

+=

minus=

=

In Problems 4 ndash 6 let

1 2 5 1 8 10 5 33 0 2 4 2 1 0 0

A Bminus minus

= = minus minus minus

4 Compute 3 2A Bminus Solution

1 2 5 1 8 10 5 3 3 6 15 3 16 20 10 63 2 3 2

3 0 2 4 2 1 0 0 9 0 6 12 4 2 0 0

13 14 25 913 2 6 12

A Bminus minus minus minus

minus = minus = minus minus minus minus minus minus minus minus minus minus

= minus 5 Compute TAB Solution

8 21 2 5 1 10 13 0 2 4 5 0

3 0

1 8 2 10 5( 5) ( 1)3 1( 2) 2( 1) 5 0 ( 1)0 0 4

3 8 0 10 2( 5) ( 4)3 3( 2) 0( 1) 2 0 ( 4)0 2 6

TAB

minus minus minus = = minus minus

sdot + sdot + minus + minus minus + minus + sdot + minus minus = = sdot + sdot + minus + minus minus + minus + sdot + minus minus

6 Compute TA B Solution

1 3 1 8 3( 2) 1 10 3( 1) 1( 5) 3 0 1 3 3 02 0 8 10 5 3 2 8 0( 2) 2 10 0( 1) 2( 5) 0 0 2 3 0 05 2 2 1 0 0 5 8 2( 2) 5 10 2( 1) 5( 5) 2 0 5 3 2 01 4 ( 1)8 ( 4)( 2) ( 1)10 ( 4)( 1) ( 1)( 5) ( 4)0 ( 1)3 ( 4)0

sdot + minus sdot + minus minus + sdot sdot + sdot minus sdot + minus sdot + minus minus + sdot sdot + sdot = minus minus sdot + minus sdot + minus minus + sdot sdot + sdot minus minus minus + minus minus minus + minus minus minus minus + minus minus + minus

2 7 5 316 20 10 6

36 48 25 150 6 5 3

=

minus minus = minus minus minus

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

SYLLABUS Course MATH 151 Linear Mathematics Catalog description

Cartesian coordinates linear equations in two variables graphing lines systems of linear equations and inequalities Gauss-Jordan elimination matrices matrix addition and multiplication matrix inversion geometric solution of linear programming problems the Simplex method duality Prerequisite MATH 118 with a grade of ldquoCrdquo or better or MATH 161 (or higher) placement

Learning outcomes

Upon successful completion of this course students will be able to

1 Graph lines linear inequalities and systems of linear inequalities in the plane

2 Determine whether a system of linear equations is independent dependent or inconsistent and solve systems of linear equations using matrices

3 Solve linear programming problems graphically and using the simplex method

Instructor Dr Arkady Kitover Office 327 Email (the best way to contact me) akitoverccpedu or akitoverhotmailcom Web Page httpfacultyccpeduFACULTYakitover The web page contains the syllabus and the reviews with complete solutions Phone (215) 751-8723 Office hours MWR 500 ndash 630 pm

Book Community College of Philadelphia Linear Math (Math 151)

Course outline Straight Lines and Linear Functions Sections 11 ndash 12 Systems of linear equations Section 21 Review 1 Test 1 Matrices and their applications Sections 23 ndash 26 Supplement linear regression and matrices Review 2 Test 2 Linear programming Chapters 3 and 4 Review 3 Test 3 Quizzes for extra credit will be given in class at the discretion of the instructor Grading I will add the results of three tests (100 points each) plus

what you earn on extra credit quizzes A 270 ndash 300 points B 240 ndash 269 points C 210 ndash 239 points D 180 ndash 209 points F 0 ndash 179 points Class Rules No food is allowed in the classroom Put your cell phones in the vibration mode before the class starts You may not use cell phones during a test You may not use electronic devices in the classroom for texting surfing the web et cetera Attendance will be taken every time the class meets people missing three classes or more without a valid reason will be dropped from the course

151 Linear Mathematics Review 1 1 The straight line goes through the points (-3 2) and (4 -1) Find the slope of the line

According to the formula 2

2 1

y ym 1

x xminus

=minus

we have 1 2 34 ( 3) 7

m minus minus minus= =

minus minus

2 Write a point-slope equation for the line from Problem 1 If we know the slope m and a point on the straight line we can write an equation in the point-slope form as If we use eg the point (-32) then we can write

( )a b(y b m x aminus = minus )

3 32 [ ( 3)] or 2 (7 7

y x y xminus = minus minus minus minus = minus + 3)

3 Write the slope-intercept equation for the line from Problem 1 We will solve the equation we got in the previous problem for y

3 3 9 14( 3) 27 7 7 7

y x x x= minus + + = minus minus + = minus +3 57 7

4 Write an equation of the line parallel to the line from Problem 1 and going through the point (13) Parallel lines have equal slopes and therefore we can write an equation of this parallel

line in the point-slope form 33 (7

y xminus = minus minus1) Solving for y we get an equation in the

slope-intercept form 3 3 3 237 7 7 7

y x x= minus + + = minus +4

5 Write an equation of the line perpendicular to the line from Problem 1 and going through the point (13) The slopes of perpendicular lines are negative reciprocals ( 1 2 1m m = minus ) and therefore the

slope of the perpendicular line is 73

and its equation in the point-slope form

is 73 ( 13

y xminus = minus ) Solving for y we get an equation in the slope-intercept form

7 7 733 3 3

y x x= minus + = +23

6 Graph the lines from Problems 1 and 5 in the same coordinate system

7 Find the point of intersection of the lines from Problems 1 and 5 We have to solve the system of equations

3 57 7

7 23 3

y x

y x

= minus +

= +

Because the left parts are equal so are the right parts 3 5 77 7 3 3

x xminus + = +2 Multiplying both parts by the common denominator 21 we get

9 15 49 14x xminus + = + or and58 1xminus = minus 158

x = After we plug this value of x into the firs

equation we get 3 1 5 287 77 58 7 406

y = minus times + = asymp

8 The straight line has the x-intercept -2 and the y-intercept 3 Write an equation of the

line We use the formula 1x ya b+ = where a and b are the x-intercept and the y-

intercept respectively In our case 12 3x y+ =

minus or 2 3y x 6minus =

9 If the price of a CD-player is $40 the demand is 4000 If the price of the same CD- player is $60 the demand is 3000 Assuming that the demand is a linear function of the price find the demand if the price is $55 First we need the slope-intercept equation of the line through the points (40 4000) and

(60 3000) The slope of the line is 3000 4000 1000 5060 40 20

m minus minus= = = minus

minus Using for example

the first point we can write an equation of the line in the point-slope form From here4000 50( 40)d minus = minus minusp 50 6000d p= minus + If

then 55p = 50 55 6000 3250d = minus times + =10 A small company produces dolls The permanent monthly expenses are $6000 and the Cost of producing a doll is $3 If the dolls sell for $10 each what is the breakeven point The cost of producing x dolls is ( ) 3 6000C x x= + The corresponding revenue is ( ) 10R x = x To find the break-even point we have to solve the equation ( ) ( )R x C x= or10 We have 3 6000x x= + 7 600x 0= or 6000 7 857x = asymp

151 Linear Mathematics Review 2 1 Solve the system

2 3 94 3 13 2 4

x y zx y zx y z

+ + =minus + = minus+ minus = minus

Solution The augmented matrix of the system is2 3 1 94 1 3 13 1 2 4

minus minus minus minus

To get rid of x in the

second equation we perform the operation 2 12R Rminus The result

is2 3 1 9 2 3 1 9

4 2 2 1 2 3 3 2 1 1 2 9 0 7 1 193 1 2 4 3 1 2 4

minus sdot minus minus sdot minus sdot minus minus sdot = minus minus minus minus minus minus

Next we get rid of x in

the third equation with the help of the operation 3 12 3R Rminus The result

is2 3 1 9 2 3 1 90 7 1 19 0 7 1 19

2 3 3 2 2 1 3 3 2( 2) 3 1 2( 4) 3 9 0 7 7 35

minus minus = minus minus sdot minus sdot sdot minus sdot minus minus sdot minus minus sdot minus minus minus

Finally to

bring the matrix to the triangular form we perform the operation 3 2R Rminus with the

result2 3 1 9 2 3 1 90 7 1 19 0 7 1 190 7 ( 7) 7 1 35 ( 19) 0 0 8 16

minus minus = minus minus minus minus minus minus minus minus minus minus minus minus

The corresponding

system is 2 3 9

7 198 16

x y zy z

z

+ + =minus + = minus

minus = minus

From the last equation we get 2z = Then we plug in this value of z into the second equation 7 2 19 7 21 3y y yminus + = minus minus = minus = Finally we plug in these values of y and z into the first equation 2 3 3 2 9 2 2 1x x x+ sdot + = = minus = minus The system has the unique solution 1 3 2x y z= minus = = Plugging in these numbers into the original system we check that the solution is correct 2 Show that the following system has no solutions

5 4 2 45 3 1715 5 3 24

x y zx y zx y z

minus + =+ minus =minus + =

Solution

2 1 3 1

3 2

5 4 2 4 5 4 2 4 5 4 2 45 3 1 17 0 7 3 13 3 0 7 3 13

15 5 3 24 15 5 3 24 0 7 3 12

5 4 2 40 7 3 130 0 0 1

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a contradiction 0 = 1 Therefore the system has no solutions 3 Show that the following system has infinitely many solutions and express its solutions in parametric form

3

2 3 5 58 7 13 21

x y zx y zx y z

minus + =+ minus =+ minus =

Solution

2 1 3 1

3 2

1 1 1 3 1 1 1 3 1 1 1 32 3 5 5 2 0 5 7 1 8 0 5 7 18 7 13 21 8 7 13 21 0 15 21 3

1 1 1 33 0 5 7 1

0 0 0 0

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a tautology 0 = 0 and we are left with two equations

35 7 1

x y zy z

minus + =minus = minus

Let us make z the parameter z t= where t can be any real number Then 7 15 7 1 5 7 1

5ty t y t y minus

minus = minus = minus =

and 7 1 7 1 15 5 7 1 2 143 35 5 5 5

t t t t tx t x tminus minus minus + minus +minus + = = minus + = = The solutions in

parametric form are 2 14

57 1

5

tx

ty

z t

+=

minus=

=

In Problems 4 ndash 6 let

1 2 5 1 8 10 5 33 0 2 4 2 1 0 0

A Bminus minus

= = minus minus minus

4 Compute 3 2A Bminus Solution

1 2 5 1 8 10 5 3 3 6 15 3 16 20 10 63 2 3 2

3 0 2 4 2 1 0 0 9 0 6 12 4 2 0 0

13 14 25 913 2 6 12

A Bminus minus minus minus

minus = minus = minus minus minus minus minus minus minus minus minus minus

= minus 5 Compute TAB Solution

8 21 2 5 1 10 13 0 2 4 5 0

3 0

1 8 2 10 5( 5) ( 1)3 1( 2) 2( 1) 5 0 ( 1)0 0 4

3 8 0 10 2( 5) ( 4)3 3( 2) 0( 1) 2 0 ( 4)0 2 6

TAB

minus minus minus = = minus minus

sdot + sdot + minus + minus minus + minus + sdot + minus minus = = sdot + sdot + minus + minus minus + minus + sdot + minus minus

6 Compute TA B Solution

1 3 1 8 3( 2) 1 10 3( 1) 1( 5) 3 0 1 3 3 02 0 8 10 5 3 2 8 0( 2) 2 10 0( 1) 2( 5) 0 0 2 3 0 05 2 2 1 0 0 5 8 2( 2) 5 10 2( 1) 5( 5) 2 0 5 3 2 01 4 ( 1)8 ( 4)( 2) ( 1)10 ( 4)( 1) ( 1)( 5) ( 4)0 ( 1)3 ( 4)0

sdot + minus sdot + minus minus + sdot sdot + sdot minus sdot + minus sdot + minus minus + sdot sdot + sdot = minus minus sdot + minus sdot + minus minus + sdot sdot + sdot minus minus minus + minus minus minus + minus minus minus minus + minus minus + minus

