T. James 1

Notebook Resource

Portfolio of Problems

Towanda James

Problem Solving, EMAT 6600

Dr. Jim Wilson

August 2, 2006

T. James 2Portfolio of Problems

1. Solve by iteration. I used an excel spreadsheet in order to do the iteration process. The items marked in red are where I noticed the answers were very close. x - .9371 or x 1.07755.

=2^A1 =D1^10-2 0.25 -2 1024-1 0.5 -1 10 1 0 01 2 1 12 4 2 1024

-0.9 0.535887 -0.9 0.348678-0.8 0.574349 -0.8 0.107374-0.95 0.517632 -0.988 0.886277-0.94 0.521233 -0.97 0.737424-0.93 0.524858 -0.96 0.664833-0.92 0.528509 -0.95 0.598737-0.935 0.523042 -0.94 0.538615-0.939 0.521594 -0.93 0.483982-0.937 0.522318 -0.935 0.510642-0.9371 0.522282 -0.939 0.5329131.07 2.099433 -0.937 0.521671.075 2.106722 -0.9371 0.5222271.076 2.108183 1.1 2.5937421.077 2.109645 1.05 1.6288951.0775 2.110376 1.06 1.7908481.07755 2.110449 1.07 1.967151

1.075 2.0610321.076 2.0802841.077 2.0996991.0775 2.1094671.07755 2.110446

Look at the graphs of the functions.

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2. Volume of a Frustum of a ConeA frustum may be formed from a cone with a circular base by cutting off the tip of the cone with a cut perpendicular to the height, forming a lower base and an upper base that are circular and parallel.Let h be the height, R the radius of the lower base, and r the radius of the upper base. One picture of the frustum is the following.

Given R, r, and h, find the volume of the frustum.

The Formula

Extensions: 1. Derive the formula.

2. What is the area of the curved surface of the frustum of a right circular cone? What is the total surface area of the frustum?

3. What is the volume of a frustum formed from square based pyramid of height h, lower base having a side length of S, and an upper base of side length s?

4. What is the total surface area of the square based frustum?

Solution:Consider the cone before it was cut. Let the height of the cut part = H.

H

h

R

T. James 4The volume of a cone is V = .

The volume of the original pre-cut cone is .The volume of the cut part is .

Now we have a subtraction problem in order to get the volume of the frustum. Consider the difference of the two cones. Subtracting the cut-off part from the original pre-cut cone would leave the bottom part of the cone called the frustum.

-

= = =

=

=

=

Now, recall the original pre-cut cone. Consider the triangles formed by the height and the bases. The triangles are similar by AA similarity and the sides are proportional.

Cross Multiplying yields,

Substitute for hr in the last step for .Recall:

=

r

T. James 5=

=

=

= , which is the formula.

The formula for the lateral area of a right circular cone is .The slant height of the original pre-cut cone can be found by using the Pythagorean theorem

.

The lateral area of the original pre-cut cone is .

The lateral area of the cut part is .

-

In order to get the total surface area of the frustum, add on the area of the base, which is a circle. The formula in order to find the area of a circle is .

Therefore, the total surface area of a frustum is - + .

I think this is a good activity to share with students. I wonder if the formula for the area of a circle makes any since to them. Consider the following activity. This is problem solving on a middle school or lower high school level. 10

8

6

4

2

-2

-4

-6

-8

-15 -10 -5 5 10 15

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A portion of this circle is covered by a shaded square. The length of a side of the shaded square is the same length as the radius of the circle. We call such a square a “radius square.”

1. How many radius squares does it take to cover the circle? (You can cut out the radius squares, cover the circle and see how many it takes to cover.) Record your data in a table with columns for the radius, area of the radius square, number of radius squares needed to cover the circle, and estimated area of the circle.

2. Do you think the same number of radius squares is needed to fill all circles? Construct a different size circle in GSP. Construct a quadrilateral interior for the radius square in the circle like the picture above. Copy your circle in word. Copy the grid and radius squares in Word. Crop pictures where needed. Record your new data in the second row in the table you made for the 1st circle. Repeat with another different circle. Do all the circles take the same number of radius squares to cover them? How many is that again?

3. Did you get a little over 3 squares to fill the circle? What number have we been studying that’s a little over 3?

4. Describe any patterns and relationships you see in your table that will allow you to predict the area of the circle from its radius square. Test you ideas on some other circular objects.

