copyright © by houghton mifflin company, all rights reserved. calculus concepts 2/e latorre,...

Copyright © by Houghton Mifflin Company, All rights reserved.Copyright © by Houghton Mifflin Company, All rights reserved.

Calculus Concepts 2/eCalculus Concepts 2/eLaTorre, Kenelly, Fetta, Harris, and CarpenterLaTorre, Kenelly, Fetta, Harris, and Carpenter

Chapter 10Chapter 10Analyzing Multivariable Change: OptimizationAnalyzing Multivariable Change: Optimization


Chapter 10 Key ConceptsChapter 10 Key Concepts• Multivariable Critical Points Multivariable Critical Points

• Multivariable OptimizationMultivariable Optimization

• Constrained OptimizationConstrained Optimization

• The Method of Least SquaresThe Method of Least Squares


Multivariable Critical PointsMultivariable Critical Points• Relative minimum: an output value smaller than Relative minimum: an output value smaller than

any of those near itany of those near it• Relative maximum: an output value larger than Relative maximum: an output value larger than

any of those near itany of those near it• Absolute minimum: an output value smaller than Absolute minimum: an output value smaller than

all of those in the regionall of those in the region• Absolute maximum: an output value larger than all Absolute maximum: an output value larger than all

of those in the regionof those in the region• Saddle point: an output that is a maximum of one Saddle point: an output that is a maximum of one

cross section and a minimum of anothercross section and a minimum of another


Multivariable Critical Points: ExampleMultivariable Critical Points: Example


Multivariable Critical Points: ExampleMultivariable Critical Points: Example

Contour curves near a saddle point curve away Contour curves near a saddle point curve away from the point.from the point.


Multivariable Critical Points: ExampleMultivariable Critical Points: ExampleThe percentage of sugar converted to olestra when the The percentage of sugar converted to olestra when the temperature is 150°C is shown below as a function of temperature is 150°C is shown below as a function of processing time and the ratio of peanut oil to sugar.processing time and the ratio of peanut oil to sugar.

The saddle point occurs when time is 11 hours and The saddle point occurs when time is 11 hours and the ratio is 11:1the ratio is 11:1


Critical Points: Exercise 10.1 #5Critical Points: Exercise 10.1 #5Make a dot on the contour plot at the approximate Make a dot on the contour plot at the approximate location of all critical points. Label points as relative location of all critical points. Label points as relative maximum, relative minimum, or saddle point.maximum, relative minimum, or saddle point.


Multivariable OptimizationMultivariable Optimization• Determinant TestDeterminant Test

– Let f be a continuous multivariable function with Let f be a continuous multivariable function with two input variables x and y. Let (a,b) be a point at two input variables x and y. Let (a,b) be a point at which the first partial derivatives of f are both 0. which the first partial derivatives of f are both 0. The determinant of the second partials matrix The determinant of the second partials matrix evaluated at the point (a,b) isevaluated at the point (a,b) is

yxxyyyxxyyyx

xyxx ffffffff

)b,a(D yxxyyyxxyyyx

xyxx ffffffff

)b,a(D

If D(a,b) > 0 and fIf D(a,b) > 0 and fxx xx < 0 at (a,b), f(a,b) is a maximum.< 0 at (a,b), f(a,b) is a maximum.

If D(a,b) > 0 and fIf D(a,b) > 0 and fxx xx > 0 at (a,b), f(a,b) is a minimum.> 0 at (a,b), f(a,b) is a minimum.

If D(a,b) < 0 at (a,b), f has a saddle point at (a,b).If D(a,b) < 0 at (a,b), f has a saddle point at (a,b).If D(a,b) = 0, the test fails.If D(a,b) = 0, the test fails.


Multivariable Optimization: ExampleMultivariable Optimization: Example The volume index for cake batter is given by The volume index for cake batter is given by

V(L, t) = -3.1LV(L, t) = -3.1L2 2 + 22.4L - 0.1t+ 22.4L - 0.1t22 + 5.3t + 5.3t when L grams of leavening is used and the cake is when L grams of leavening is used and the cake is baked at 177°C for t minutes. Find the maximum baked at 177°C for t minutes. Find the maximum volume possible and the conditions needed to volume possible and the conditions needed to achieve that volume. achieve that volume.