2 7 5 316 20 10 6

36 48 25 150 6 5 3

=

minus minus = minus minus minus

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

Book Community College of Philadelphia Linear Math (Math 151)

Course outline Straight Lines and Linear Functions Sections 11 ndash 12 Systems of linear equations Section 21 Review 1 Test 1 Matrices and their applications Sections 23 ndash 26 Supplement linear regression and matrices Review 2 Test 2 Linear programming Chapters 3 and 4 Review 3 Test 3 Quizzes for extra credit will be given in class at the discretion of the instructor Grading I will add the results of three tests (100 points each) plus

what you earn on extra credit quizzes A 270 ndash 300 points B 240 ndash 269 points C 210 ndash 239 points D 180 ndash 209 points F 0 ndash 179 points Class Rules No food is allowed in the classroom Put your cell phones in the vibration mode before the class starts You may not use cell phones during a test You may not use electronic devices in the classroom for texting surfing the web et cetera Attendance will be taken every time the class meets people missing three classes or more without a valid reason will be dropped from the course

151 Linear Mathematics Review 1 1 The straight line goes through the points (-3 2) and (4 -1) Find the slope of the line

According to the formula 2

2 1

y ym 1

x xminus

=minus

we have 1 2 34 ( 3) 7

m minus minus minus= =

minus minus

2 Write a point-slope equation for the line from Problem 1 If we know the slope m and a point on the straight line we can write an equation in the point-slope form as If we use eg the point (-32) then we can write

( )a b(y b m x aminus = minus )

3 32 [ ( 3)] or 2 (7 7

y x y xminus = minus minus minus minus = minus + 3)

3 Write the slope-intercept equation for the line from Problem 1 We will solve the equation we got in the previous problem for y

3 3 9 14( 3) 27 7 7 7

y x x x= minus + + = minus minus + = minus +3 57 7

4 Write an equation of the line parallel to the line from Problem 1 and going through the point (13) Parallel lines have equal slopes and therefore we can write an equation of this parallel

line in the point-slope form 33 (7

y xminus = minus minus1) Solving for y we get an equation in the

slope-intercept form 3 3 3 237 7 7 7

y x x= minus + + = minus +4

5 Write an equation of the line perpendicular to the line from Problem 1 and going through the point (13) The slopes of perpendicular lines are negative reciprocals ( 1 2 1m m = minus ) and therefore the

slope of the perpendicular line is 73

and its equation in the point-slope form

is 73 ( 13

y xminus = minus ) Solving for y we get an equation in the slope-intercept form

7 7 733 3 3

y x x= minus + = +23

6 Graph the lines from Problems 1 and 5 in the same coordinate system

7 Find the point of intersection of the lines from Problems 1 and 5 We have to solve the system of equations

3 57 7

7 23 3

y x

y x

= minus +

= +

Because the left parts are equal so are the right parts 3 5 77 7 3 3

x xminus + = +2 Multiplying both parts by the common denominator 21 we get

9 15 49 14x xminus + = + or and58 1xminus = minus 158

x = After we plug this value of x into the firs

equation we get 3 1 5 287 77 58 7 406

y = minus times + = asymp

8 The straight line has the x-intercept -2 and the y-intercept 3 Write an equation of the

line We use the formula 1x ya b+ = where a and b are the x-intercept and the y-

intercept respectively In our case 12 3x y+ =

minus or 2 3y x 6minus =

9 If the price of a CD-player is $40 the demand is 4000 If the price of the same CD- player is $60 the demand is 3000 Assuming that the demand is a linear function of the price find the demand if the price is $55 First we need the slope-intercept equation of the line through the points (40 4000) and

(60 3000) The slope of the line is 3000 4000 1000 5060 40 20

m minus minus= = = minus

minus Using for example

the first point we can write an equation of the line in the point-slope form From here4000 50( 40)d minus = minus minusp 50 6000d p= minus + If

then 55p = 50 55 6000 3250d = minus times + =10 A small company produces dolls The permanent monthly expenses are $6000 and the Cost of producing a doll is $3 If the dolls sell for $10 each what is the breakeven point The cost of producing x dolls is ( ) 3 6000C x x= + The corresponding revenue is ( ) 10R x = x To find the break-even point we have to solve the equation ( ) ( )R x C x= or10 We have 3 6000x x= + 7 600x 0= or 6000 7 857x = asymp

151 Linear Mathematics Review 2 1 Solve the system

2 3 94 3 13 2 4

x y zx y zx y z

+ + =minus + = minus+ minus = minus

Solution The augmented matrix of the system is2 3 1 94 1 3 13 1 2 4

minus minus minus minus

To get rid of x in the

second equation we perform the operation 2 12R Rminus The result

is2 3 1 9 2 3 1 9

4 2 2 1 2 3 3 2 1 1 2 9 0 7 1 193 1 2 4 3 1 2 4

minus sdot minus minus sdot minus sdot minus minus sdot = minus minus minus minus minus minus

Next we get rid of x in

the third equation with the help of the operation 3 12 3R Rminus The result

is2 3 1 9 2 3 1 90 7 1 19 0 7 1 19

2 3 3 2 2 1 3 3 2( 2) 3 1 2( 4) 3 9 0 7 7 35

minus minus = minus minus sdot minus sdot sdot minus sdot minus minus sdot minus minus sdot minus minus minus

Finally to

bring the matrix to the triangular form we perform the operation 3 2R Rminus with the

result2 3 1 9 2 3 1 90 7 1 19 0 7 1 190 7 ( 7) 7 1 35 ( 19) 0 0 8 16

minus minus = minus minus minus minus minus minus minus minus minus minus minus minus

The corresponding

system is 2 3 9

7 198 16

x y zy z

z

+ + =minus + = minus

minus = minus

From the last equation we get 2z = Then we plug in this value of z into the second equation 7 2 19 7 21 3y y yminus + = minus minus = minus = Finally we plug in these values of y and z into the first equation 2 3 3 2 9 2 2 1x x x+ sdot + = = minus = minus The system has the unique solution 1 3 2x y z= minus = = Plugging in these numbers into the original system we check that the solution is correct 2 Show that the following system has no solutions

5 4 2 45 3 1715 5 3 24

x y zx y zx y z

minus + =+ minus =minus + =

Solution

2 1 3 1

3 2

5 4 2 4 5 4 2 4 5 4 2 45 3 1 17 0 7 3 13 3 0 7 3 13

15 5 3 24 15 5 3 24 0 7 3 12

5 4 2 40 7 3 130 0 0 1

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a contradiction 0 = 1 Therefore the system has no solutions 3 Show that the following system has infinitely many solutions and express its solutions in parametric form

3

2 3 5 58 7 13 21

x y zx y zx y z

minus + =+ minus =+ minus =

Solution

2 1 3 1

3 2

1 1 1 3 1 1 1 3 1 1 1 32 3 5 5 2 0 5 7 1 8 0 5 7 18 7 13 21 8 7 13 21 0 15 21 3

1 1 1 33 0 5 7 1

0 0 0 0

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a tautology 0 = 0 and we are left with two equations

35 7 1

x y zy z

minus + =minus = minus

Let us make z the parameter z t= where t can be any real number Then 7 15 7 1 5 7 1

5ty t y t y minus

minus = minus = minus =

and 7 1 7 1 15 5 7 1 2 143 35 5 5 5

t t t t tx t x tminus minus minus + minus +minus + = = minus + = = The solutions in

parametric form are 2 14

57 1

5

tx

ty

z t

+=

minus=

=

In Problems 4 ndash 6 let

1 2 5 1 8 10 5 33 0 2 4 2 1 0 0

A Bminus minus

= = minus minus minus

4 Compute 3 2A Bminus Solution

1 2 5 1 8 10 5 3 3 6 15 3 16 20 10 63 2 3 2

3 0 2 4 2 1 0 0 9 0 6 12 4 2 0 0

13 14 25 913 2 6 12

A Bminus minus minus minus

minus = minus = minus minus minus minus minus minus minus minus minus minus

= minus 5 Compute TAB Solution

8 21 2 5 1 10 13 0 2 4 5 0

3 0

1 8 2 10 5( 5) ( 1)3 1( 2) 2( 1) 5 0 ( 1)0 0 4

3 8 0 10 2( 5) ( 4)3 3( 2) 0( 1) 2 0 ( 4)0 2 6

TAB

minus minus minus = = minus minus

sdot + sdot + minus + minus minus + minus + sdot + minus minus = = sdot + sdot + minus + minus minus + minus + sdot + minus minus

6 Compute TA B Solution

1 3 1 8 3( 2) 1 10 3( 1) 1( 5) 3 0 1 3 3 02 0 8 10 5 3 2 8 0( 2) 2 10 0( 1) 2( 5) 0 0 2 3 0 05 2 2 1 0 0 5 8 2( 2) 5 10 2( 1) 5( 5) 2 0 5 3 2 01 4 ( 1)8 ( 4)( 2) ( 1)10 ( 4)( 1) ( 1)( 5) ( 4)0 ( 1)3 ( 4)0

sdot + minus sdot + minus minus + sdot sdot + sdot minus sdot + minus sdot + minus minus + sdot sdot + sdot = minus minus sdot + minus sdot + minus minus + sdot sdot + sdot minus minus minus + minus minus minus + minus minus minus minus + minus minus + minus

2 7 5 316 20 10 6

36 48 25 150 6 5 3

=

minus minus = minus minus minus

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

151 Linear Mathematics Review 1 1 The straight line goes through the points (-3 2) and (4 -1) Find the slope of the line

According to the formula 2

2 1

y ym 1

x xminus

=minus

we have 1 2 34 ( 3) 7

m minus minus minus= =

minus minus

2 Write a point-slope equation for the line from Problem 1 If we know the slope m and a point on the straight line we can write an equation in the point-slope form as If we use eg the point (-32) then we can write

( )a b(y b m x aminus = minus )

3 32 [ ( 3)] or 2 (7 7

y x y xminus = minus minus minus minus = minus + 3)

3 Write the slope-intercept equation for the line from Problem 1 We will solve the equation we got in the previous problem for y

3 3 9 14( 3) 27 7 7 7

y x x x= minus + + = minus minus + = minus +3 57 7

4 Write an equation of the line parallel to the line from Problem 1 and going through the point (13) Parallel lines have equal slopes and therefore we can write an equation of this parallel

line in the point-slope form 33 (7

y xminus = minus minus1) Solving for y we get an equation in the

slope-intercept form 3 3 3 237 7 7 7

y x x= minus + + = minus +4

5 Write an equation of the line perpendicular to the line from Problem 1 and going through the point (13) The slopes of perpendicular lines are negative reciprocals ( 1 2 1m m = minus ) and therefore the

slope of the perpendicular line is 73

and its equation in the point-slope form

is 73 ( 13

y xminus = minus ) Solving for y we get an equation in the slope-intercept form

7 7 733 3 3

y x x= minus + = +23

6 Graph the lines from Problems 1 and 5 in the same coordinate system

7 Find the point of intersection of the lines from Problems 1 and 5 We have to solve the system of equations

3 57 7

7 23 3

y x

y x

= minus +

= +

Because the left parts are equal so are the right parts 3 5 77 7 3 3

x xminus + = +2 Multiplying both parts by the common denominator 21 we get

9 15 49 14x xminus + = + or and58 1xminus = minus 158

x = After we plug this value of x into the firs

equation we get 3 1 5 287 77 58 7 406

y = minus times + = asymp

8 The straight line has the x-intercept -2 and the y-intercept 3 Write an equation of the

line We use the formula 1x ya b+ = where a and b are the x-intercept and the y-

intercept respectively In our case 12 3x y+ =

minus or 2 3y x 6minus =

9 If the price of a CD-player is $40 the demand is 4000 If the price of the same CD- player is $60 the demand is 3000 Assuming that the demand is a linear function of the price find the demand if the price is $55 First we need the slope-intercept equation of the line through the points (40 4000) and

