2012 Excellence in Mathematics Contest

Team Project Level I

(Precalculus and above)

School Name:

Group Members:

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Reference Sheet

Formulas and Facts

You may need to use some of the following formulas and facts in working through this project. You may not need

to use every formula or each fact.

A bh 2 2C l w 2A r

Area of a rectangle Perimeter of a rectangle Area of a circle

2C r 1

2A bh 2 1

2 1

y ym

x x

Circumference of a circle Area of a triangle Slope

2 2 2a b c 5280 feet = 1 mile 3 feet = 1 yard

Pythagorean Theorem

16 ounces = 1 pound 2.54 centimeters = 1 inch 2

0 04.9h t v t h

2

0 016h t v t h

1 kilogram = 2.2 pounds 1 meter = 39.3701 inches 1 gigabyte = 1000 megabytes

1 mile = 1609 meters 1 gallon = 3.8 liters 1 square mile = 640 acres

1 sq. yd. = 9 sq. ft 1 cu. ft. of water = 7.48 gallons 1 ml = 1 cu. cm.

2V r h V Area of Base height 34

3V r

Volume of cylinder Volume Volume of a sphere

Lateral SA = hr 2 a

acbbx

2

42

cos

sintan

Lateral surface area of cylinder Quadratic Formula

This team project is taken from The Mathematics Teacher, volume 105, Number 5, December 2011, National

Council of Teachers of Mathematics (nctm.org)

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TEAM PROJECT Level I

2012 Excellence in Mathematics Contest ____________________________________________________________________________________________

The Team Project is a group activity in which the students are presented an open ended, problem situation

relating to a specific theme. The team members are to solve the problems and write a narrative about the theme

which answers all the mathematical questions posed. Teams are graded on accuracy of mathematical content,

clarity of explanations, and creativity in their narrative. We encourage the use of a graphing calculator.

During a visit with his sister’s family, Ron

Lancaster was shown an unusual bottle of

Coca-Cola® that consisted of a sphere

with a cap. Placing this bottle beside a can

of Pepsi cola revealed the contrast (see

photograph). Ron’s nephew Matt

challenged Ron to pick the container that

held the most liquid without touching

either or making any measurements. Ron

studied the bottle and can from a distance,

picked the one he thought had the greater

volume, and then found out he was

wrong. Ron then mailed his colleague

Doug Wilcock the spherical bottle with

this challenge: Set it beside a Pepsi can

and choose the container with the greater

volume. Not only did Doug pick the right

container, but he also devised the following set of questions related to the bottle and the can. Your task in this team

project is to respond to these questions as clearly and accurately as possible. Have fun!

Part 1 – Begin the Exploration

If you were given the same challenge as Ron Lancaster, which container would you choose as having the greater

volume? That is, without making any measurements or calculations, what does your gut instinct say? Explain.

All the following calculations involve quantities that were measured. Because the measurements are not exact, we

should be aware that all the following answers are approximations.

We are certain that students will choose either the Coca-Cola bottle or the Pepsi can or that they decided that the

volumes were the same. Look at their explanation and judge accordingly.

4

Part 2 – With Measurements and Calculations

1. We can think of the Pepsi can as being a cylinder with a top in the shape of a frustum (A frustum is the portion

of a cone or pyramid that remains after its upper part has been cut off by a plane parallel to its base. See the

figure below). Use the measurements provided to determine the volume of the can.

The volume of the cylinder is given by 2V r h . Since the height is 10 cm and the radius is 3.25 cm, it follows

that the volume is approximately 331.8 cm3. The next step is to calculate the volume of the frustum. To do so, we

need to find x in the figure. Using similar triangles, we get

1.3

2.6 3.25

x x .

Solving, we find that x = 5.2. Therefore, the volume of the frustum is 2 21

( ) ( )3

V R H r h , giving a volume of

approximately 35.1 cm3. Thus, the total volume is 366.9 cm

3. The Pepsi can says that it holds 355 cm

3. Our answer

is slightly larger because we assumed that the can is completely filled, but it is not.

5

Part 2 continued…

2. We now request that you use calculus to find the volume of

the Coca-Cola bottle. If your team does not have a member

with the necessary calculus background, then estimate the

volume of the bottle by assuming it is a perfect sphere.

a. A cross-section view of half the Coca-Cola bottle along

with its measurements is shown in the figure. Use calculus

to estimate the volume of a solid of revolution that models

the spherical bottle.

Referring to the figure and applying the Pythagorean Theorem,

we find that the rounded length of AB is 4.2 cm. We will

determine the volume V of the truncated sphere by thinking of

it as a solid of revolution. With the x-axis at the tabletop, the

general equation of the semicircle with a flat bottom that will

be revolved around the y-axis is

2 2( )x r y a

We use circular disks, replacing a with 4.2 and using 4.2 + 4.3

= 8.5 as the upper limit of the integrand, so that we have the

following:

8.5 22 2

0

8.5

2 2

0

8.53

0

4.6 ( 4.2)

4.6 ( 4.2)

( 4.2)21.16

3

128.66 404.2

V y dy

y dy

yy

Therefore, the volume is approximately 404.2 cm3.

