This week’s Riddler Classic asks:

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?Note:If you solve this riddle by baking an optimal pie, you automatically win.

### Solution

First, happy Pi Day! Now on to new business. Setting up this problem, we are trying to maximize the volume of pie filling that can fit inside the cylindrical pie crust. The volume of filling will be:

\[V=\pi r^2\cdot h\]

with \(V\) being the volume of pie filling, \(pi r^2\) is the surface area of the base of the pie crust, and \(h\) the height of the walls. We are going to maximize \(V\) with respect to \(r\), but first we have to constrain \(h\) in terms of \(r\). The sheet of pie crust, before being shaped into a pie, will have some area \(A\), and from it we must extract two circular sections with area \(A_b=\pi r^2\) and also the rectangular wall which will have area \(A_h=2\pi r\cdot h\). This gives

\[A=2\pi r^2+2\pi r\cdot h.\]

From this, we can extract that

\[h=\frac{A-2\pi r^2}{2\pi r}\]

which will be substituted into our volume equation:

\[V=\pi r^2\cdot\left(\frac{A-2\pi r^2}{2\pi r}\right).\]

We need to maximize \(V\) with respect to \(r\), which we will do by taking the partial derivative and setting it equal to 0, to find the maximum

\[\frac{\partial V}{\partial r}=\frac{A}{2}-3\pi r^2=0.\]

Now, we just need to find the value of \(\pi r^2\) (the area of the circle) in terms of \(A\) to find the fraction of \(A\) used to make the base. With some simple arithmetic, we get that

\[\pi r^2=\frac{A}{6},\]

and that the fraction is **1/6**.