(60 3000) The slope of the line is 3000 4000 1000 5060 40 20

m minus minus= = = minus

minus Using for example

the first point we can write an equation of the line in the point-slope form From here4000 50( 40)d minus = minus minusp 50 6000d p= minus + If

then 55p = 50 55 6000 3250d = minus times + =10 A small company produces dolls The permanent monthly expenses are $6000 and the Cost of producing a doll is $3 If the dolls sell for $10 each what is the breakeven point The cost of producing x dolls is ( ) 3 6000C x x= + The corresponding revenue is ( ) 10R x = x To find the break-even point we have to solve the equation ( ) ( )R x C x= or10 We have 3 6000x x= + 7 600x 0= or 6000 7 857x = asymp

151 Linear Mathematics Review 2 1 Solve the system

2 3 94 3 13 2 4

x y zx y zx y z

+ + =minus + = minus+ minus = minus

Solution The augmented matrix of the system is2 3 1 94 1 3 13 1 2 4

minus minus minus minus

To get rid of x in the

second equation we perform the operation 2 12R Rminus The result

is2 3 1 9 2 3 1 9

4 2 2 1 2 3 3 2 1 1 2 9 0 7 1 193 1 2 4 3 1 2 4

minus sdot minus minus sdot minus sdot minus minus sdot = minus minus minus minus minus minus

Next we get rid of x in

the third equation with the help of the operation 3 12 3R Rminus The result

is2 3 1 9 2 3 1 90 7 1 19 0 7 1 19

2 3 3 2 2 1 3 3 2( 2) 3 1 2( 4) 3 9 0 7 7 35

minus minus = minus minus sdot minus sdot sdot minus sdot minus minus sdot minus minus sdot minus minus minus

Finally to

bring the matrix to the triangular form we perform the operation 3 2R Rminus with the

result2 3 1 9 2 3 1 90 7 1 19 0 7 1 190 7 ( 7) 7 1 35 ( 19) 0 0 8 16

minus minus = minus minus minus minus minus minus minus minus minus minus minus minus

The corresponding

system is 2 3 9

7 198 16

x y zy z

z

+ + =minus + = minus

minus = minus

From the last equation we get 2z = Then we plug in this value of z into the second equation 7 2 19 7 21 3y y yminus + = minus minus = minus = Finally we plug in these values of y and z into the first equation 2 3 3 2 9 2 2 1x x x+ sdot + = = minus = minus The system has the unique solution 1 3 2x y z= minus = = Plugging in these numbers into the original system we check that the solution is correct 2 Show that the following system has no solutions

5 4 2 45 3 1715 5 3 24

x y zx y zx y z

minus + =+ minus =minus + =

Solution

2 1 3 1

3 2

5 4 2 4 5 4 2 4 5 4 2 45 3 1 17 0 7 3 13 3 0 7 3 13

15 5 3 24 15 5 3 24 0 7 3 12

5 4 2 40 7 3 130 0 0 1

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a contradiction 0 = 1 Therefore the system has no solutions 3 Show that the following system has infinitely many solutions and express its solutions in parametric form

3

2 3 5 58 7 13 21

x y zx y zx y z

minus + =+ minus =+ minus =

Solution

2 1 3 1

3 2

1 1 1 3 1 1 1 3 1 1 1 32 3 5 5 2 0 5 7 1 8 0 5 7 18 7 13 21 8 7 13 21 0 15 21 3

1 1 1 33 0 5 7 1

0 0 0 0

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a tautology 0 = 0 and we are left with two equations

35 7 1

x y zy z

minus + =minus = minus

Let us make z the parameter z t= where t can be any real number Then 7 15 7 1 5 7 1

5ty t y t y minus

minus = minus = minus =

and 7 1 7 1 15 5 7 1 2 143 35 5 5 5

t t t t tx t x tminus minus minus + minus +minus + = = minus + = = The solutions in

parametric form are 2 14

57 1

5

tx

ty

z t

+=

minus=

=

In Problems 4 ndash 6 let

1 2 5 1 8 10 5 33 0 2 4 2 1 0 0

A Bminus minus

= = minus minus minus

4 Compute 3 2A Bminus Solution

1 2 5 1 8 10 5 3 3 6 15 3 16 20 10 63 2 3 2

3 0 2 4 2 1 0 0 9 0 6 12 4 2 0 0

13 14 25 913 2 6 12

A Bminus minus minus minus

minus = minus = minus minus minus minus minus minus minus minus minus minus

= minus 5 Compute TAB Solution

8 21 2 5 1 10 13 0 2 4 5 0

3 0

1 8 2 10 5( 5) ( 1)3 1( 2) 2( 1) 5 0 ( 1)0 0 4

3 8 0 10 2( 5) ( 4)3 3( 2) 0( 1) 2 0 ( 4)0 2 6

TAB

minus minus minus = = minus minus

sdot + sdot + minus + minus minus + minus + sdot + minus minus = = sdot + sdot + minus + minus minus + minus + sdot + minus minus

6 Compute TA B Solution

1 3 1 8 3( 2) 1 10 3( 1) 1( 5) 3 0 1 3 3 02 0 8 10 5 3 2 8 0( 2) 2 10 0( 1) 2( 5) 0 0 2 3 0 05 2 2 1 0 0 5 8 2( 2) 5 10 2( 1) 5( 5) 2 0 5 3 2 01 4 ( 1)8 ( 4)( 2) ( 1)10 ( 4)( 1) ( 1)( 5) ( 4)0 ( 1)3 ( 4)0

sdot + minus sdot + minus minus + sdot sdot + sdot minus sdot + minus sdot + minus minus + sdot sdot + sdot = minus minus sdot + minus sdot + minus minus + sdot sdot + sdot minus minus minus + minus minus minus + minus minus minus minus + minus minus + minus

2 7 5 316 20 10 6

36 48 25 150 6 5 3

=

minus minus = minus minus minus

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

6 Graph the lines from Problems 1 and 5 in the same coordinate system

7 Find the point of intersection of the lines from Problems 1 and 5 We have to solve the system of equations

3 57 7

7 23 3

y x

y x

= minus +

= +

Because the left parts are equal so are the right parts 3 5 77 7 3 3

x xminus + = +2 Multiplying both parts by the common denominator 21 we get

9 15 49 14x xminus + = + or and58 1xminus = minus 158

x = After we plug this value of x into the firs

equation we get 3 1 5 287 77 58 7 406

y = minus times + = asymp

8 The straight line has the x-intercept -2 and the y-intercept 3 Write an equation of the

line We use the formula 1x ya b+ = where a and b are the x-intercept and the y-

intercept respectively In our case 12 3x y+ =

minus or 2 3y x 6minus =

9 If the price of a CD-player is $40 the demand is 4000 If the price of the same CD- player is $60 the demand is 3000 Assuming that the demand is a linear function of the price find the demand if the price is $55 First we need the slope-intercept equation of the line through the points (40 4000) and

(60 3000) The slope of the line is 3000 4000 1000 5060 40 20

m minus minus= = = minus

minus Using for example

the first point we can write an equation of the line in the point-slope form From here4000 50( 40)d minus = minus minusp 50 6000d p= minus + If

then 55p = 50 55 6000 3250d = minus times + =10 A small company produces dolls The permanent monthly expenses are $6000 and the Cost of producing a doll is $3 If the dolls sell for $10 each what is the breakeven point The cost of producing x dolls is ( ) 3 6000C x x= + The corresponding revenue is ( ) 10R x = x To find the break-even point we have to solve the equation ( ) ( )R x C x= or10 We have 3 6000x x= + 7 600x 0= or 6000 7 857x = asymp

151 Linear Mathematics Review 2 1 Solve the system

2 3 94 3 13 2 4

x y zx y zx y z

+ + =minus + = minus+ minus = minus

Solution The augmented matrix of the system is2 3 1 94 1 3 13 1 2 4

minus minus minus minus

To get rid of x in the

second equation we perform the operation 2 12R Rminus The result

is2 3 1 9 2 3 1 9

4 2 2 1 2 3 3 2 1 1 2 9 0 7 1 193 1 2 4 3 1 2 4

minus sdot minus minus sdot minus sdot minus minus sdot = minus minus minus minus minus minus

Next we get rid of x in

the third equation with the help of the operation 3 12 3R Rminus The result

is2 3 1 9 2 3 1 90 7 1 19 0 7 1 19

2 3 3 2 2 1 3 3 2( 2) 3 1 2( 4) 3 9 0 7 7 35

minus minus = minus minus sdot minus sdot sdot minus sdot minus minus sdot minus minus sdot minus minus minus

Finally to

bring the matrix to the triangular form we perform the operation 3 2R Rminus with the

result2 3 1 9 2 3 1 90 7 1 19 0 7 1 190 7 ( 7) 7 1 35 ( 19) 0 0 8 16

minus minus = minus minus minus minus minus minus minus minus minus minus minus minus

The corresponding

system is 2 3 9

7 198 16

x y zy z

z

+ + =minus + = minus

minus = minus

From the last equation we get 2z = Then we plug in this value of z into the second equation 7 2 19 7 21 3y y yminus + = minus minus = minus = Finally we plug in these values of y and z into the first equation 2 3 3 2 9 2 2 1x x x+ sdot + = = minus = minus The system has the unique solution 1 3 2x y z= minus = = Plugging in these numbers into the original system we check that the solution is correct 2 Show that the following system has no solutions

5 4 2 45 3 1715 5 3 24

x y zx y zx y z

minus + =+ minus =minus + =

Solution

2 1 3 1

3 2

5 4 2 4 5 4 2 4 5 4 2 45 3 1 17 0 7 3 13 3 0 7 3 13

15 5 3 24 15 5 3 24 0 7 3 12

5 4 2 40 7 3 130 0 0 1

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a contradiction 0 = 1 Therefore the system has no solutions 3 Show that the following system has infinitely many solutions and express its solutions in parametric form

3

2 3 5 58 7 13 21

x y zx y zx y z

minus + =+ minus =+ minus =

Solution

2 1 3 1

3 2

1 1 1 3 1 1 1 3 1 1 1 32 3 5 5 2 0 5 7 1 8 0 5 7 18 7 13 21 8 7 13 21 0 15 21 3

1 1 1 33 0 5 7 1

0 0 0 0

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a tautology 0 = 0 and we are left with two equations

35 7 1

x y zy z

minus + =minus = minus

Let us make z the parameter z t= where t can be any real number Then 7 15 7 1 5 7 1

5ty t y t y minus

minus = minus = minus =

and 7 1 7 1 15 5 7 1 2 143 35 5 5 5

t t t t tx t x tminus minus minus + minus +minus + = = minus + = = The solutions in

parametric form are 2 14

57 1

5

tx

ty

z t

+=

minus=

=

In Problems 4 ndash 6 let

1 2 5 1 8 10 5 33 0 2 4 2 1 0 0

A Bminus minus

= = minus minus minus

4 Compute 3 2A Bminus Solution

1 2 5 1 8 10 5 3 3 6 15 3 16 20 10 63 2 3 2

3 0 2 4 2 1 0 0 9 0 6 12 4 2 0 0

13 14 25 913 2 6 12

A Bminus minus minus minus

minus = minus = minus minus minus minus minus minus minus minus minus minus

= minus 5 Compute TAB Solution

8 21 2 5 1 10 13 0 2 4 5 0

3 0

1 8 2 10 5( 5) ( 1)3 1( 2) 2( 1) 5 0 ( 1)0 0 4

3 8 0 10 2( 5) ( 4)3 3( 2) 0( 1) 2 0 ( 4)0 2 6

TAB

minus minus minus = = minus minus

sdot + sdot + minus + minus minus + minus + sdot + minus minus = = sdot + sdot + minus + minus minus + minus + sdot + minus minus

