Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2 R}{\sqrt{3}} .$ Also find the maximum volume.

(A) Let $r$ and $h$ be the radius and height of the cylinder inscribed in a sphere of radius $R$.
From the geometry of the sphere,we have the relation $R^2 = r^2 + (h/2)^2$,which implies $r^2 = R^2 - \frac{h^2}{4}$.
The volume $V$ of the cylinder is given by $V = \pi r^2 h = \pi (R^2 - \frac{h^2}{4}) h = \pi R^2 h - \frac{\pi h^3}{4}$.
Differentiating with respect to $h$,we get $\frac{dV}{dh} = \pi R^2 - \frac{3\pi h^2}{4}$.
Setting $\frac{dV}{dh} = 0$,we have $\pi R^2 = \frac{3\pi h^2}{4}$,which gives $h^2 = \frac{4R^2}{3}$,so $h = \frac{2R}{\sqrt{3}}$.
To check for maximum,$\frac{d^2V}{dh^2} = -\frac{6\pi h}{4} = -\frac{3\pi h}{2}$. Since $h > 0$,$\frac{d^2V}{dh^2} < 0$,confirming the volume is maximum at $h = \frac{2R}{\sqrt{3}}$.
The maximum volume is $V = \pi (R^2 - \frac{1}{4} \cdot \frac{4R^2}{3}) \cdot \frac{2R}{\sqrt{3}} = \pi (R^2 - \frac{R^2}{3}) \cdot \frac{2R}{\sqrt{3}} = \pi (\frac{2R^2}{3}) \cdot \frac{2R}{\sqrt{3}} = \frac{4\pi R^3}{3\sqrt{3}}$.