VVLL = -6.2L + 22.4 = 0 = -6.2L + 22.4 = 0 L L 3.6 grams of leavening 3.6 grams of leavening

VVtt = -0.2t + 5.3 = 0 = -0.2t + 5.3 = 0 t t 26.5 minutes baking time 26.5 minutes baking time

V(3.6, 26.5) V(3.6, 26.5) 110.7. 110.7.

Note: V = 100 is the uncooked batter volume.Note: V = 100 is the uncooked batter volume.


Multivariable Optimization: ExampleMultivariable Optimization: Example

The cross sections of V at L = 3.6 and t = 26.5 are The cross sections of V at L = 3.6 and t = 26.5 are both concave down which suggests the critical both concave down which suggests the critical point is a maximum.point is a maximum.


Multivariable Optimization: ExampleMultivariable Optimization: Example We must verify the volume is a maximum. We must verify the volume is a maximum. At (3.6, 26.5, 110.7),At (3.6, 26.5, 110.7),

VVLLLL = -6.2 = -6.2

VVtttt = -0.2 = -0.2

VVLt Lt = 0= 0

VVtLtL= 0 = 0

24.1)0)(0()2.0)(2.6(

2.0002.6

)5.26,6.3(D

24.1)0)(0()2.0)(2.6(

2.0002.6

)5.26,6.3(D

D(3.6, 26.5) > 0 and VD(3.6, 26.5) > 0 and VLL LL < 0 so (3.6, 26.5, 110.7) is < 0 so (3.6, 26.5, 110.7) is

a maximum.a maximum.


Optimization: Exercise 10.2 #9Optimization: Exercise 10.2 #9A restaurant mixes ground beef that costs $b per A restaurant mixes ground beef that costs $b per pound with pork sausage that costs $p per pound to pound with pork sausage that costs $p per pound to make a meat mixture used on its pizza. The make a meat mixture used on its pizza. The quarterly revenue from its pizza sales is given byquarterly revenue from its pizza sales is given byR(b,p) = 14b - 3bR(b,p) = 14b - 3b22 - bp - 2p - bp - 2p22 + 12p thousand dollars. + 12p thousand dollars.Find the prices at which the restaurant should try to Find the prices at which the restaurant should try to purchase the beef and sausage in order to purchase the beef and sausage in order to maximize quarterly revenue.maximize quarterly revenue.

RRbb = 14 - 6b - p = 0 = 14 - 6b - p = 0

RRpp = -b - 4p + 12 = 0 = -b - 4p + 12 = 0

Solving the system of equations gives b Solving the system of equations gives b $1.91 and $1.91 and p p $2.52. The critical point is (1.91, 2.52, 28.52). $2.52. The critical point is (1.91, 2.52, 28.52).


Optimization: Exercise 10.2 #9Optimization: Exercise 10.2 #9At (1.91, 2.52, 28.52),At (1.91, 2.52, 28.52),RRbbbb = -6 = -6

RRbp bp = -1= -1

RRpb pb == -1-1

RRpp pp = -4= -4

23)1)(1()4)(6(

4116

)52.2,91.1(D

23)1)(1()4)(6(

4116

)52.2,91.1(D

D(1.91, 2.52) > 0 and RD(1.91, 2.52) > 0 and Rbb bb < 0 so (1.91, 2.52, 28.52), < 0 so (1.91, 2.52, 28.52),

is a maximum.is a maximum.


Constrained OptimizationConstrained Optimization• The minimum (or maximum) output of f The minimum (or maximum) output of f

subject to g(x,y) = c can be determined by subject to g(x,y) = c can be determined by finding the smallest (or largest) value M for finding the smallest (or largest) value M for which the constraint curve g(x,y) = c and the which the constraint curve g(x,y) = c and the contour curve f(x,y) = M touch and then contour curve f(x,y) = M touch and then determining the point (xdetermining the point (x00, y, y00) where these two ) where these two

curves meet.curves meet.