6 Compute TA B Solution

1 3 1 8 3( 2) 1 10 3( 1) 1( 5) 3 0 1 3 3 02 0 8 10 5 3 2 8 0( 2) 2 10 0( 1) 2( 5) 0 0 2 3 0 05 2 2 1 0 0 5 8 2( 2) 5 10 2( 1) 5( 5) 2 0 5 3 2 01 4 ( 1)8 ( 4)( 2) ( 1)10 ( 4)( 1) ( 1)( 5) ( 4)0 ( 1)3 ( 4)0

sdot + minus sdot + minus minus + sdot sdot + sdot minus sdot + minus sdot + minus minus + sdot sdot + sdot = minus minus sdot + minus sdot + minus minus + sdot sdot + sdot minus minus minus + minus minus minus + minus minus minus minus + minus minus + minus

2 7 5 316 20 10 6

36 48 25 150 6 5 3

=

minus minus = minus minus minus

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

9 If the price of a CD-player is $40 the demand is 4000 If the price of the same CD- player is $60 the demand is 3000 Assuming that the demand is a linear function of the price find the demand if the price is $55 First we need the slope-intercept equation of the line through the points (40 4000) and

(60 3000) The slope of the line is 3000 4000 1000 5060 40 20

m minus minus= = = minus

minus Using for example

the first point we can write an equation of the line in the point-slope form From here4000 50( 40)d minus = minus minusp 50 6000d p= minus + If

then 55p = 50 55 6000 3250d = minus times + =10 A small company produces dolls The permanent monthly expenses are $6000 and the Cost of producing a doll is $3 If the dolls sell for $10 each what is the breakeven point The cost of producing x dolls is ( ) 3 6000C x x= + The corresponding revenue is ( ) 10R x = x To find the break-even point we have to solve the equation ( ) ( )R x C x= or10 We have 3 6000x x= + 7 600x 0= or 6000 7 857x = asymp

151 Linear Mathematics Review 2 1 Solve the system

2 3 94 3 13 2 4

x y zx y zx y z

+ + =minus + = minus+ minus = minus

Solution The augmented matrix of the system is2 3 1 94 1 3 13 1 2 4

minus minus minus minus

To get rid of x in the

second equation we perform the operation 2 12R Rminus The result

is2 3 1 9 2 3 1 9

4 2 2 1 2 3 3 2 1 1 2 9 0 7 1 193 1 2 4 3 1 2 4

minus sdot minus minus sdot minus sdot minus minus sdot = minus minus minus minus minus minus

Next we get rid of x in

the third equation with the help of the operation 3 12 3R Rminus The result

is2 3 1 9 2 3 1 90 7 1 19 0 7 1 19

2 3 3 2 2 1 3 3 2( 2) 3 1 2( 4) 3 9 0 7 7 35

minus minus = minus minus sdot minus sdot sdot minus sdot minus minus sdot minus minus sdot minus minus minus

Finally to

bring the matrix to the triangular form we perform the operation 3 2R Rminus with the

result2 3 1 9 2 3 1 90 7 1 19 0 7 1 190 7 ( 7) 7 1 35 ( 19) 0 0 8 16

minus minus = minus minus minus minus minus minus minus minus minus minus minus minus

The corresponding

system is 2 3 9

7 198 16

x y zy z

z

+ + =minus + = minus

minus = minus

From the last equation we get 2z = Then we plug in this value of z into the second equation 7 2 19 7 21 3y y yminus + = minus minus = minus = Finally we plug in these values of y and z into the first equation 2 3 3 2 9 2 2 1x x x+ sdot + = = minus = minus The system has the unique solution 1 3 2x y z= minus = = Plugging in these numbers into the original system we check that the solution is correct 2 Show that the following system has no solutions

5 4 2 45 3 1715 5 3 24

x y zx y zx y z

minus + =+ minus =minus + =

Solution

2 1 3 1

3 2

5 4 2 4 5 4 2 4 5 4 2 45 3 1 17 0 7 3 13 3 0 7 3 13

15 5 3 24 15 5 3 24 0 7 3 12

5 4 2 40 7 3 130 0 0 1

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a contradiction 0 = 1 Therefore the system has no solutions 3 Show that the following system has infinitely many solutions and express its solutions in parametric form

3

2 3 5 58 7 13 21

x y zx y zx y z

minus + =+ minus =+ minus =

Solution

2 1 3 1

3 2

1 1 1 3 1 1 1 3 1 1 1 32 3 5 5 2 0 5 7 1 8 0 5 7 18 7 13 21 8 7 13 21 0 15 21 3

1 1 1 33 0 5 7 1

0 0 0 0

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a tautology 0 = 0 and we are left with two equations

35 7 1

x y zy z

minus + =minus = minus

Let us make z the parameter z t= where t can be any real number Then 7 15 7 1 5 7 1

5ty t y t y minus

minus = minus = minus =

and 7 1 7 1 15 5 7 1 2 143 35 5 5 5

t t t t tx t x tminus minus minus + minus +minus + = = minus + = = The solutions in

parametric form are 2 14

57 1

5

tx

ty

z t

+=

minus=

=

In Problems 4 ndash 6 let

1 2 5 1 8 10 5 33 0 2 4 2 1 0 0

A Bminus minus

= = minus minus minus

4 Compute 3 2A Bminus Solution

1 2 5 1 8 10 5 3 3 6 15 3 16 20 10 63 2 3 2

3 0 2 4 2 1 0 0 9 0 6 12 4 2 0 0

13 14 25 913 2 6 12

A Bminus minus minus minus

minus = minus = minus minus minus minus minus minus minus minus minus minus

= minus 5 Compute TAB Solution

8 21 2 5 1 10 13 0 2 4 5 0

3 0

1 8 2 10 5( 5) ( 1)3 1( 2) 2( 1) 5 0 ( 1)0 0 4

3 8 0 10 2( 5) ( 4)3 3( 2) 0( 1) 2 0 ( 4)0 2 6

TAB

minus minus minus = = minus minus

sdot + sdot + minus + minus minus + minus + sdot + minus minus = = sdot + sdot + minus + minus minus + minus + sdot + minus minus

6 Compute TA B Solution

1 3 1 8 3( 2) 1 10 3( 1) 1( 5) 3 0 1 3 3 02 0 8 10 5 3 2 8 0( 2) 2 10 0( 1) 2( 5) 0 0 2 3 0 05 2 2 1 0 0 5 8 2( 2) 5 10 2( 1) 5( 5) 2 0 5 3 2 01 4 ( 1)8 ( 4)( 2) ( 1)10 ( 4)( 1) ( 1)( 5) ( 4)0 ( 1)3 ( 4)0

sdot + minus sdot + minus minus + sdot sdot + sdot minus sdot + minus sdot + minus minus + sdot sdot + sdot = minus minus sdot + minus sdot + minus minus + sdot sdot + sdot minus minus minus + minus minus minus + minus minus minus minus + minus minus + minus

2 7 5 316 20 10 6

36 48 25 150 6 5 3

=

minus minus = minus minus minus

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

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AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

151 Linear Mathematics Review 2 1 Solve the system

2 3 94 3 13 2 4

x y zx y zx y z

+ + =minus + = minus+ minus = minus

Solution The augmented matrix of the system is2 3 1 94 1 3 13 1 2 4

minus minus minus minus

To get rid of x in the

second equation we perform the operation 2 12R Rminus The result

is2 3 1 9 2 3 1 9

4 2 2 1 2 3 3 2 1 1 2 9 0 7 1 193 1 2 4 3 1 2 4

minus sdot minus minus sdot minus sdot minus minus sdot = minus minus minus minus minus minus

Next we get rid of x in

the third equation with the help of the operation 3 12 3R Rminus The result

is2 3 1 9 2 3 1 90 7 1 19 0 7 1 19

2 3 3 2 2 1 3 3 2( 2) 3 1 2( 4) 3 9 0 7 7 35

minus minus = minus minus sdot minus sdot sdot minus sdot minus minus sdot minus minus sdot minus minus minus

Finally to

bring the matrix to the triangular form we perform the operation 3 2R Rminus with the

result2 3 1 9 2 3 1 90 7 1 19 0 7 1 190 7 ( 7) 7 1 35 ( 19) 0 0 8 16

minus minus = minus minus minus minus minus minus minus minus minus minus minus minus

The corresponding

system is 2 3 9

7 198 16

x y zy z

z

+ + =minus + = minus

minus = minus

From the last equation we get 2z = Then we plug in this value of z into the second equation 7 2 19 7 21 3y y yminus + = minus minus = minus = Finally we plug in these values of y and z into the first equation 2 3 3 2 9 2 2 1x x x+ sdot + = = minus = minus The system has the unique solution 1 3 2x y z= minus = = Plugging in these numbers into the original system we check that the solution is correct 2 Show that the following system has no solutions

5 4 2 45 3 1715 5 3 24

x y zx y zx y z

minus + =+ minus =minus + =

Solution

2 1 3 1

3 2

5 4 2 4 5 4 2 4 5 4 2 45 3 1 17 0 7 3 13 3 0 7 3 13

15 5 3 24 15 5 3 24 0 7 3 12

5 4 2 40 7 3 130 0 0 1

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a contradiction 0 = 1 Therefore the system has no solutions 3 Show that the following system has infinitely many solutions and express its solutions in parametric form

3

2 3 5 58 7 13 21

x y zx y zx y z

minus + =+ minus =+ minus =

Solution

2 1 3 1

3 2

1 1 1 3 1 1 1 3 1 1 1 32 3 5 5 2 0 5 7 1 8 0 5 7 18 7 13 21 8 7 13 21 0 15 21 3

1 1 1 33 0 5 7 1

0 0 0 0

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a tautology 0 = 0 and we are left with two equations

35 7 1

x y zy z

minus + =minus = minus

Let us make z the parameter z t= where t can be any real number Then 7 15 7 1 5 7 1

5ty t y t y minus

minus = minus = minus =

and 7 1 7 1 15 5 7 1 2 143 35 5 5 5

t t t t tx t x tminus minus minus + minus +minus + = = minus + = = The solutions in

parametric form are 2 14

57 1

5

tx

ty

z t

+=

minus=

=

In Problems 4 ndash 6 let

1 2 5 1 8 10 5 33 0 2 4 2 1 0 0

A Bminus minus

= = minus minus minus

4 Compute 3 2A Bminus Solution

1 2 5 1 8 10 5 3 3 6 15 3 16 20 10 63 2 3 2

3 0 2 4 2 1 0 0 9 0 6 12 4 2 0 0

13 14 25 913 2 6 12

A Bminus minus minus minus

minus = minus = minus minus minus minus minus minus minus minus minus minus

= minus 5 Compute TAB Solution

8 21 2 5 1 10 13 0 2 4 5 0

3 0

1 8 2 10 5( 5) ( 1)3 1( 2) 2( 1) 5 0 ( 1)0 0 4

3 8 0 10 2( 5) ( 4)3 3( 2) 0( 1) 2 0 ( 4)0 2 6

TAB

minus minus minus = = minus minus

sdot + sdot + minus + minus minus + minus + sdot + minus minus = = sdot + sdot + minus + minus minus + minus + sdot + minus minus

6 Compute TA B Solution

1 3 1 8 3( 2) 1 10 3( 1) 1( 5) 3 0 1 3 3 02 0 8 10 5 3 2 8 0( 2) 2 10 0( 1) 2( 5) 0 0 2 3 0 05 2 2 1 0 0 5 8 2( 2) 5 10 2( 1) 5( 5) 2 0 5 3 2 01 4 ( 1)8 ( 4)( 2) ( 1)10 ( 4)( 1) ( 1)( 5) ( 4)0 ( 1)3 ( 4)0

sdot + minus sdot + minus minus + sdot sdot + sdot minus sdot + minus sdot + minus minus + sdot sdot + sdot = minus minus sdot + minus sdot + minus minus + sdot sdot + sdot minus minus minus + minus minus minus + minus minus minus minus + minus minus + minus

