Problem Statement
Show that the height of the right circular cylinder of maximum volume that can be inscribed in a given sphere of radius $R$ is $\frac{2R}{\sqrt{3}}$. Also find the maximum volume.
Verified Solution & Marking Scheme
Set up Geometric Relation between Radius and Height
Let the cylinder have radius $r$ and height $h$.
From the cross-section of the sphere through the axis of the cylinder, by Pythagoras theorem:
$r^2 + \left(\frac{h}{2}\right)^2 = R^2 \implies r^2 = R^2 - \frac{h^2}{4}$
Express Volume V as a Single-Variable Function of h
The volume of the cylinder is $V = \pi r^2 h$.
Substituting $r^2$:
$V(h) = \pi \left( R^2 - \frac{h^2}{4} \right) h = \pi R^2 h - \frac{\pi}{4} h^3$
Differentiate and Find Stationary Points
$\frac{dV}{dh} = \pi R^2 - \frac{3\pi}{4} h^2$
Setting $\frac{dV}{dh} = 0$ for critical points:
$\pi R^2 = \frac{3\pi}{4} h^2 \implies h^2 = \frac{4R^2}{3} \implies h = \frac{2R}{\sqrt{3}}$
Second Derivative Test and Maximum Volume Calculation
$\frac{d^2V}{dh^2} = -\frac{6\pi}{4} h = -\frac{3\pi}{2} h$
At $h = \frac{2R}{\sqrt{3}}$, $\frac{d^2V}{dh^2} = -\frac{3\pi}{2} \left(\frac{2R}{\sqrt{3}}\right) = -\sqrt{3}\pi R < 0$, which confirms maximum volume.
Now find maximum volume:
$r^2 = R^2 - \frac{1}{4}\left(\frac{4R^2}{3}\right) = R^2 - \frac{R^2}{3} = \frac{2R^2}{3}$
$V_{\text{max}} = \pi r^2 h = \pi \left(\frac{2R^2}{3}\right) \left(\frac{2R}{\sqrt{3}}\right) = \frac{4\pi R^3}{3\sqrt{3}}$