The PCTM Magazine

College Mathematics Corner

by Mary Ann Matras

The Wyatt Burp Soda Company

The Wyatt Burp Soda Company wants to make liter-size soda cans; that is, the Burp Company wants to make cylindrical cans that will hold 1 liter (1000 cm³) of soda. Being ecology- and economy-minded, the Burp Company wants to use a minimum amount of material. What dimensions will enable the use of minimum material?

In days of old, problems like the Burp problem, typical of optimization problems from calculus I, were posed only in forms that were easily solved by hand. Problem parameters were chosen to factor or to take an easy square or cube root rather than for realism. Today this problem can be solved by hand, by graphing calculator or by using a symbolic manipulation utility. Using technology, the parameters can be more realistic and the problems can be solved earlier in the curriculum.

Setting Up the Model:

The first step is solving this optimization problem is the same for all solution methods; we need to make a mathematical model of the surface area of a soda can. Figure 1 gives a physical model of a flattened can. In order to make the soda can, a circular-shaped top and bottom and a rectangular-shaped side must be used. The area of the circle would be pi*r², and the area of the rectangle would be 2*pi*r*h, so a function for the surface area of the whole can might be:
f(r,h) = 2*pi*r² + 2*pi*r*h

Since this function depends on two variables, it is necessary to use the fact that the can holds 1000 cm³. This means

V = pi*r²*h = 1000 cm³

Solving for h, we get

h = 1000 / (pi*r²)

This can be substituted into the first function to get a function of r:

f(r) = 2*pi*r² + 2000/r

The question is now, "what values of r will minimize the amount of metal used?"

Method One:

This is a traditional hand solution method. To find the minimum of the surface area function, we find its derivative and set it equal to zero.

f'(r) = 4*pi*r - 2000/r²

= (4*pi*r³ - 2000) /r ²

When will f'(r) = 0? It will be zero when the numerator is zero.
4*pi*r³ - 2000 = 0

r³ = 500/pi

so r is approximately 5.419 cm.

Method Two

A symbolic manipulation utility such as Mathematica or the TI-92 can be used to solve this problem using essentially the same steps as method one. See Figure Two for a solution from Mathematica.

Method Three

A graphing calculator can be used to graph f(r) and trace can used to find r. Again r is approximately 5.4 cm. Since taking the derivative is not involved, students with graphing calculator skills in algebra I or II can use this solution method.

Completing The Problem

The problem asks for the optimal dimensions for the can as a solution. All three of our methods have given us an optimal r of approximately 5.4 cm. Substituting that r into the equation
h = 1000 / (pi*r²)
gives an approximate h of 10.8 cm. Our solution means that the optimal soda can that hold a liter of soda will have a radius of 5.4 cm and a height of 10.8 cm. This can would be rather squat and square-looking since its diameter and height are the same. (For reference, a 12-ounce soda can has radius of approximately 3 cm and height of about 12 cm.)

Things To Think About

Why does a 12-ounce can have the shape it does? What would be the optimal dimensions of a two liter soda can? Can you find the step in the "hand" method that would require a calculator, a slide rule or incredible skill in taking the cube root? Why haven't any manufacturers followed the lead of the Wyatt Burp Company? Isn't it great that modern technology allows us to do all this?

Figure 1

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