From Michael Lester comes a delectable dilemma about soup:
I have a large, hemispherical piece of bread with a radius of 1 foot. I make a bread bowl by boring out a cylindrical hole with radius r, centered at the top of the hemisphere and extending all the way to the flat bottom crust.
What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?
The volume of a cylinder is
$$V = \pi R^2H$$Substituting $R = \sqrt{1-H^2}$ yields:
\begin{align*} V &= \pi \left(\sqrt{1-H^2}\right)^2 H \\ &= \pi \left(H-H^3)\right) \\ \end{align*}In order to maximize volume, set $\dfrac{dV}{dH} = 0$:
\begin{align*} \dfrac{dV}{dH} &= \pi \left(1-3H^2\right) \\ 0 &= 3H^2-1\\ H &=\boxed{\sqrt{\frac{1}{3}}} \end{align*}Substituting into $R = \sqrt{1-H^2}$ yields
$$\boxed{R =\sqrt{\frac{2}{3}}}$$The radius for maximum volume is $\boxed{R =\sqrt{\frac{2}{3}}}$.
Instead of a hemisphere, now suppose my bread is a sphere with a radius of 1 foot.
Again, I make a bowl by boring out a cylindrical shape with radius r, extending all the way to (but not through) the curved bottom crust of the bread. The central axis of the hole must pass through the center of the sphere.
What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?
The total volume is the volume of the cylinder and the spherical cap.
$$V = \pi R^2(2H-2) + \frac{1}{3}\pi h^2(3R-h)$$
Where $h = 2-H$ and $R = \sqrt{1-(H-1)^2} = \sqrt{2H-H^2}$. Substituting yields:
$$V = \pi \left(\sqrt{2H-H^2}\right)^2(2H-2) + \frac{1}{3}\pi (2-H)^2 \left(3\sqrt{2H-H^2}-(2-H)\right)$$
$$V = \pi(-2H^3+6H^2-4H) + \frac{1}{3}\pi \left(3(2-H)^2\sqrt{2H-H^2} - (2-H)^3\right)$$
$$V = \frac{\pi}{3}\left((-6H^3+18H^2-12H) + \left((3H^2-12H+12)\sqrt{2H-H^2} - (-H^3+6H^2-12H+8)\right)\right)$$
$$V = \frac{\pi}{3}\left(-5H^3+12H^2-8 + (3H^2-12H+12)\sqrt{2H-H^2}\right)$$
Take the derivative:
\begin{align*}
\frac{dV}{dH} &= \frac{\pi}{3}\left(-15H^2+24H + (3H^2-12H+12)\cdot \frac{1}{2} \frac{2-2H}{\sqrt{2H-H^2}} + (6H-12)\sqrt{2H-H^2}\right) \\
0 &= -5H^2+8H + \frac{(H-2)^2(1-H)}{\sqrt{2H-H^2}} + 2(H-2)\sqrt{2H-H^2}\\
0 &= -5H^2+8H + \frac{(H-2)^2(1-3H)}{\sqrt{2H-H^2}}\\
\end{align*}
The maximum value occurs at $H \approx 1.43811$.
$$R = \sqrt{1.43811(2-1.43811)} \approx \boxed{0.8989 }$$