5. How can you find the area of a circle if you know the radius?

6. How can you find the radius of a circle if you know the area?

7. Calculate the expression (5) (3.14) .How does this expression relate to the area of the 1st circle you worked with? Now calculate the expression using the symbol on the calculator. Which expression is more accurate?

T. James 73. PROBLEM: Average Rate

Danielsville is 20 miles from Jim's home in Athens. He Drives 60 mph going to a meeting (he is almost late) but coming home the weather is bad and drives 30 mph. What is his average speed for the time he is on the road?

A. Discuss why 45 mph is not correct. Going to D'ville took 1/3 of an hour. Coming home took 2/3 of an hour. So the total 40 miles took one hour. The speed is 40 mph.

B. Use d = r t

to verify and develop an understanding of how to find average rates.

Solution:The average rate is over the same distance each way. The average rate is not the arithmetic mean of the two rates. That is, the rate is not

Distance is rate times time. Since Jim drove two different rates, we need to calculate going and returning.

Going Returning

It took hr traveling at 60 mph and hr traveling at 30 mph for the entire trip. The total distance was 40 miles that took one hour. The speed, therefore, is 40 mph.

and For this problem, r = 60 and r = 30

and r =

r = d = rt2d = r ( ) r = 40

2d = r( + ) The average is 40 mph.

2 = r( )

Note: This is the harmonic mean of r and r because it is the arithmetic mean of the inverse.

r =

T. James 84. Volume of a land Fill;

Curved Path, Trapezoid Cross-SectionThe following e-mail was sent to me:

I am an engineering geologist. I am involved in a case where a landowner filled in a stream with a road, several bridges, and an island. The local regulatory agency would like to estimate the volume of fill placed in the stream. That's my job. I have been able to calculate most of the volume by slicing the fill into cross sections and calculating the area of cross section and then multiplying the area for whatever length I decide to evaluate. My problem is where some cross sections are attached at a pivot point and I have a angular "length" to evaluate. For example how would you calculate the volume where two trapezoid cross sections of different areas that have a common pivot point and are separated by 50 degrees ? Or more easily, what is the volume of a trapezoid that sweeps across an angle of 50 degrees?No wonder I am a Geologist, couldn't handle the math ! Thanks for yourhelp.---------------------Interpret what he is asking for and devise an easily understood strategy for estimating the volume he is looking for.There is a pivot point P and a distance R from the pivot point to the center of the trench. We have a trapezoid cross section with bases a and b and height h. The pivot angle is 50 degrees.

Devise a spreadsheet that can be used to estimate the volume given input of the dimensions of the trapezoid, the distance from the pivot point to the center of the trench, and the pivot angle.

Possible Solution:h = height of the trapezoida = base 1 of the trapezoidb = base 2 of the trapezoidR = distance from the pivot point to the center of the trench

= pivot angleV = volume of a land fill

Area of trapezoid =

Arc Length = R ( )

Volume = R ( ) =

Spreadsheet on the next page.

T. James 9 h a b R V =A1+2 =B1+4 =C1+3 D2-5 E1-2 PI()/360*A1*(B1+C1)*D1*E1

12 20 10 100 50 15707.9614 24 13 97 48 2104716 28 16 92 46 25999.5418 32 19 87 44 30666.3420 36 22 82 42 34863.322 40 25 77 40 38435.6428 44 28 72 38 48134.2330 48 31 67 36 49885.3532 52 34 62 14 20845.6534 56 12 57 12 13800.3936 60 15 52 10 12252.2110 64 18 47 8 2690.612 68 21 42 6 2348.65514 72 24 37 24 10415.0116 76 27 32 22 10124.5918 80 30 27 20 9330.5320 84 33 22 18 8086.45922 88 36 17 16 6475.31124 92 39 12 14 4609.34526 96 42 7 12 2630.141

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5. HERON'S FORMULA

See Also: Problem Solving with Heron's Formula

IntroductionHeron's formula for the area of a triangle with sides of length a, b, c is

where

Problem:Develop a proof of Heron's Formula for the area of a triangle.

Algebraic Proof:

h is perpendicular to the base. So using Pythagorean theorem we know that Also, c = p+q, q = c-p and p = c-q.

Since q = c-p, then q =(c-p) and q =c -2cp +p . Adding h to each side gives

h +q =h =c -2cp +p . Substituting, b =a -2cp+c solving for p we get,

Now since . Substitute and get an expression in terms of a, b and c.

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6. Square Root of 2

Prove that the Square Root of 2 is an irrational number.