2 7 5 316 20 10 6

36 48 25 150 6 5 3

=

minus minus = minus minus minus

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

Solution

2 1 3 1

3 2

5 4 2 4 5 4 2 4 5 4 2 45 3 1 17 0 7 3 13 3 0 7 3 13

15 5 3 24 15 5 3 24 0 7 3 12

5 4 2 40 7 3 130 0 0 1

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a contradiction 0 = 1 Therefore the system has no solutions 3 Show that the following system has infinitely many solutions and express its solutions in parametric form

3

2 3 5 58 7 13 21

x y zx y zx y z

minus + =+ minus =+ minus =

Solution

2 1 3 1

3 2

1 1 1 3 1 1 1 3 1 1 1 32 3 5 5 2 0 5 7 1 8 0 5 7 18 7 13 21 8 7 13 21 0 15 21 3

1 1 1 33 0 5 7 1

0 0 0 0

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus minus minus minus minus minus minus minus

minus minus minus minus

The last equation is a tautology 0 = 0 and we are left with two equations

35 7 1

x y zy z

minus + =minus = minus

Let us make z the parameter z t= where t can be any real number Then 7 15 7 1 5 7 1

5ty t y t y minus

minus = minus = minus =

and 7 1 7 1 15 5 7 1 2 143 35 5 5 5

t t t t tx t x tminus minus minus + minus +minus + = = minus + = = The solutions in

parametric form are 2 14

57 1

5

tx

ty

z t

+=

minus=

=

In Problems 4 ndash 6 let

1 2 5 1 8 10 5 33 0 2 4 2 1 0 0

A Bminus minus

= = minus minus minus

4 Compute 3 2A Bminus Solution

1 2 5 1 8 10 5 3 3 6 15 3 16 20 10 63 2 3 2

3 0 2 4 2 1 0 0 9 0 6 12 4 2 0 0

13 14 25 913 2 6 12

A Bminus minus minus minus

minus = minus = minus minus minus minus minus minus minus minus minus minus

= minus 5 Compute TAB Solution

8 21 2 5 1 10 13 0 2 4 5 0

3 0

1 8 2 10 5( 5) ( 1)3 1( 2) 2( 1) 5 0 ( 1)0 0 4

3 8 0 10 2( 5) ( 4)3 3( 2) 0( 1) 2 0 ( 4)0 2 6

TAB

minus minus minus = = minus minus

sdot + sdot + minus + minus minus + minus + sdot + minus minus = = sdot + sdot + minus + minus minus + minus + sdot + minus minus

6 Compute TA B Solution

1 3 1 8 3( 2) 1 10 3( 1) 1( 5) 3 0 1 3 3 02 0 8 10 5 3 2 8 0( 2) 2 10 0( 1) 2( 5) 0 0 2 3 0 05 2 2 1 0 0 5 8 2( 2) 5 10 2( 1) 5( 5) 2 0 5 3 2 01 4 ( 1)8 ( 4)( 2) ( 1)10 ( 4)( 1) ( 1)( 5) ( 4)0 ( 1)3 ( 4)0

sdot + minus sdot + minus minus + sdot sdot + sdot minus sdot + minus sdot + minus minus + sdot sdot + sdot = minus minus sdot + minus sdot + minus minus + sdot sdot + sdot minus minus minus + minus minus minus + minus minus minus minus + minus minus + minus

2 7 5 316 20 10 6

36 48 25 150 6 5 3

=

minus minus = minus minus minus

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

In Problems 4 ndash 6 let

1 2 5 1 8 10 5 33 0 2 4 2 1 0 0

A Bminus minus

= = minus minus minus

4 Compute 3 2A Bminus Solution

1 2 5 1 8 10 5 3 3 6 15 3 16 20 10 63 2 3 2

3 0 2 4 2 1 0 0 9 0 6 12 4 2 0 0

13 14 25 913 2 6 12

A Bminus minus minus minus

minus = minus = minus minus minus minus minus minus minus minus minus minus

= minus 5 Compute TAB Solution

8 21 2 5 1 10 13 0 2 4 5 0

3 0

1 8 2 10 5( 5) ( 1)3 1( 2) 2( 1) 5 0 ( 1)0 0 4

3 8 0 10 2( 5) ( 4)3 3( 2) 0( 1) 2 0 ( 4)0 2 6

TAB

minus minus minus = = minus minus

sdot + sdot + minus + minus minus + minus + sdot + minus minus = = sdot + sdot + minus + minus minus + minus + sdot + minus minus

6 Compute TA B Solution

1 3 1 8 3( 2) 1 10 3( 1) 1( 5) 3 0 1 3 3 02 0 8 10 5 3 2 8 0( 2) 2 10 0( 1) 2( 5) 0 0 2 3 0 05 2 2 1 0 0 5 8 2( 2) 5 10 2( 1) 5( 5) 2 0 5 3 2 01 4 ( 1)8 ( 4)( 2) ( 1)10 ( 4)( 1) ( 1)( 5) ( 4)0 ( 1)3 ( 4)0

sdot + minus sdot + minus minus + sdot sdot + sdot minus sdot + minus sdot + minus minus + sdot sdot + sdot = minus minus sdot + minus sdot + minus minus + sdot sdot + sdot minus minus minus + minus minus minus + minus minus minus minus + minus minus + minus

2 7 5 316 20 10 6

36 48 25 150 6 5 3

=

minus minus = minus minus minus

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

7 Let1 21 2

A = minus

Compute the inverse matrix 1Aminus

Solution Recall that if a b

Ac d

=

and det 0A ad bc= minus ne then the matrix A is invertible

and 1 1 d bA

c aad bcminus minus = minusminus

In our case 1 2 2( 1) 4ad bcminus = sdot minus minus = and

1 2 2 1 2 1 21 1 1 1 4 1 44

Aminus minus minus = =

8 Let1 1 11 2 31 4 9

A =

Compute 1Aminus

Solution We will solve the problem in two ways ndash by Gauss ndash Jordan elimination and using cofactors (when solving a similar problem on the test you can solve it the way you like) (a) Gauss ndash Jordan elimination

2 1 3 1

3 2 3

1 1 1 1 0 0 1 1 1 1 0 01 2 3 0 1 0 0 1 2 1 1 01 4 9 0 0 1 1 4 9 0 0 1

1 1 1 1 0 0 1 1 1 1 0 00 1 2 1 1 0 3 0 1 2 1 1 0 20 3 8 1 0 1 0 0 2 2 3 1

1 1 10 1 20 0 1

R R R R

R R R

rarr rarr

rarr rarr

minus minus minus

minus minus minus minus minus

2 3 1 3

1 2

1 0 0 1 1 1 1 0 01 1 0 2 0 1 0 3 4 1

1 3 2 1 2 0 0 1 1 3 2 1 2

1 1 0 0 3 2 1 2 1 0 0 3 5 2 1 20 1 0 3 4 1 0 1 0 3 4 1 0 0 1 1 3 2 1 2 0 0 1 1 3 2 1 2

R R R R

R R

rarr rarr

rarr

minus minus minus minus minus minus minus

minus minus minus minus minus minus minus minus minus

We got that the inverse matrix is

1

3 5 2 1 2 6 5 113 4 1 6 8 2 2

1 3 2 1 2 2 3 1Aminus

minus minus = minus minus = minus minus minus minus

It is a good idea to check our answer

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

1

6 5 1 1 1 11 6 8 2 1 2 32

2 3 1 1 4 9

6 1 ( 5)1 1 1 6 1 ( 5)2 1 4 6 1 ( 5)3 1 91 ( 6)1 8 1 ( 2) 1 ( 6)1 8 2 ( 2)4 ( 6)1 8 3 ( 2)92

2 1 ( 3)1 1 1 2 1 ( 3)2 1 4 2 1 ( 3)3 1 9

2 0 01 0 2 02

0 0 2

A Aminus

minus = minus minus = minus sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

= minus + sdot + minus sdot minus + sdot + minus minus + sdot + minus = sdot + minus + sdot sdot + minus + sdot sdot + minus + sdot

=1 0 00 1 0 0 0 1

=

(b) Method of cofactors We will use the notationa b

ad bcc d

= minus Recall also the rule of

signs for cofactors+ minus +minus + minus+ minus +

The matrix of cofactors CA can be computed in the

following way 2 3 1 3 1 24 9 1 9 1 4

1 1 1 1 1 14 9 1 9 1 4

1 1 1 1 1 12 3 1 3 1 2

2 9 3 4 (1 9 3 1) 1 4 2 1 6 6 2(1 9 1 4) 1 9 1 1 (1 4 1 1) 5 8 3 1 3 1 2 (1 3 1 1) 1 2 1 1 1 2 1

CA

minus

= minus minus = minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

= minus sdot minus sdot sdot minus sdot minus sdot minus sdot = minus minus sdot minus sdot minus sdot minus sdot sdot minus sdot minus

Next we compute the matrix ( )6 5 16 8 2

2 3 1

TCAminus

= minus minus minus

This matrix is proportional to the

matrix 1Aminus The product ( )6 5 1 1 1 1 2 0 06 8 2 1 2 3 0 2 0

2 3 1 1 4 9 0 0 2

TCA Aminus

= minus minus = minus

Therefore

( )1

3 5 2 1 21 3 4 1 2

1 3 2 1 2

TCA Aminus

minus = = minus minus minus

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

9 AN INPUT-OUTPUT MODEL FOR A THREE-SECTOR ECONOMY Consider the economy consisting of three sectors agriculture (A) manufacturing (M) and transportation (T) with an input-output matrix given by

04 01 0101 04 0302 02 02

A M TAMT

a Find the gross output of goods needed to satisfy a consumer demand for $200 million worth of agricultural products $100 million worth of manufactured products and $60 million worth of transportation b Find the value of goods and transportation consumed in the internal process of production in order to meet this gross output

Solution (a) The demand matrix D is20010060

D =

The matrix X of the gross output

required to satisfy this demand is given by the formula 1( )X I A Dminus= minus To find X we will

first compute1 0 0 04 01 01 06 01 010 1 0 01 04 03 01 06 030 0 1 02 02 02 02 02 08

I Aminus minus

minus = minus = minus minus minus minus

To find 1( )I A minusminus it is convenient to write 01I A Bminus = where6 1 11 6 32 2 8

Bminus minus

= minus minus minus minus

Recall that 1 1 1 1(01 ) (01) 10B B Bminus minus minus minus= = Therefore we will first find the matrix 1Bminus We will use the method of cofactors Let us compute the matrix of cofactors CB

6 3 1 3 1 62 8 2 8 2 2

1 1 6 1 6 12 8 2 8 2 2

1 1 6 1 6 16 3 1 3 1 6

6 8 ( 3)( 2) [( 1)8 ( 3)( 2)] ( 1)( 2) 6( 2)[( 1)8 ( 1)( 2)] 6 8 ( 1)( 2) [6( 2) ( 1)( 2)]( 1)( 3) ( 1)6 [6( 3)

CB

minus minus minus minus minus minus minus minus minus

minus minus minus minus = minus minus =

minus minus minus minus minus minus minus minus minus minus minus minus minus sdot minus minus minus minus minus minus minus minus minus minus minus minus

= minus minus minus minus minus sdot minus minus minus minus minus minus minus minusminus minus minus minus minus minus minus

42 14 1410 46 14

( 1)( 1)] 6 6 ( 1)( 1) 9 19 35

= minus minus sdot minus minus minus

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

The transpose to this matrix is ( )42 10 914 46 1914 14 35

TCB =

Next

( )6 1 1 42 10 91 6 3 14 46 192 2 8 14 14 35

6 42 1 14 1 14 6 10 1 46 1 14 6 9 1 19 1 35 21 42 6 14 3 14 1 10 6 46 3 14 1 9 6 19 3 352 42 2 14 8 14 2 10 2 46 8 14 2 9 2 19 8 35

TCB Bminus minus

= minus minus = minus minus

sdot minus sdot minus sdot sdot minus sdot minus sdot sdot minus sdot minus sdot = minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot = minus sdot minus sdot + sdot minus sdot minus sdot + sdot minus sdot minus sdot + sdot

