Problem Statement
Show that the right circular cylinder of maximum volume that can be inscribed in a sphere of fixed radius $R$ has height $h = \frac{2R}{\sqrt{3}}$, and find this maximum volume.
Verified Solution & Marking Scheme
Set up geometric relationship between cylinder radius r, height h, and sphere R
Let the inscribed cylinder have radius $r$ and height $h$. By the Pythagorean theorem on the right triangle from the center of the sphere:
$$r^2 + \left(\frac{h}{2}\right)^2 = R^2 \implies r^2 = R^2 - \frac{h^2}{4}$$
Volume of the cylinder is:
$$V = \pi r^2 h = \pi \left( R^2 - \frac{h^2}{4} \right) h = \pi R^2 h - \frac{\pi}{4}h^3$$
Differentiate V with respect to h and find stationary points
$$\frac{dV}{dh} = \pi R^2 - \frac{3\pi}{4}h^2$$
Setting $\frac{dV}{dh} = 0$:
$$\pi R^2 = \frac{3\pi}{4}h^2 \implies h^2 = \frac{4R^2}{3} \implies h = \frac{2R}{\sqrt{3}}$$
Verify Maximum with Second Derivative Test
$$\frac{d^2V}{dh^2} = -\frac{6\pi}{4}h = -\frac{3\pi}{2}h$$
Since $h > 0$, $\frac{d^2V}{dh^2} < 0$, confirming a local maximum.
Evaluate Maximum Volume
$$V_{\max} = \pi \left( R^2 - \frac{1}{4}\left(\frac{4R^2}{3}\right) \right) \left( \frac{2R}{\sqrt{3}} \right) = \pi \left( \frac{2R^2}{3} \right) \left( \frac{2R}{\sqrt{3}} \right) = \frac{4\pi R^3}{3\sqrt{3}}$$