Can You Make the Biggest Bread Bowl?

This Week's Fiddler

We want to find a relationship between our chosen hole radius $r$ and the height of the cylinder formed from said hole. Note that as $r$ increases, $h$ decreases, since you need to cut off more of the top of the bowl in order to achieve a larger radius.

A cross-section of the bread bowl will show that $r$ and $h$ are $\cos\theta$ and $\sin\theta$, respectively, for any angle $\theta$ from $0$ to $\frac{\pi}{2}$. Since the radius of the sphere is 1 foot, this means that $r^2+h^2=1$ square foot; hence, $h=\sqrt{1-r^2}$.

The volume of the cylinder is $V=\pi r^2h=\pi r^2\sqrt{1-r^2}$. We want to find the local maximum of $V$ given $r$, so we should find $r$ ($0\le r\le1$) such that $\frac{dV}{dr}=0$.

$$\begin{align*} \frac{dV}{dr}&=\frac{d}{dr}\Bigl(\pi r^2\sqrt{1-r^2}\Bigr) \\[1em] &=2\pi r\sqrt{1-r^2}-\frac{\pi r^3}{\sqrt{1-r^2}} \\[1em] &=0 \\[1em] 2\pi r\sqrt{1-r^2}&=\frac{\pi r^3}{\sqrt{1-r^2}} \\[1em] r^2&=\frac{2}{3}&(0\le r\le1) \\[1em] r&=\frac{\sqrt6}{3} \end{align*}$$

So the ideal radius for maximizing soup volume is $r=\frac{\sqrt6}{3}\approx0.8165$ feet, with a height of $h=\frac{\sqrt3}{3}\approx0.5774$ feet and a final volume of $V=\frac{2\sqrt3}{9}\pi\approx1.2092$ cubic feet.

Answer: sqrt(6)/3 feet

Extra Credit

Hopefully I'm not missing anything, but it seems the extra credit scenario is simply two hemispheres from the previous scenario stuck together. It's effectively the same optimization problem, just with a cylinder twice the height.

Thus, it's still the case that $r=\frac{\sqrt6}{3}\approx0.8165$ feet, this time with a height of $h=\frac{2\sqrt3}{3}\approx1.1547$ feet and a final volume of $V=\frac{4\sqrt3}{9}\pi\approx2.4184$ cubic feet.

Answer: sqrt(6)/3 feet