24 0 00 224 00 0 224

Therefore

( )1

42 10 91 1 14 46 19

224 22414 14 35

TCB Bminus

= =

and

1 1

42 10 9 42 10 910 5( ) 10 14 46 19 14 46 19224 112

14 14 35 14 14 35I A Bminus minus

minus = = =

Finally

1

42 10 9 200 42 200 10 100 9 605 5( ) 14 46 19 100 14 200 46 100 19 60

112 11214 14 35 60 14 200 14 100 35 60

9940 443755 8540 38125

1126300 28125

X I A Dminus

sdot + sdot + sdot = minus = = sdot + sdot + sdot = sdot + sdot + sdot

= =

It means that in order to satisfy the demand the gross product in agriculture should be $44375 million in manufacturing - $38125 million and in transportation - $28125 million (b) The answer to part b is given by the formula

44375 200 2437538125 100 2812528125 60 22125

X D minus = minus =

whence the value of the goods internally consumed

is in agriculture ndash $24375 million in manufacturing ndash $28125 million and in transportation ndash $22125 million

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

10 Solve Problem 1 on page 8 in the supplement Display the scatter plot and find the regression line for the points (3 5) (4 6) (5 8) (6 10) and (7 9) Display the graph of the regression line on the scatter plot Solution We are looking for an equation of the regression line in the form y mx b= + If we could find the line going exactly through all the five points above then m and b would satisfy the following system of linear equations

3 54 65 86 107 9

m bm bm bm bm b

+ =+ =+ =+ =+ =

In the matrix form this system can be written as AX B= where 5 3 16 4 1

and 8 5 110 6 19 7 1

mX B A

b

= = =

But the points are not on the same straight line and the system AX B= has no exact solutions We will get the best approximate solution if we solve the system T TA AX A B= Let us compute the matrices involved in this system

2 2 2 2 2

3 14 1

3 4 5 6 7 3 4 5 6 7 5 1

1 1 1 1 1 1 1 1 1 16 17 1

135 253 4 5 6 7 3 4 5 6 7

25 53 4 5 6 7 1 1 1 1 1

56

3 4 5 6 7 3 5 4 6 5 8 6 10 7 98

1 1 1 1 1 5 6 8109

T T

T

A A A

A B

= = =

+ + + + + + + + = = + + + + + + + +

sdot + sdot + sdot + sdot + sdot = = + + +

202

10 9 38

= +

To solve the system T TA AX A B= notice that the inverse matrix ( ) 1TA Aminus

can be computed

as ( ) 1 5 25 5 25 01 051 1 25 135 25 135 05 27135 5 25 25 50

TA Aminus minus minus minus = = = minus minus minussdot minus sdot

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

Therefore ( ) 1 01 05 202 202 19 12( )

05 27 38 101 1026 16T TX A A A B

minus minus minus = = = = minus minus +

We get 12 16m b= = and the slope-intercept equation of the regression line is 12 16y x= +

The computer generated graph below shows the points and the regression line on the same screen

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

151 Linear Mathematics Review 3 Solve problems 1 and 2 using the geometric approach 1 A manufacturer has 240 370 and 180 pounds of wood plastic and steel respectively Product A requires 1 3 and 2 pounds of wood plastic and steel per unit respectively product B requires 3 4 and 1 pound per unit respectively How many units of each product should be made to maximize gross income in each of the following cases (a) Product A sells for $40 and B for $60 (b) Product A sells for $40 and B for $50 Solution First we will set our problem as a problem of linear programming Suppose that x units of product A and y units of product B are produced Then we have to maximize the gross income (a) 40 60G x y= + (b) 40 50G x y= + subject to the following constraints (wood) 3 240(plastic) 3 4 370(steel) 2 180

x yx yx y

+ le+ le+ le

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions below the lines 3 240 3 4 370 x y x y+ = + = and 2 180x y+ = For the first line the x-intercept is 240 and the y-intercept is 2403=80 For the second line the x-intercept is 3703=123+13 and the y-intercept is 3704=925 For the third line the x-intercept is 1802=90 and the y-intercept is 180 A computer generated graph of the region is shown below

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

The region has five vertices A B C D E We can see at once that the vertices

A B C have coordinates (0 0) (0 80) and (90 0) respectively To find the coordinates of the vertex D we have to solve the system of equations

3 2403 4 370

x yx y+ =+ =

We will solve it by elimination multiplying both parts of the first equation by 3 we get 3 9 7203 4 370

x yx y+ =+ =

Subtracting the second equation from the first one we have 5 350y = whence 70y = After we plug in this value of y into the first equation we get 210 240x + = and 30x = So the vertex D is at (30 70)

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

We will find the vertex E by solving the system 2 180

3 4 370x y

x y+ =+ =

We multiply both parts of the first equation by 4 8 4 7203 4 370

x yx y+ =+ =

and subtract the equations We get 5 350x = whence 70x = and from the first equation we see that 40y = So the vertex E is at (70 40) The next table shows the values of the objective function G at each vertex in case (a) and in case (b) Vertex x y ( ) 40 60a G x y= + ( ) 40 50b G x y= + A 0 0 0 0 B 0 80 4800 4000 C 30 70 5400 4700 D 70 40 5200 4800 E 90 0 3600 3600

In case (a) we have the best solution when 30 units of product A and 70 units of product B are produced generating the gross income of $5400

In case (b) we have the best solution when 70 units of product A and 40 units of product B are produced generating the gross income of $4800

2 Suppose that 8 12 and 9 units of proteins carbohydrates and fats respectively are the minimum weekly requirements for a person Food A contains 2 6 and 1 unit of proteins carbohydrates and fats respectively per pound food B contains 1 1 and 3 units respectively per pound How many pounds of each food type should be made to minimize cost while meeting the requirements in each of the following cases (a) Food A costs $340 per pound and B costs $160 per pound (b) Food A costs $300 per pound and B costs $160 per pound

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

Solution First we will set our problem as a problem of linear programming Suppose that x pounds of food A and y pounds of food B are made Then we have to minimize the cost (a) 340 160C x y= + (b) 300 160G x y= + subject to the following constraints (proteins) 2 8(carbohydrates) 6 12(fats) 3 9

x yx yx y

+ ge+ ge+ ge

and the standard constraints 00

xygege

We will now graph the above system of inequalities The inequalities show that our region is in the first quadrant and is the intersection of the regions above the lines 2 8 6 12 and 3 9x y x y x y+ = + = + = For the first line the x-intercept is 82=4 and the y-intercept is 8 For the second line the x-intercept is 126=2 and the y-intercept is 12 For the third line the x-intercept is 9 and the y-intercept is 93=3 A computer generated graph of the region is shown below

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

Our region has four vertices A B C D Vertex A has coordinates (0 12) and vertex D ndash coordinates (9 0) To find the coordinates of vertex B we solve the system

6 122 8

x yx y+ =+ =

Subtracting the second equation from the first one we get 4 4x = whence 1x = and 6y = To find the coordinates of vertex C we solve the system

3 92 8x y

x y+ =+ =

We multiply both parts of the first equation by 2 2 6 182 8

x yx y+ =+ =

And subtract the equations Then 5 10y = and 2y = Plugging this value of y for example into the second equation we get 2 2 6x + = and 3x = The next table shows the values of the objective function C at each vertex in case (a) and in case (b)

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

Vertex x y ( ) 340 160a C x y= + ( ) 300 160b C x y= + A 0 12 1920 1920 B 1 6 1300 1260 C 3 2 1340 1220 D 9 0 3060 2700

In case (a) we have the best solution when 1 pound of food A and 6 pounds of food B are made at the cost of $1300

In case (b) we have the best solution when 3 pounds of food A and 2 pounds of food B are made at the cost of $1220 Solve problems 3 and 4 using the simplex method 3 Maximize 1 2 32z x x x= + + subject to

1 2 3

1 2 3

1 2 3

2 3 80 60

0 0 0

x x xx x xx x x

+ + le+ + lege ge ge

This is a standard maximization problem of linear programming (see definition on page 220) We introduce two slack variables 1 2 380 2 3u x x x= minus minus minus and 1 2 360v x x x= minus minus minus Thus we obtain the following system of equations

1 2 3

1 2 3

1 2 3

1 2 3

2 3 80 60

2 00 0 0 0 0

x x x ux x x vz x x xx x x u v

+ + + =+ + + =

minus minus minus =ge ge ge ge ge

We start with the initial solution 1 2 3 0 0 80 60x x x z u v= = = = = = Notice that in this solution the basic variables 1 2 3 x x x are passive (equal to 0) and the slack variables u and v are active The initial tableau below corresponds to this solution

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

1x 2x 3x u v z u 2 3 1 1 0 0 80 v 1 1 1 0 1 0 60 z -1 -1 -2 0 0 1 0 We look for the pivot The smallest number in the z-row is -2 in the 3x -column so the pivot will be in this column Next we look at the positive ratios of numbers in the last column to the right and the numbers in the pivot column The ratio in the u-row is 801=80 and the ratio in the v-row is 601=60 Therefore the smallest ratio is in the v-row and the pivot is on the intersection of the v-row and the 3x -column It means that in the next solution the slack variable v will be passive (equal to 0) and the basic variable 3x will be active The pivot is in the bold print in the table above It is equal to 1 so we can start at once to eliminate the elements above and below the pivot We do it by performing the operations 1 2R Rminus and 3 22R R+ The result is the next tableau 1x 2x 3x u v z u 1 2 0 1 -1 0 20

3x 1 1 1 0 1 0 60 z 1 1 0 0 2 1 120 Because all the elements in the z-row are non-negative we got the best solution In this solution only one of the basic variables - 3x is active and the solution is

1

2

3

0060

120

xxxz

====

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

4 A candy store plans to make three mixtures I II and III out of three different candies A B and C Mixture I sells for $1 per pound and contains 30 of candy A 10 of

candy B and 60 of candy C Mixture II sells for $250 per pound and contains 50 of candy A 30 of candy B and 20 of candy C Mixture III sells for $3 per pound and

contains 60 of candy A 30 of candy B and 10 of candy C If the owner has available 900 pounds of candy A 750 pounds of candy B and 425 pounds of candy C how many pounds of each mixture should be made up in order to maximize revenue

First we have to write our problem as a standard maximization problem of linear programming Suppose x pounds of mixture I y pounds of mixture II and z pounds of mixture III are made up We have to maximize the objective function (revenue)

25 3R x y z= + + subject to the following constraints

(Candy A) 03 05 06 900x y z+ + le (Candy B) 01 03 03 750x y z+ + le (Candy C) 06 02 01 425x y z+ + le

and of course the standard constraints 0 0 0x y zge ge ge

We introduce three slack variables u v and w and write the following system of equations

03 05 06 90001 03 03 75006 02 01 425

25 3 0

x y z ux y z vx y z w

R x y z

+ + + =+ + + =+ + + =

minus minus minus =

From this system we construct the initial tableau below x y z u v w R u 03 05 06 1 0 0 0 900 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0 The smallest value in the R-row ndash negative 3 is in the z-column The corresponding ratios are 42501=4250 75003=2500 90006=1500 The smallest ratio corresponds to the u-row and the pivot is 06 We have to make the pivot equal to 1 so we perform the operation 1 06R divide The result is x y z u v w R u 05 56 1 53 0 0 0 1500 v 01 03 03 0 1 0 0 750 w 06 02 01 0 0 1 0 425 R -1 -25 -3 0 0 0 1 0

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

Next we eliminate the elements below the pivot with the help of the operations 2 1 3 103 01R R R Rminus minus and 4 13R R+ Keep in mind that the slack variable u becomes passive and its place is taken by the basic variable z x y z u v w R z 05 56 1 53 0 0 0 1500 v -005 005 0 -05 1 0 0 300 w 055 760 0 -16 0 1 0 275 R 05 0 0 0 0 0 1 4500 We have no negative numbers in the R-row therefore we got the best solution

0015004500

xyzR

====

Use duality to solve Problem 5 5 Minimize 1 2 38 6 7z x x x= + + subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 11