Discussion: That is, prove that this number can not be expressed as a rational fraction in lowest terms, where the greatest common divisor of a and b is 1.

For what positive integers N does your proof generalize to show that

is irrational?

Why? or Why not?

Show irrational.

Solution:

1. Assume that √2 is a rational number. This would mean that there exist integers a and b such that a / b = √2.

2. Then √2 can be written as an irreducible fraction a / b such that a and b are coprime integers and (a / b)2 = 2.

3. It follows that a2 / b2 = 2 and a2 = 2 b2. 4. Therefore a2 is even because it is equal to 2 b2 which is obviously even. 5. It follows that a must be even (odd square numbers have odd square roots and even

square numbers have even square roots). 6. Because a is even, there exists a k that fulfills: a = 2k. 7. We insert the last equation of (3) in (6): (2k)2 = 2b2 is equivalent to 4k2 = 2b2 is equivalent

to 2k2 = b2. 8. Because 2k2 is even it follows that b2 is also even which means that b is even because

only even numbers have even squares. 9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated

in (2). Since we have found a contradiction the assumption (1) that √2 is a rational number must be false. The opposite is proven. √2 is irrational.This proof can be generalized to show that any root of any natural number is either a natural number or irrational.

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7. PROBLEM: The Box Problem Use the Arithmetic Mean -- Geometric Mean Inequality to find the maximum volume of

a box made from a 25 by 25 square sheet of cardboard by removing a small square from each corner and folding up the sides to form a lidless box.

Determine what shape boxes could be created by this method from the 25 by 25 square sheet to hold a volume of 200 cu. units. 400 cu units? 800 cu units?

Generalize. Use the AM-GM Inequality to discuss the maximum volume of a box formed from an n X n square sheet of cardboard.

Why will the AM-GM Inequality not be a useful tool when the sheet of cardboard is 20 by 25?

Solution:

Let’s name the length of the corners that were cut out x. With the corners (x) cut out, each side has length 25-2x. The height of the box is x. Thus a box formed would have volume

= . Using the AM-GM inequality [ or for

this problem ]

=

So, . Therefore, the maximum volume would be 1157.407407 cubic units. It would be equal to 1157.407407 cubic units when 4x=25-2x, 6x=25 and x = = 4. . To get V=200 cubic units

a. 0.338 x 24.324 x 24.324b. 10.296 x 4.408 x 4.408

T. James 14obtained by solving x(25-2x) = 200To get V=400 cubic units

a. .7207 x 6.59 x 23.5586b. 9.204 x 6.592 x 6.592

obtained by solving x(25-2x) =400

To get V = 800 cubic units

a. 1.7215 x 21.557 x 21.557b. 7.2464 x 10.5072 x 10.5072

obtained by solving x(25-2x) =800

These are rounded so it won’t be exactly 200, 400 or 800.

To generalize, use AM-GM inequality to discuss the maximum volume of a box formed from an n by n square sheet.

V=x(n-2x)(n-2x)

= (4x)(n-2x)(n-2x) =

So, the max volume would be cubic units, with equality when 4x=n-2x, 6x=n and x=

The AM-GM Inequality or for this problem states that equality is reached if and only if a=b or in

this case a=b=c. If we have a 20 by 25 sheet of cardboard, the volume would be V=x(20-2x)(25-2x) = (4x) (20-2x)(25-2x)Let a=4x and b=20-2x and c=25-2x

Can we satisfy the equality condition of a=b=c?4x=20-2x=25-2x4x=20-2x 4x=25-2x6x=20 6x=25 x= x=

For a 20 x 25 sheet of cardboard, the AM-GM inequality would not be a useful tool because 4x = 20-2x = 25-2x will not hold.

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8. PROBLEM: Maximum area -- triangle.

Use the Arithmetic Mean -- Geometric Mean Inequality to show that the maximum area of a triangular region with a given perimeter is attained when the triangle is equilateral.

Solution:

Semi-perimeter =S=

2S=a+b+c

The area using Heron’s Formula is A=

Using the AM-GM Inequality

(s-a)(s-b)(s-c)

=

= substituting 2s for a + b + c

= =

(s-a)(s-b)(s-c) with equality when s-a = s-b = s-c, since a+b+c is a constant, then S=

is also a constant so a=b=c which mean an equilateral triangle. A=s

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9. PROBLEM: Minimum of x + 1/x.