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Part 2 continued…

b. The answer for part a. over-estimates the actual volume of

the contents of the Coca-Cola bottle. The reason: The base

is not flat but indented to provide more stability. A side

view of the indentation is shown in the figure. Measuring

indicates that 0.96AF cm and 1.9AE cm. What is the

volume of the indented section?

The figure shows the indentation at the bottom of the bottle. To

determine the bottle’s volume, we use the measurements shown in

figure 9.

Let r represent the lengths of GF and GE. Note that r is the radius

of a circle from which the indentation is formed. Using the

Pythagorean Theorem in triangle AEG, we have (r – 0.96)2 + 1.92 = r

2. Solving, we find that r ≈ 2.36 cm. We can

determine the volume of the indentation by again considering it a solid rotated about the y-axis.

20.96

2 2 3

0

2.36 ( 1.4) 5.9 cmV y dy

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Part 2 continued…

c. If we take the calculated volume of the spherical bottle and subtract the volume of the indentation, we get

the approximate net volume of the bottle. Doing so, can we reach a conclusion about the relative sizes of the

spherical bottle and the cylindrical can? Explain.

With the tabletop as the x-axis, the volume of the indentation is

20.96

2 2

0

2.36 ( 1.4)V y dy

Evaluating the integral, we find that V ≈ 5.9 cm3. Subtracting this result from our answer to problem 3, we find that

the volume is 398.3 cm3. Therefore, the volume of the round bottle is greater than the volume of the cylindrical

can. In case you are wondering, Coca- Cola labels the bottle as 400 ml (400 cm3).

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Part 3 – Selling Soft Drinks

Suppose that Coca-Cola were to sell these special bottles in packages of six arranged as shown in the photograph

(two rows of three bottles).

1. If we define area efficiency as the ratio of the area of the bottles to the area of the container that will hold the

bottles (see the figure below), how efficient is the rectangular six-pack? Express your answer as a simple ratio.

The area of the bottles (their “footprint”) is 398.86 cm2. The area of the rectangular container is 507.84 cm

2. The

ratio is 0.785.

Alternatively, we can say that the bottles’ footprint is 26 r , whereas the container’s area is 6r • 4r = 24r2. Thus,

the ratio is simply 4

.

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Part 3 – continued…

2. A second way to consider efficiency is to consider the ratio of the volume of the liquid in the containers to the

volume of the packages (volumetric efficiency). The volume of soda in each bottle is 400 ml (400 cm3). The

bottles are 11.4 cm high. What is the volumetric efficiency of the rectangular six-pack?

For the volumetric efficiency, the volume of the bottles is 2400 cm3.The volume of the container is 5789.38 cm

3,

so the volumetric efficiency is 0.415.

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Part 3 – continued…

3. Another way of packaging the six bottles is to arrange

them in the shape of a triangle (see photograph 3).

(a) Determine the area efficiency of the triangular six-

pack (see fig. 5).

The package that will contain this triangular six-pack has a

side of length 4 2 3 4 2 3r r r . Since it is an

equilateral triangle, we calculate its area as 2 3

4A s .

Thus, the desired ratio is

2

2

6 6

7 3 127 3 12

r

r

.

This result is approximately 0.781, a configuration slightly

less efficient than the standard six-pack.

(b) Determine the volumetric efficiency of the triangular six-

pack.

The volumetric efficiency is likewise slightly less efficient than that of the rectangular six-pack.

The total volume of the container is 24.6 7 3 12 11.4 5819.37 cm3. This result gives a ratio of about 0.412.

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Part 3 – continued…

4. A creative idea for packaging the six bottles is to stack them in pairs to form barbells (see photograph 4).

(a) What is the area efficiency of the barbell six-pack? In

answering this question, remember that there are six bottles,

not simply the three we see when we look at a plan of the

package, as shown in figure 6.

If we look at the bottom of the container, its area is 2 4 3 6r .

The area of the bases of the six bottles is again 26 r , so the area

efficiency could be argued to be

61.458

4 3 6

.

If we take the more traditional approach of simply looking at the

three bottles that form the base of the barbells, the ratio is about

0.729.

(b) What is the volumetric efficiency of the barbell six-pack?

For the volumetric efficiency, we need the volume of the container. It

will be 24.6 4 3 6 22.8 6237.19V cm3. This gives an

efficiency of around 0.385.

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Part 3 – continued…

5. Suppose that Coca-Cola decided to sell these unusual

bottles in a highly original pack of seven, arranged as

shown in photograph 5.

(a) What is the area efficiency of the heptahex-pack (see

fig. 7)?

The heptahex-pack container has an area of 218 3 r . The

answer can be determined by taking six of the triangles that

make up the heptahex-pack (see fig. 10).

Since CB = r and triangle ABC is a 30°- 60°-90° right triangle,

AB = 3r . Thus, the side of the triangle is 2 3 r , and the

triangle’s area is 23 3r . Thus, the hexagon has area

218 3r , and the area ratio is 2

2

70.705

18 3

r

r

.

(b) What is the volumetric efficiency of the heptahex-pack?

The volume will be 218 4.6 3 11.4 7520.62 cm

3. This

gives a volumetric efficiency ratio of 0.372.

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Part 3 – continued…

6. What type of package might you use for eight bottles?

A possible design is shown.