# 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