Use the Arithmetic Mean -- Geometric Mean Inequality to show that

Solution:

Using graphing calculator, we see that the minimum value for x, y>0, is y = f(x)=2.Using AM-GM inequality, we know that

Thus, the minimum value for f(x) = is 2.

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10. PROBLEM: Find AM-GM Problems

Find maximization or minimization problems for which the A.M.-G.M. inequality can be used as a tool. Many of the traditional max/min problems from calculus are candidates (except that they might not be interesting . . .) Create a file of good problems.

1. Two nonnegative numbers sum is 10. How large can their product be?2. A piece of wire 100cm long is going to be cut into several pieces and used to

construct a skeleton of a rectangular box with a square base. What are the dimensions of the box with the largest volume?

3. Federalist Express limits the size of a mailable parcel. The longest side plus the girth (the perimeter of a cross-section perpendicular to the longest side) may not exceed 108 inches. Which mailable cube has the largest volume?

11. Problem: Comparison of areas in square of sides a+b

Interpret the comparison of the areas of the total square and the five shown parts.

Solution:

The total area of the square = (a+b) . The area of each unshaded rectangular part = ab. The total area of the unshaded part = 4ab. We also have a shaded square part which we will denote as I. Thus, (a+b) =4ab + I. (a+b) = 4ab if the area of I = 0.

So, (a+b) 4ab (a+b) 2 AM-GM Inequality!

T. James 18 (a+b) 4ab with equality iff a=b.

12. PROBLEM: Maximum of f(x) = (1-x)(1+x)(1+x)

Find the maximum of

in the interval [0, 1].

A. Graph with TI-81, TI-82, or a computer graphing tool.

B. Use AM-GM Inequality.

C. Use calculus

AM-GM Inequality

Find the maximum of f(x) = (1-x)(1+x)(1+x)

= (2-2x)(1+x)(1+x)

= = = 1.1851852x 0

T. James 19with equality when1 + x = 2 – 2x, 3x =1, x = .33333333Calculus

F(x) = (1-x)(1+x) =(1-x)(1+2x+x ) =1+2x+ x -x-2 x -x =1 + x - x - x

f’(x) = 0 gives the maximum value of the functionf’(x) =1-2x-3 xsolving f’(x) =0,1-2x-3 x = 0x =-1 or x=Since we are restricted to [0,1], x=-1 is not allowed. So the maximum value of the function F(x) = (1-x)(1+x) is when x= ( , 1.185185)

13.Given segment AB with parallel intersecting segments AE and BC on the same side of AB. Construct the segments AC and BE, intersecting at F and construct the segment FG parallel to AE and BC with G on AB.

Prove:

Given:

and

Adding the two equations together yields,

T. James 20

Substituting AB for GB+AG, since AB=GB+AG, and factoring yields,

Dividing both sides by AB yields,

Simplifying the right-hand side yields,

=

Thus: =

14. PROBLEM: Arithmetic Mean -- Geometric Mean Inequality

Find 5 different demonstrations (proofs) of the Arithmetic Mean -- Geometric Mean inequality for two non-negative variables

A stands for Arithmetic Mean, H stands for Harmonic Mean and G stands for Geometric Mean. a + b =

diameter A =radius.

x y

y

x

y x

y

x

T. James 21

It’s given that segments a+b form the diameter of the circle and that A is the radius. Since it takes two radii to equal a diameter, 2A = a+b. Solving for A we get, A = . A is an arithmetic mean of a and b because it is an average of a+b. Now look at the big right triangle. H + the dashed line (the hypotenuse of the right triangle) = r because it is a segment from the center of the circle to the edge of the circle. The short leg of the right triangle is r – b. By using the Pythagorean Theorem, we can solve for G, . This simplifies to , but 2r = a+b because it is give that a+b is a diameter and 2 radii = a diameter. Now we have

. Simplifying that we get . So, . This shows the G is the geometric mean of a and b because it is the square root of the product a and b. The geometric mean between two positive numbers a and b is the positive number x where . Solving for x the geometric mean is , which is what G equals. The harmonic mean between a and b is x if is the arithmetic mean of and . Let’s take the arithmetic mean of and .

We have . Simplifying that we have . Then we must invert which gives . Now that

we now that is the harmonic means of a and b, how do we know that it is the length of the line segment H? Now for the inequalities, H G A. From a geometric point of view, the radius will always be the biggest segment in a circle coming from the center. G is parallel to A in a different spot of a semi-circle it has to be smaller. H is a leg where G is the hypotenuse, H must be smaller than G.

