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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Benny is cutting himself a slice of pizza of radius
The perimeter of the pizza slice, which is a sector of a circle with
p = 2 * r + r * ?theta?,
The area of the pizza slice is
This is our constraint, because we are given that
A = (1/2) * ?theta? * r^2.
and substituting this into the perimeter equation we get
?theta? = 2 * A / r^2,
Now all we have to do is find the
p = 2 * r + r * (2 * A / r^2) = 2 * r + 2 * A / r.
'deriv(p, r) = 2 - 2 * A / r^2 = 0,
r^2 = A,
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
On the other hand, we can't make
p = (2 + ?theta?) * r = (2 + 2 * 'pi) * 'root(A/'pi)
p = 'quot(2 + 2*'pi, 'root('pi)) * 'root(A).
Comparing this with the above, we find that
Now it is safe to conclude that the minimum perimeter