Constrained OptimizationConstrained Optimization

c)y,x(g

gλf

gλf

yy

xx

c)y,x(g

gλf

gλf

yy

xx

• To find the optimal solution subject to the To find the optimal solution subject to the constraint we must solve the system of constraint we must solve the system of equations:equations:

• The Lagrange MultiplierThe Lagrange Multiplier

y

y

x

x

g

f

g

fλ

y

y

x

x

g

f

g

fλ


Constrained Optimization: Example Constrained Optimization: Example A matrix manufacturing process has the Cobb-A matrix manufacturing process has the Cobb-Douglas production function ofDouglas production function of

f(L,K) = 48.1Lf(L,K) = 48.1L0.60.6KK0.4 0.4 mattressesmattresseswhere L represents the number of worker hours (in where L represents the number of worker hours (in thousands) and K represents the amount invested in thousands) and K represents the amount invested in capital (in thousands of dollars). A plant manager capital (in thousands of dollars). A plant manager has $98,000 to be invested in capital or spent on has $98,000 to be invested in capital or spent on labor and the current average wage of an employee labor and the current average wage of an employee is $8 per hour. Write the budget constraint equation is $8 per hour. Write the budget constraint equation and find the capital and labor expenditures that will and find the capital and labor expenditures that will result in the greatest number of mattresses being result in the greatest number of mattresses being produced subject to the budget constraint.produced subject to the budget constraint.


Constrained Optimization: Example Constrained Optimization: Example The budget constraint equation isThe budget constraint equation is

g(L, K) = 8L + K thousand dollarsg(L, K) = 8L + K thousand dollarswhere L and K are as previously described.where L and K are as previously described.The following equations must be satisfied at the The following equations must be satisfied at the point at which maximum production occurs.point at which maximum production occurs.

98KL8c)y,x(g)3(

)1(λKL24.19K

gλ

K

f)2(

)8(λKL86.28L

gλ

L

f)1(

6.06.0

4.04.0

98KL8c)y,x(g)3(

)1(λKL24.19K

gλ

K

f)2(

)8(λKL86.28L

gλ

L

f)1(

6.06.0

4.04.0


Constrained Optimization: Example Constrained Optimization: Example Solving the system yields K = 39.2 and L = 7.35. Solving the system yields K = 39.2 and L = 7.35. Therefore, labor costs are (7.35 thousand worker Therefore, labor costs are (7.35 thousand worker hours)($8 per hour) = $58,800. The capital hours)($8 per hour) = $58,800. The capital investment is $39,200. The production level is investment is $39,200. The production level is f(7.35, 39.2) = 690.608 f(7.35, 39.2) = 690.608 690 mattresses. 690 mattresses.

To verify that a maximum occurred, we consider To verify that a maximum occurred, we consider nearby points. Observe f(7.3, 39.6) = 690.585 and nearby points. Observe f(7.3, 39.6) = 690.585 and f(7.4, 38.8) = 690.584 which are both less than the f(7.4, 38.8) = 690.584 which are both less than the extreme value. So a maximum occurred.extreme value. So a maximum occurred.


Constrained Optimization: Exercise 10.3 #5Constrained Optimization: Exercise 10.3 #5

Find the optimal point of f(r,p) = 2rFind the optimal point of f(r,p) = 2r22 + rp - p + rp - p22 + p + p under the constraint g(r,p) = 2r + 3p = 1. Identify the under the constraint g(r,p) = 2r + 3p = 1. Identify the output as a maximum or minimum using close points.output as a maximum or minimum using close points.

1p3r2c)p,r(g)3(

)3(λ1p2rp

gλ

p

f)2(

)2(λpr4r

gλ

r

f)1(

1p3r2c)p,r(g)3(

)3(λ1p2rp

gλ

p

f)2(

)2(λpr4r

gλ

r

f)1(

Solving the system of equation yields, p = 0.375 Solving the system of equation yields, p = 0.375 and r = -0.0625. We must determine if this is a and r = -0.0625. We must determine if this is a maximum or minimum.maximum or minimum.