Geometric Proof:

A =(x+y)

We see that the area of each of the 4 rectangles within each square is xy.

Thus, (x+y) 4xy (x+y) 2

AM GM

I

xy

xy

xy

xy

T. James 22

(x+y) =4xy if the area of I = 0. For the area of I to equal 0, x must equal y. Algebraic Proof

(x-y) 0 for all x,y > 0, with equality iff x =y.

x -2xy+y 0x +4xy-2xy+y 4xy x +2xy+y 4xy(x-y) 4xyx+y 2

AM GM

Let’s name the blue circle P and the red circle O. Given two circles, O & P, with radii b & a respectively.

(a-b) + x = (a+b) x = (a+b) - (a-b)

x = a + 2ab + b - [a - 2ab + b ] x = 4ab x = x = 2

a+b x (a+b is the hypotenuse and the longest side of the triangle) and x = 2 substituting a+b 2

x

x

a-b

T. James 23

AM GM

15. Triangle Mid-Segment TheoremTake any triangle ABC. Find the mid-point D of AB and the mid-point E or AC.

Click here to open a GSP file. Explore with measures of DE and BC as any vertex is moved.One of the most useful theorems in geometry of plane triangles is the Triangle Mid-Segment Theorem:

In any triangle, a segment joining the midpoints of any two sides will be parallel to the third side and half its length.

Prove the Triangle Mid-Segment Theorem.

Solution:

I extended DE and constructed EF DE and CF AD.

AD DB and AE EC by definition of midpoint.AED CEF since they are vertical angles

T. James 24EF DE by construction

AED CEF by SASA ECF by CPCTC

as constructedAD FC CPCTCDB FC transitive propertyBCFE is a parallelogram by definitionDE BC since opposite sides of a parallelogram are parallelDF BC since opposite sides of a parallelogram are DE+EF=DF segment addition postulateDE+DE=DF substitution2DE=DF substitutionDE= DF division property of equalityRecall DF=BC, so ED= BCThus ED BC and ED= BC

16. Problem: Square roots relation

Characterize the set of solutions of the equation

for positive integers x, y, and z.

The set of solutions of the equation for x,y,z > 0 is the set of perfect squares.

Any of the perfect squares can be substituted for x, y and z. Irrational numbers wil not work, because they can not be expressed as where b can not equal 0.

17. Problem: Sum of Unit Fractions

Consider all fractions of the form

where n is a positive integer. Which of these fractions may be expressed as the sum of two different unit fractions (reciprocals) of integers?

T. James 25I only found a couple of fractions of the form , n>0, that can be expressed as the sum of two different unit fractions.

N=3 N =5

nxy=(n+1)(y+x)nxy=ny+nx+y+xnxy-ny-nx=x+yn(xy-y-x)=x+yn(xy-(x+y))=x+yn=

18. Problem: Million Drops of Water

If you had a million drops of water, would you be more likely to drink it?

take a bath in it?

swim in it?

I would drink the water if it were clean, because as hot as it is this summer I am always thirsty.

19. PROBLEM: Volume of a Can

What is the volume of a 12 ounce can?

In other words, how may cubic inches in a 12 oz can?

Make a guess.

I don’t know 40 cubic inches.

Measure the radius and height of a soda can and compute an estimate.

T. James 26 V= = (2in) 5in 63 cubic inches in an 8 oz can. Approximately 95 cubic inches in a

12 oz can.How many cubic centimeters are in a 12 ounce can? V= = (5.5cm) 11.5cm 1092 cubic centimeters

20. PROBLEM: Big Tires

My Ford Bronco was fitted at the factory with 30 inch diameter tires. That means its speedometer is calibrated for 30 inch diameter tires. I "enhanced" the vehicle with All Terrain tires that have a 31 inch diameter. How will this change the speedometer readings? Specifically, assuming the speedometer was accurate in the first place, what should I make the speedometer read as I drive with my 31 inch tires so that the actual speed is 55 mph?

Forego the discussion of whether it is safe to drive a Bronco that fast . . . One would expect some use of ratios here . . .

, x = 56.83 mph

21. Problem: Ladder and Box

A ladder 5 meters long leans against a wall, reaching over the top of a box that is 1meter on each side. The box is against the wall. What is the maximum height on thewall that the ladder can reach? The side view is:

Assume the wall is perpendicular to the floor. Use your calculator to find themaximum height to the nearest .01 meter.

Possible Solution:

T. James 27