2012 Excellence in Mathematics Contest Team Project Level I (Precalculus and above) School Name: Group Members:

Transcription

1 01 Excellence in Mathematics Contest Team Project Level I (Precalculus and above) School Name: Group Members:

2 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 C l w A r Area of a rectangle Perimeter of a rectangle Area of a circle 1 y y C r A bh m x x Circumference of a circle Area of a triangle Slope 1 1 a b c 580 feet = 1 mile feet = 1 yard Pythagorean Theorem 16 ounces = 1 pound.54 centimeters = 1 inch h 4.9t v0t h0 h 16t v0t h0 1 kilogram =. pounds 1 meter = inches 1 gigabyte = 1000 megabytes 1 mile = 1609 meters 1 gallon =.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. 4 V r h V Area of Base height V r Volume of cylinder Volume Volume of a sphere Lateral SA = r h Lateral surface area of cylinder b b 4ac x a Quadratic Formula tan sin cos This team project is taken from The Mathematics Teacher, volume 105, Number 5, December 011, National Council of Teachers of Mathematics (nctm.org)

3 TEAM PROJECT Level I 01 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 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 V r h. Since the height is 10 cm and the radius is.5 cm, it follows that the volume is approximately 1.8 cm. 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 x x Solving, we find that x = 5.. Therefore, the volume of the frustum is V ( ) ( R H r h), giving a volume of approximately 5.1 cm. Thus, the total volume is 66.9 cm. The Pepsi can says that it holds 55 cm. Our answer is slightly larger because we assumed that the can is completely filled, but it is not. 4

5 Part continued. 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. 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 x r ( y a) We use circular disks, replacing a with 4. and using = 8.5 as the upper limit of the integrand, so that we have the following: 8.5 V 4.6 ( y 4.) dy ( y 4.) ( y 4.) 1.16y Therefore, the volume is approximately 404. cm. dy

6 Part 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 AF 0.96 cm and AE 1.9 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) = r. Solving, we find that r.6 cm. We can determine the volume of the indentation by again considering it a solid rotated about the y-axis V.6 ( y 1.4) dy 5.9 cm 6

7 Part 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 ( 1.4) V y dy Evaluating the integral, we find that V 5.9 cm. Subtracting this result from our answer to problem, we find that the volume is 98. cm. 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 cm ). 7

8 Part 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 cm. The area of the rectangular container is cm. The ratio is Alternatively, we can say that the bottles footprint is the ratio is simply. 4 6 r, whereas the container s area is 6r 4r = 4r. Thus, 8

9 Part continued. 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 cm ). 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 400 cm.the volume of the container is cm, so the volumetric efficiency is

10 Part continued. Another way of packaging the six bottles is to arrange them in the shape of a triangle (see photograph ). (a) Determine the area efficiency of the triangular sixpack (see fig. 5). The package that will contain this triangular six-pack has a side of length 4r r r 4. Since it is an equilateral triangle, we calculate its area as Thus, the desired ratio is 6r 6. r A s. 4 This result is approximately 0.781, a configuration slightly less efficient than the standard six-pack. (b) Determine the volumetric efficiency of the triangular sixpack. The volumetric efficiency is likewise slightly less efficient than that of the rectangular six-pack. The total volume of the container is cm. This result gives a ratio of about

11 Part 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 4 6 The area of the bases of the six bottles is again efficiency could be argued to be r. 6 r, so the area 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 (b) What is the volumetric efficiency of the barbell six-pack? For the volumetric efficiency, we need the volume of the container. It V cm. This gives an will be efficiency of around

12 Part 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 18 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 right triangle, AB = r. Thus, the side of the triangle is r, and the triangle s area is 18r r, and the area ratio is. Thus, the hexagon has area 7 r 18r (b) What is the volumetric efficiency of the heptahex-pack? The volume will be cm. This gives a volumetric efficiency ratio of

13 Part continued 6. What type of package might you use for eight bottles? A possible design is shown. 1