Can You Make the Biggest Bread Bowl?

You have a hemispherical loaf of bread of radius $1$ foot, flat side down. You hollow out a bread bowl by boring a cylindrical hole of radius $r$, centred on the vertical axis through the top of the dome and reaching all the way down to the flat bottom crust. What radius $r$ maximises the volume of soup the bowl can hold?

The Fiddler, Zach Wissner-Gross, October 18, 2024(original post)

Solution

The bowl is the cylindrical hole, and the dome sets how deep it can go. A point on the cylinder’s wall sits at distance $r$ from the axis and meets the curved crust at height $z=\sqrt{1-r^2}$ above the flat base; above that the bread is gone. So the cylinder reaches height $\sqrt{1-r^2}$, and the soup volume is $$V(r)=\pi r^2\sqrt{1-r^2}.$$ Maximising $r^4(1-r^2)$ gives $4r^3-6r^5=0$, hence $r^2=\tfrac23$, so $$\boxed{\,r=\sqrt{\tfrac23}\,}\approx0.8165\ \text{ft},$$ and the bowl holds $V=\pi\cdot\tfrac23\cdot\tfrac{1}{\sqrt3}=\tfrac{2\pi\sqrt3}{9}\approx1.209$ cubic feet.

The computation

Maximise the soup volume itself: search $r$ to minimise $-\pi r^2\sqrt{1-r^2}$ on $(0,1)$.

import numpy as np
from scipy.optimize import minimize_scalar
r = minimize_scalar(lambda r: -r**2 * np.sqrt(1 - r**2),
                    bounds=(0, 1), method='bounded').x
print(r, np.sqrt(2 / 3))                        # 0.81650  0.81650

Extra Credit

Now the bread is a full sphere of radius $1$. You bore a cylinder whose axis passes through the centre, reaching down to but not through the curved bottom crust. What radius maximises the soup?

Solution

The cylinder of radius $r$ now fits between the lower and upper crust, from height $-\sqrt{1-r^2}$ to $+\sqrt{1-r^2}$, so its usable depth doubles to $2\sqrt{1-r^2}$ and $$V(r)=2\pi r^2\sqrt{1-r^2}.$$ The constant factor of two does not move the maximiser, so the optimal radius is again $$\boxed{\,r=\sqrt{\tfrac23}\,},$$ and the sphere simply holds twice the soup, $\tfrac{4\pi\sqrt3}{9}\approx2.418$ cubic feet. The best proportions of the bowl do not care whether the loaf is a dome or a full sphere.

The computation

The same search on the doubled volume returns the same radius.

r = minimize_scalar(lambda r: -2 * r**2 * np.sqrt(1 - r**2),
                    bounds=(0, 1), method='bounded').x
print(r, np.sqrt(2 / 3))                        # 0.81650  0.81650