# Solution to the Pizza Problem

Here's a solution to the pizza problem for Alvirne's AP Calculus Class, submitted by Shane Eaton. I've also written a solution to problem 28, the other problem on Alvirne's AP Calculus page.

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## Problem

by Shane Eaton

Benny is cutting himself a slice of pizza of radius r and central angle ?theta?. Find r and ?theta? if he would like to cut a piece with a minimum perimeter and the area A of his slice is fixed.

## Solution

by Ka-Ping Yee

The perimeter of the pizza slice, which is a sector of a circle with radius r and central angle ?theta?, consists of two straight lines and and arc, so its length is

p = 2 * r + r * ?theta?,
where ?theta? is measured in radians. This is what we want to minimize.

The area of the pizza slice is

A = (1/2) * ?theta? * r^2.
This is our constraint, because we are given that A is fixed. The relationship between r and ?theta? under this constraint is
?theta? = 2 * A / r^2,
and substituting this into the perimeter equation we get
p = 2 * r + r * (2 * A / r^2) = 2 * r + 2 * A / r.
Now all we have to do is find the r that corresponds to the smallest possible p. Extreme values of p will occur when the derivative of p is zero, that is,
'deriv(p, r) = 2 - 2 * A / r^2 = 0,

or r^2 = A,

which yields r = 'root(A), ?theta? = 2.

At this point, the perimeter is

p = (2 + ?theta?) * r = 4 * 'root(A)

Simply setting the derivative to zero does not guarantee that we have the minimum, but merely that we have one extreme. It's important to check that we haven't missed any other possibilities. In this case, we can see that 2 * r^2 = A has only one solution for r. Now let's examine the boundary cases: as r grows indefinitely large, we can waste as much perimeter as we want while making the slice very thin to keep A constant. So the upper limit is not a minimum for p.

On the other hand, we can't make r indefinitely small, because at a certain point we will need the entire pizza to make up the area A. At that point r would be 'root(A/'pi) and ?theta? would be 2*'pi, yielding

p = (2 + ?theta?) * r = (2 + 2 * 'pi) * 'root(A/'pi)

or p = 'quot(2 + 2*'pi, 'root('pi)) * 'root(A).

Comparing this with the above, we find that 'quot(2 + 2*'pi, 'root('pi)) .approxeq 4.67, which is indeed larger than 4, so this value is larger than the perimeter length we had earlier.

Now it is safe to conclude that the minimum perimeter is 4*'root(A), and it must occur when r = 'root(A), ?theta? = 2.

copyright © by Ping (e-mail) updated Sun 22 Sep 1996 at 22:04 JST
since Sun 22 Sep 1996