Constrained Optimization: Exercise 10.3 #5Constrained Optimization: Exercise 10.3 #5

f(-0.0625, 0.375) = 0.218750000f(-0.0625, 0.375) = 0.218750000

We select values of r close to -0.625 and use We select values of r close to -0.625 and use g(r,p) = 1 to find the associated p value.g(r,p) = 1 to find the associated p value.

f(-0.062, 0.37467) = 0.218750222f(-0.062, 0.37467) = 0.218750222f(-0.063, 0.37533) = 0.218750222f(-0.063, 0.37533) = 0.218750222

Evaluating f at these points we observe that f has a Evaluating f at these points we observe that f has a minimum (subject to the constraint) at r = -0.625 and minimum (subject to the constraint) at r = -0.625 and p = 0.375.p = 0.375.


Constrained Optimization: Exercise 10.3 #5Constrained Optimization: Exercise 10.3 #5The contour plot confirms the result. The contour plot confirms the result. f(-0.0625, 0.375) = 0.218750000f(-0.0625, 0.375) = 0.218750000f(-0.062, 0.37467) = 0.218750222f(-0.062, 0.37467) = 0.218750222f(-0.063, 0.37533) = 0.218750222f(-0.063, 0.37533) = 0.218750222


Method of Least SquaresMethod of Least Squares• The method of least squares is the procedure The method of least squares is the procedure

used to find the best-fitting line based on the used to find the best-fitting line based on the criterion that the sum of squared errors is as criterion that the sum of squared errors is as small as possible. The linear model obtained small as possible. The linear model obtained by this method is called the least-squares line.by this method is called the least-squares line.


Method of Least Squares: ExampleMethod of Least Squares: ExampleThe number of investment choices for people who The number of investment choices for people who invest in 401(k) plans at their place of employment is invest in 401(k) plans at their place of employment is given by the data. A scatter plot and linear model given by the data. A scatter plot and linear model are also shown.are also shown.

Year401(k) Plans

1990

1991

1992

1993

3.5

4.0

4.2

4.8


Method of Least Squares: ExampleMethod of Least Squares: Example

The deviation values (dThe deviation values (d11, d, d22, d, d33, d, d44) measure the error ) measure the error

between the model and the data points. We need to between the model and the data points. We need to minimize the sum of the squared deviations.minimize the sum of the squared deviations.


Method of Least Squares: ExampleMethod of Least Squares: Example

22

22

24

23

22

21

)]b3a(8.4[)]b2a(2.4[

)]b1a(0.4[)]b0a(5.3[

dddd

22

22

24

23

22

21

)]b3a(8.4[)]b2a(2.4[

)]b1a(0.4[)]b0a(5.3[

dddd

22

22

)ba38.4()ba22.4(

)ba4()b5.3()b,a(f

22

22

)ba38.4()ba22.4(

)ba4()b5.3()b,a(f

Finding fFinding faa and f and fbb, setting them equal to zero, and , setting them equal to zero, and

solving the resultant system of equations yields solving the resultant system of equations yields a = 0.41 and b = 3.51. Since fa = 0.41 and b = 3.51. Since faaaa(0.41, 3.51) > 0 and (0.41, 3.51) > 0 and

the determinant of the second partials matrix is 224 the determinant of the second partials matrix is 224 the error is a minimum.the error is a minimum.


Method of Least Squares: Exercise 10.4 #3Method of Least Squares: Exercise 10.4 #3

Find the linear function that best fits the data.Find the linear function that best fits the data.

x y

0

10

20

3

2

1

2

22

24

23

22

21

)]b20a(1[

)]b10a(2[)]b0a(3[

dddd

2

22

24

23

22

21

)]b20a(1[

)]b10a(2[)]b0a(3[

dddd

2

22

)ba201(

)ba102()b3()b,a(f

2

22

)ba201(

)ba102()b3()b,a(f

ffaa = 1000a + 60b - 80 = 0 = 1000a + 60b - 80 = 0

ffbb = 60a + 6b - 12 = 0 = 60a + 6b - 12 = 0

Solving the system yields, a = -0.1 and b = 3.Solving the system yields, a = -0.1 and b = 3.


Method of Least Squares: Exercise 10.4 #3Method of Least Squares: Exercise 10.4 #3

x y

0

10

20

3

2

1

So y = -0.1x + 3 So y = -0.1x + 3

Since f(-0.1, 3) = 0, the linear model Since f(-0.1, 3) = 0, the linear model is a perfect fit.is a perfect fit.

copyright © by houghton mifflin company, all rights reserved. calculus concepts 2/e latorre,...

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