Answer
$\frac{2\pi{R^{3}}}{9\sqrt 3}$
Work Step by Step
$h^{2}+r^{2}$ = $R^{2}$
$V$ = $\frac{\pi}{3}{r^{2}}{h}$ = $\frac{\pi}{3}(R^{2}h-h^{3})$
$V'(h)$ = $\frac{\pi}{3}(R^{2}-3h^{2})$
$V'(h)$ = $0$
$\frac{\pi}{3}(R^{2}-3h^{2})$ = $0$
$h$ = $\frac{R}{\sqrt 3}$
This gives an absolute maximum since
$V'(h)$ $\gt$ $0$ for $0$ $\lt$ $h$ $\lt$ $\frac{R}{\sqrt 3}$
$V'(h)$ $\lt$ $0$ for $h$ $\gt$ $\frac{R}{\sqrt 3}$
The maximum volume is
$V\left(\frac{R}{\sqrt 3}\right)$ = $\frac{\pi}{3}\left(\frac{R^{3}}{\sqrt 3}-\frac{R^{3}}{3\sqrt 3}\right)$ = $\frac{2\pi{R^{3}}}{9\sqrt 3}$