2 3 30 0 0

x x xx x xx x x

x x x

+ + geminus + + ge

minus + ge minusge ge ge

This is a standard minimization problem of linear programming (see definition on page 249) Using the method described in our book we will formulate and solve the dual problem which is a standard maximization problem The following table contains all the information about original problem In the first three rows we see the coefficients in the constraints and their right parts in the last row ndash coefficients of the objective function

1 2 1 1 -1 1 2 1 2 -3 1 -3 8 6 7

Reflecting the table above about the main diagonal we get the table for the dual problem

1 -1 2 8 2 1 -3 6 1 2 1 7 1 1 -3

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

We will write now the dual problem Maximize 1 2 33z y y yprime = + minus subject to

1 2 3

1 2 3

1 2 3

1 2 3

2 82 3 6

70 0 0

y y yy y y

y y yy y y

minus + le+ minus le+ + le

ge ge ge

We will solve the dual problem using the simplex method We introduce the slack variables (which we name as the variables in the original problem

1 2 3 and x x x ) and write the system

1 2 3 1

1 2 3 2

1 2 3 3

1 2 3

2 82 3 6

73 0

y y y xy y y x

y y y xz y y y

minus + + =+ minus + =+ + + =prime minus minus + =

The initial tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 2 1 -3 0 1 0 0 6 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0

We see that the smallest elements of zrsquo-row are in y1 and y2-columns The smallest positive ratio of the elements in last column to the right and the elements in these two columns is 62=3 Hence the pivot is on the intersection of y1-column and v-row We make the pivot equal to one by performing the operation 2 2R divide 1y 2y 3y x1 x2 x3 zrsquo x1 1 -1 2 1 0 0 0 8 x2 1 12 -32 0 12 0 0 3 x3 1 1 1 0 0 1 0 7 zrsquo -1 -1 3 0 0 0 1 0 Next we eliminate elements above and below the pivot by performing the operations 1 2 3 2 4 2 R R R R R Rminus minus + The slack variable v becomes passive and the basic variable y1 becomes active

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 1 12 -32 0 12 0 0 3 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 There is a negative number in the zrsquo-row which means that we did not reach the best solution yet Let us look for the new pivot The only negative number in the zrsquo-row is in y2-column The corresponding positive ratios are 3 (1 2) 6divide = and 4 (1 2) 8divide = Therefore the pivot is on the intersection of the y2-column and the y1-row First let us make the pivot equal to 1using the operation 2 2R times 1y 2y 3y x1 x2 x3 zrsquo x1 0 -32 72 1 -12 0 0 5 y1 2 1 -3 0 1 0 0 6 x3 0 12 52 0 -12 1 0 4 zrsquo 0 -12 32 0 12 0 1 3 Next we eliminate the elements above and below the pivot The corresponding operations are 1 2 3 2 4 2(3 2) (1 2) (1 2)R R R R R R+ minus + The variable y1 becomes passive and y2 becomes active The resulting tableau is 1y 2y 3y x1 x2 x3 zrsquo x1 3 0 -1 1 1 0 0 14

2y 2 1 -3 0 1 0 0 6 x3 -1 0 4 0 -1 1 0 3 zrsquo 1 0 0 0 1 0 1 6 We got the solution of the dual problem and we can read the values of the variables

1 2 3 and x x x of the original problem from the last tableau in the zrsquo-row in the u v and w-columns Also the value of z equals to the value of zrsquo We get

1

2

3

010

6

xxxz

====

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

28 Business industry and governments are interested in an~wers to such questions as How many cell phones will be needed in the next three years What will be the population growth of Boulder Colorado over the next five years Will the Midway School District need to build an additional elementary school within three years How much will the number of 18-wheccr~ on 1-35 increase over the next decade

Answers to these kinds of questions are important because it may take years to build the manufacturing plants schools or highway~ that will be needed In some cases information from past years can indicate a trend from which reasonshyable estimates of future growth can be obtained For example the problem of air quality has been addressed by cities in recent years The citicgt of Los Angeles and Long Beach California have made significant progress a~ ~hown by the following table which gives the number of days during the year that the Air Quality Index exceeded 100 As the index exceeds 100 the environment becomes unhealthy for sensitive groups

Year Days

1994 139

1995 113

1996 94 1997 60 1998 56

1999 27

When we plot this information using the year (1 for 1994 2 for 1995 and so on) for the x-value and the number of days for the y-value we get the graph of Figure 2-12

We call a graph of points such as this a scatter plot Notice that the points have a general approximate linear downward trend even though the points do not lie on a line Even when the pattern of data points is not exactly linear it may be useshyful to approximate their trend with a line so that we can estimate future behavior

There may be cases when we believe the variables arc related in a linear manshyner hut the data deviate from a line because (1) the data collected may not be acshycurate or (2) the assumption of a linear relationship is not valid In either case it may be useful to find the line that best approximates the trend and use it to obshytain additional information and make predictions

Figure 2-13 shows a line that approximates the trend of the points in Figshyure 2-12

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

2

150

bull

94 9S 96 97 98 99 Year

FIGURE 2-12

150

100

0

C - j E Z

50

97 98 99 100 Year

FIGURE 2-13

100

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

icent yeniii4lPis

-d 4

bull

d3

l8 linw hgrmion 3

P4 FIGURE 2-14

Although the line seems to be a reasonable repre~entation of the general trend we would like to know if this is the best approximation of the trend Mathematicians use the least squares line also called the regression line for the line that best fits the data In Figure 2-13 the line shown is the least squares or regression line y = -219x + 158 You have not been told how to find that equation Before doing so lets look at the idea behind the least squares procedure

In Figure 2-14 we look at a simple case where we have drawn a line in the general direction of the trend of the scatter plot of the four pOtnts PI Pz ~ and P4 bull For each of the points we have indicated the vertical distance from each point to the line and labeled the distances db dz dJ and d4bull The distances to points above the line will be positive and the distances to the points below the line will be negative The basic idea of the least square~ procedure is to find a line that somehow minimizes the entirety of these distance~ To do so find the line that gives the smallest possible sum of the squares of the ds - that is the line y mx + b that makes

d + d + d)2 + di

the least value possible

We find the values of m and b of the regression line y = mx + b from a sysshytem of two equations in the variables m and b Lets illustrate the procedure usshying the points (25) (37) (59) and (6 11) The scatter plot IS shown in Figshyure 2-15

The system takes the form

Am + Bb = C Dm + Eb = F

bull

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

4 (IIAPHR Z linear Systfms

b FIGURE 2-15

bull to

bull 8

bull 6

bull 4

2

I I I )I m

2 3 4 5 6

where

A = the sum of the squares of the x-coordinates in this case 22 + 32 + 52 + 62 = 74

B = the sum of the x-coordinates of the given points in this case 2 + 3 + 5 + 6 = 16

C = the sum of the products of the x- and y-coordinates of the given points in this case (2 X 5) + (3 X 7) + (5 X 9) + (6 X 11) = 142

D = B the sum of the x-coordinates 16

E = the number of given points in this case four points

F = the sum of the y-coordinates in this caseS + 7 + 9 + 11 = 32

Thus the solution to the system

74m + 16b = 142 16m + 4b = 32

gives the coefficients of the regression line that best fits the four given points The solution is m = 14 and b = 24 (be sure you can find the solution) and the linshyear regression line is y = l4x + 24 (see Figure 2-16)

The linear function y = mx + b is the least squares Hne for the points (Xlgt Yl) (X2 Y2) (xm Yn) when m and b are solutions of the system of equations

(X2 + xl + + x2)m + (XI + X2 + + xn)b = (XIYI + X2Y2 + + XnYn)

(Xl + X2 + + Xn)m + nb = (Yl + Y2 + + Yn)

Least Squares line

We now show you two ways to organize the data that makes it easier to keep track of the computations needed to find the coefficients of the system

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

28 linear hqr~ssioll 5

y FIGURE 2-16

_--t_--l_--II_--_1shy1 shy ~--I__bull X

2 3 4 5 6

Method I Form a table like the followmg

x X Z

2 3 5 6

5 7

9 II

4

9 25 36

10 2J 45 66

Sum 16 32 74 42

This gives the system 74m + 16b = 142 16m + 4b = 32

Method II We find the augmented matrix A of the system of equations with a matrix product For this example it is

2 1 5]= [2 3 5 6J 3 1 7 = [74 16 142J All 1 1 5 1 9 16 4 32[

6 1 11

We state the general case as follows Given the points (XI YI) (xz Yz) (xn Yn) the augmented matrix M of the system

Am + Bb = C Dm + Eb = F

whose solution gives the least squares line of best fit for the given points is the product

M = [~I X2 xnJ[~ ~ ~~l 1middot 1

Xn y

This method is useful when using a calculator with matrix operations bull

t

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

6 (PHR Z linw Ssttms

Lets use the data on the number of computers in use in the United States and work an example

The computer industry estimated that the number of computers in use in the United States for the years 1991 through 1995 was the following

Year Computers in Use (millions)

1991 620 1992 682

1993 765

1994 858

1995 962

Find the scatter plot and least squares line y = mx + b where x is the year (x = 1 for 1991 x = 2 for 1992 etc) and y is the number of computers in use

SOLUTION Figure 2-17 shows the scatter plot

We will find the coefficients to the system of equations using both methods illustrated

Method I

2x y x xy

1 620 620 2 682 4 l364

3 765 9 2295 4 858 16 3432 5 962 25 4810

Sum 15 3887 55 12521

y FIGURE 2-17

100

bull 90

bull 80

bull 70 bull

x 2 3 4 5

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

l8 tmdm 7

y FIGURE 2-18

100

90

80

70

T bull x 2 3 4

The system of equations is

55m + 15b = 12521 15m + 5b 3887

Multiplying the second equation by 3 we have the system

SSm + ISb = 12521 4Sm + ISb = 11661 (Subtract)

10m = 86 m =86

3887 15(86) b = 5 = 5194

The least squares line is y = 86x + 5194 its graph on the scatter plot is above Method II Using matrices the augmented matrix of the lea~t squares sysshy

tem is

1 620

2 3 4 682 _ [55 15 12521 ] 1A = [~ ~JI 1

765 - 151 1 1 5 3887858 1

5 1 962

Using this augmented matrix we obtain the same line as m method I y = 86x + 5194 bull

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

USHgt IIIlI

8 (HAPUR l linear Systems

28 X

Find the coefficients of the regression line to two decimiddot mals unless otherwise noted

Display the carter plot and find the regression line 1 for the points (3 5) (4 6) (5 8) (610) and (79) Display the graph or the regression line on the scatshyter plot

Display the gtcaUer plot and find the regression line 2 for the points (2 XL t4 6) (65) and (83) Disshyplay the graph of the regression line on the scatter plot

Display the scatter rIot and flOd the regression line 3 for the point~ (4 R) (575) (76) and (935) Display the graph of (he regression line on the scatshyter plot

Display the scatter plot and find the regression line 4 for the points (1012) (il 15) (1214) (13 16) and (1418) (a) Display the graph of the regression line on the

scatter plot (b) Set the window iO the x-range includes 20 and

find the point on the regression line for x = 20

5 The percentage of the United States population that IB uses the internet is given for selected years from

1997 through 2001

Year of US Population

1lt)97 222

1998 327

2000 444

2001 539

(a) Use x = 0 for 1997 x == 1 for 1998 and so on and l = percentage of population to find the regre~sion line for the data

(b) Use the regngtltslon line to estimate the percentshyage of user in 2005

(c) Use the regnision line to estimate when the percentage will reach 100

Display the scatter plot and find the regression line ~~ for the points (127 1i0) (95171) (153145) WI 6

and (191 122) Diiplay the graph of the regression line on the scatter plot

7 The following table gives the life expectancy at

bull birth for selected years of females born in thebull United States

Year To Age

1940 652 1960 731 1980 774

1996 791

2000 791

(a) Using x = 0 for 1940 x = 20 for 1960 and so on find the least squares line that fits these data

(b) Using the least squares line find the estimated life expectancy of a female born in 2010

