It's Pi Day, Pi Day, Gotta get down on Pi Day!

This Sunday, March 14, is Pi Day! To celebrate, you are planning to bake a pie. You have a sheet of crust laid out in front of you. After baking, your pie crust will be a cylinder of uniform thickness (or rather, thinness) with delicious filling inside.

To maximize the volume of your pie, what fraction of your crust should you use to make the circular base (i.e., the bottom) of the pie?

Assume that you have a thin sheet of area $A$ that is malleable and reformable, such that we can form the crust around any particular circular cylinder with surface area less than or equal to $A.$ In order to avoid any Monty Python-esque Bridge of Death ambiguity issues with the air speed of unladen one-crusted or two-crusted pies, we shall get the answer for each by defining the surface are of a $k$-crusted cylindrical pie as $$S_k(r,h) = 2\pi r h + k \pi r^2,$$ for $k = 1, 2.$ Then the solution will be $\frac{\pi (r^*_k)^2}{A},$ where $(r^*_k, h^*_k)$ is a solution to \begin{align*} \max_{r, h} \,\, & \pi r^2 h\\ \text{s.t.} \,\, & S_k(r,h) \leq A\\ & r, h \geq 0 \end{align*}

We can reduce this problem by noting that the surface area inequality will be active at any solution and then eliminating on of the variables. For instance, let $h_k = h_k(r)$ be defined such that $S_k(r,h_k(r)) = A,$ that is $$h_k = \frac{A - k \pi r^2}{2\pi r}.$$ Then, since the volume reduces to $V_k(r) = \pi r^2 h_k(r) = \frac{1}{2} r (A - k \pi r^2) = \frac{A}{2} r - \frac{k}{2} \pi r^3,$ the optimization problem becomes $\max \{ V_k(r) \mid r \geq 0 \},$ which we can solve using elementary calculus methods: $$\frac{d}{dr} V_k(r^*_k) = \frac{A}{2} - \frac{3k}{2} \pi (r^*_k)^2 = 0 \,\, \Rightarrow r^*_k = \sqrt{\frac{A}{3k\pi}}.$$

Therefore, the circular base of the maximal $k$-crusted cylindrical pies should be $\frac{1}{3k}$ of the overall area of the pie crust sheet.