Can You Make a Toilet Paper Roll?

You have a parallelogram of cardboard with side lengths $2$ and $6$ units and interior angles of $30^\circ$ and $150^\circ$. By swirling two opposite edges together, you wrap it into the lateral surface of a right cylinder, a little cardboard tube. What is the volume of the cylinder you can make from this piece of cardboard?

The Fiddler, Zach Wissner-Gross, September 20, 2024(original post)

Solution

Wrapping the parallelogram into a right cylinder turns one pair of opposite edges into the seam and the other pair into the two circular rims. Join the two short edges and the long edges (length $6$) become the rims: the circumference is $6$, so the radius is $6/2\pi=3/\pi$, and the height is the perpendicular distance across, $2\sin30^\circ=1$, giving volume $\pi(3/\pi)^2(1)=9/\pi$. Join the long edges instead and the circumference is $2$, the radius $1/\pi$, the height $6\sin30^\circ=3$, for volume $\pi(1/\pi)^2(3)=3/\pi$. So the two tubes have volumes $$\boxed{\ \frac{9}{\pi}\ }\approx2.865 \qquad\text{and}\qquad \boxed{\ \frac{3}{\pi}\ }\approx0.955 .$$ Both have lateral surface area $2\pi rh=6$, exactly the parallelogram’s area $6\cdot2\sin30^\circ$, a useful check.

The computation

Build each tube from the geometry: the joined edge fixes nothing, the perpendicular edge becomes the rim’s circumference (hence the radius), and the slanted height is the side times $\sin\theta$.

import numpy as np
theta = np.radians(30)
def tube(rim_edge, slant_edge):                     # rim_edge wraps into the circle
    r = rim_edge / (2 * np.pi)
    h = slant_edge * np.sin(theta)
    return np.pi * r**2 * h
print(tube(6, 2), tube(2, 6))                        # 2.8648  0.9549  =  9/pi, 3/pi

Extra Credit

Let $V$ be the average volume of the two cylinders you can make from a parallelogram of area $1$. What is the smallest possible $V$?

Solution

For a parallelogram of sides $p,q$ and area $A=pq\sin\theta$, joining the $q$-edges gives radius $p/2\pi$ and height $A/p$, hence volume $pA/(4\pi)$; the other joining gives $qA/(4\pi)$. With $A=1$ the two volumes are $p/(4\pi)$ and $q/(4\pi)$, averaging $(p+q)/(8\pi)$. The constraint $pq\sin\theta=1$ forces $pq\ge1$ (equality only at a right angle), and then $p+q\ge2\sqrt{pq}\ge2$ (equality only at $p=q$). So the minimum is $$\boxed{\ \frac{1}{4\pi}\ }\approx0.0796,$$ achieved by the unit square, whose two cylinders are identical.

The computation

Minimise the average volume over all area-one parallelograms (eliminating $q$ through the area constraint); the optimum is the unit square.

from scipy.optimize import minimize
def avg_volume(x):
    p, th = x; q = 1 / (p * np.sin(th))             # area = p q sin(theta) = 1
    return (p + q) / (8 * np.pi)
r = minimize(avg_volume, [1.3, np.radians(70)], bounds=[(0.3, 3), (0.3, np.pi/2)])
print(round(r.fun, 4), round(1 / (4*np.pi), 4))     # 0.0796  0.0796