(c) Use the least squares line to estimate when a newborns life expectancy will reach 100

8 The following table gives the suicide rate (number of suicides per 100000 population) for selected bull

years of the 15- to 24-year-old age group in the United States

Year Rate

1970 88

1980 123 1990 132 1996 120

1999 103

Use x == 0 for 1970 x == 10 for 1980 and so on (a) Find the least squares line [or these data (b) Use the least squares line to estimate the suicide

rate for the year 2015

amp9 The average monthly temperature (in Fahrenheit) iii in Springfield 1I1i~ois for January April and July is

January 246 Apnl 533 and July 765 (0) Find the scatter plot for the data (b) Find the regression line for the data (c) Graph the regression line on the scatter plot

Does the line seem to fit the data well (d) Use the regression line to estimate the average

monthly temperature for October (e) Is the October estimate realistic Whyor why not

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

tL0 The number of miles that passenger cars travel annushyIII ally in the United States is given for selected years

Miles Year (billions)

1960 0587 1965 0723

1970 0917

1975 1034

1980 1112

1985 1247

1990 1408

1996 1468

1999 1569

(a) Display the scatter plot for the data (b) Find the regression line for the data with the

coefficients to three decimal places (c) Based on the regression line when will the anshy

nual mileage reach 2 billion miles

tL1 The following table gives the birth rates per 1000 forIII Israel and the United States for five different years

Birth Rate Year Israel US

1975 282 140 1980 241 162 1985 235 157 1990 222 167

1998 200 144 2002 189 141

(a) Usingx = ofor 1975 x = 5 for 1980 and soon find the birth-rate regression lines for pach country

(b) Using the regression lines estimate when if ever the birth rate of the two countries will be the same

tI12 The percentage of the Us adult population who III smoked is given for selected years

Overall Males Females Year Population () () ()

1974 371 431 321 1980 332 376 293 1985 301 326 279 1990 255 284 228 1994 255 282 231 1999 233 252 216

l6 [xmim 9

~

(a) Find the regression Ill1c for the percentage of

males who smoke U~C x 0 for 1974 x = 6 for 1980 and so on

(b) Using the equation of the line estImate the pershycentage of males who will smoke In 2005

(c) Based on the regre5gtlon line will the percentage of males who smoke ever reach zero If so when Why do you thmk tlw 1~ or IS not realistic

(d) Find the regression line for thc percentage of feshymales who smoke

(e) Based on the regresamplon Imes will the percentshyage of females who make equal the percentage of males who smoke If ~o whcn

(f) Find the regression line for the percentage of the overall populatIOn who moke

(g) For which of the three groups IS the decline the greatest

13 Throughout a persons working career a portion of salary is withheld for SOCial ~ccunty Upon retireshyment a worker become eligIble for Social Security benefits The average monthly SOCial Security benefits for selected year arc

Year 1975 1980 19N5 1990 1995 2000

Benefits $146 $321 $412 $550 $672 $845

(a) Using these data find the least squares regresshysion line giving monthly benefits as a function of years since 1975 (x = 0 ror 1975)

(b) Use the linear function 0 estimate the average monthly benefit for 20 IO For 2030

14 The United Stated nationdl debt (m blllions of dolshylars) for selected years from i 980 IS shown in the folshylowing table

Year 1980 1985 1990 1995 2000

Debt 9302 19459 32333 49740 56742

(8) Find the least square regression hnear function giving national debt as a function of years since 1980 (x = 0 for 1980)

(b) Use the regression line to estimate the national debt in 2005 In 2015

(c) Assume 300 million (03 billion) people in the United States in 2015 What is the estimated per capita debt

15 The value of the Amencan dollar dechned during the second half of the 20th century The table below shows the number of dollars required to equal the 1975 dollar value For example it cost $320 in 2000 to purchase what cost $100 in 1975

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

10 (IIAPHR Z linw Syst~ms

Amount required to equal $1 in 1975 Year 1975 1980 1985 1990 1995 2000 2002

Dollar 100 153 200 243 283 320 329

(a) Find the least squaregt regression linear function giving the dollar required as a function of years since 1975 (x = 0 for 1975)

(b) Use the linear function to estimate the number of dollars required in 2010

(c) Use the linear function to estimate when it will require $1000 to equal the 1975 dollar

16 The poverty level for a family of four two parents with two children under lit for selected years from 1990 is

Year 1990

Poverty Threshold for a Family of Four

1995 2000 2002

Threshold $13300 $15500 $17500 $18200

(a) Find the least square linear regression function for threhold level lt)lt a function of years since 1990

(b) Use the linear function to estimate when the poverty threhold level will reach $25000

17 The United States per capita income for selected years is

Per Capita Income Year 1980 1985 1990 1995 2001

Income $77R7 $11013 $14387 $17227 $22851

(a) Find the least square~ linear regression function for per capita income as a function of years since 19RO

(b) Use the linear fllnc ion to estimate the per capita income kve for 2008

6nlXMil44 bull~1Jlrlf6l i 1

REGRESSION LINES

(c) Use the linear function to estimate when the per capita income level will reach $25000

18 Credit card companies aggressively compete for credit card holders The percentage of Americans having credit cards has increased in recent years as shown in the following table

Year

Percentage of Americans Having a Credit Card

1989 1992 1995 1998

Percentage 560 624 664 675

(8) Use these data to determine the linear least squares regression function with percentage having a credit card as a function of years since 1989 (x 0 for 1989)

(b) Use the linear function to estimate the percentshyage having a credit card in 1980

(c) Use the linear function to estimate the percentshyage having a credit card in 2005

(d) According to the function when will 100 of Americans have a credit card Is this reasonable

19 The median income for a four-person family for seshylected years is

Median Income Year 1988 1990 1992 1994 1996 1998

Income $39050 $41150 $44250 $47010 $51500 $56050

(a) Use these data to find the least squares linear function with income as a function of years since 1988

(b) Use the linear function to estimate the median income for 1985 (The actual was $32800)

(c) Use the linear function to estimate the median income for 2000 (The actual was $65500)

(d) Based on the linear function when will the meshydian income reach $75000

A TI graphing calculator can be used to find the equation of t~e regression line We illustrate with the points (25) (46) (67) and (79)

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

1 Enter Data Enter the points with the x-coordinates in the list L1 and the correspondmg y-coordinates inL2

2 Set the Horizontal and Vertical Seales Set Xmin Xmax scI Ymin Ymax and Yscl using IWINDOW Iin the same way it is used to set the screen for graphing functions

3 Define the Scatterplot Press ISTAT PLOT Ilt1 PLOTgt IENTERI You will see a screen sinrilar to the following

Selectscatterplot

This sequence of commands will give the following screens

LinRe9(ax+b) bull LinRe9(sx+b) LEDIT IDKlt TESTS1ll1-var-5tals L~ 2T2-Var Statso 3 Mecl-Mecl4 LinRe9(ax+b)SIGluadRe96CubicRe97jQuartRe9

The last screen indicates that the least squares line is y = 07288x + 3288

EXERCISES 1 Find the least squares line for the points (1522) (17 25) and (1827)

2 Find the least squares line for the points (21 44) (2440) (2638) and (30 32)

3 Find the least squares line for the points (32 57) (41 63) (5367) and (6072)

SCATTERPLOT The scatterplot of a set of points may be obtained by the following steps

Z8 [xmim 11

LinRe9 I=ax+b a= n88135593b328S135593

bull

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

12 (HAPHR Z Linm Systems

USING U(U

TO DRAW AND FIND ALINEAR REGRESSION LINE We can use features of EXCEL to obtain a scatter plot of points that are given find the equation of the regresshysion line and graph the line We illustrate with the points (13) (25) (49) (5 8)

bull For the given points enter x in cells A2A5 and y in cells B2B5

bull Select the Chart Wizard icon in the Ruler

Curt 1ftstlS here

t WIIuI_ Hel pound Ll -JH~jH01- bull -1 __1E

bull Under Chart type select XY Scatter then dick on the first graph in the first column in Chart Subtype

bull Click Next Be sure the cursor is in the Data Range box Select the cells containing the data points A2B5 in this case

~~

Ilm ISIIeel$AI2$B$5

_I~ Qu

bull Click Next and you will seeaplotofthe datapqm1$~gt You can now enter somenailies~ditibelsi

bull Click on Titlesat~~())OfthediaIogmiddotbolt bull In Chart titleent~tiltcentmune Ofthei~

-shybull

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

bullbull_ u (SOrl)

Oetol11 c=

l 8 blrfius 13

bull Remove the legend by clicking on Legend at the top of the dialog box and removmg the check mark byShow Legend

~-l ~o BottGM gt 6 OrO)) bull i Rlq bull Olen bull _

bull Click Finish

bull Under the Chart menu select Add Trendline

bull Under Type in the dialog box that appears select Linear

bull Under Options in the dialog box click Display equation on chart

bull IAddl

~middotT_____

oSell bull IlI=J GilIltpl _ horl

o IlItpl tmiddotod vol rI

bull Click OK

The line and its equation y = 14x + 205 are shown on the chart

CII

Regresston ~4X205

Ot~--~---~~~~~~~~~~~~~bull ~ ~

gt ~ shy-

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

AnoilcompanyoperatestworefineriesRefineryIhasanoutputof200100and100barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyRefineryIIhasanoutputof100200and600barrelsoflow-medium-andhigh-gradeoilperdayrespectivelyThecompany

wishestoproduceatleast10001400and3000barrelsoflowmedium-andhigh-gradeoilItcosts$200perdaytooperaterefineryIand$300perdaytooperaterefineryIIHowmanydayseachrefinery

shouldbeoperatedtofulfillthetaskataminimumcostLetxandybethenumberofdaysrefineryIandrefineryIIareoperatedrespectivelyWeneedtominimizetheobjectivefunction C = 200x + 300y subjectto

200x +100y ge1000100x + 200y ge1400100x + 600y ge 3000x ge 0 y ge 0

Toconstructthedualstandardmaximizationproblemwestartwiththetable

x y Constant200 100 1000100 200 1400100 600 3000200 300

Thenthetransposedmatrixprovides

u v w Constant200 100 100 200100 200 600 3001000 1400 3000

ThedualproblemnowcanbewrittenasMaximizeC = 1000u +1400v + 3000w Subjectto

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

200u +100v +100w le 200100u + 200v + 600w le 300

Theinitialtableauxis u v w x y Crsquo x 200 100 100 1 0 0 200y 100 200 600 0 1 0 300Crsquo -1000 -1400 -3000 0 0 1 0Thepivotisattheintersectionofw-columnandy-rowTomakepivotequalto1wedividethesecondrowby600 u v w x y Crsquo x 200 100 100 1 0 0 200y 16 13 1 0 1600 0 12Crsquo -1000 -1400 -3000 0 0 1 0Nextweperformtheoperations R1 minus100R2 and R3 + 3000R2 u v w x y Crsquo x 200 100 100 1 0 0 200w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Thenewpivotisattheintersectionofu-columnandx-rowWedividethepivotrowby200 u v w x y Crsquo x 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo -500 -400 0 0 5 1 1500Nextweperformtheoperations R2 minus (1 6)R1 and R3 + 500R1

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem

Thenewpivotisattheintersectionofv-columnandw-rowWemultiplythepivotrowby3 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 0 -150 250 52 5 1 2000Nextweperformtheoperation R3 +150R2 u v w x y Crsquo x 1 12 12 1200 0 0 1w 12 1 3 0 1200 0 32Crsquo 75 0 700 52 234 1 2225Becauseallthenumbersinthelastrowarenonnegativewehavesolvedthedualproblemandcanobtainthesolutionoftheoriginalproblem

x = 5 2y = 23 4C = 2225

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

u v w x y Crsquo u 1 12 1200 0 0 1w 16 13 1 0 1600 0 12Crsquo 0 -150 250 52 5 1 2000

  • SYLLABUS
  • Review1
  • Review2
  • Review3
  • Regression
    • Regression1pdf
    • Regression2
      • A minimization problem