Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and semi-vertical angle $\alpha$ is one-third that of the cone and the greatest volume of the cylinder is $\frac{4}{27} \pi h^{3} \tan^{2} \alpha$.

(N/A) Let the height of the cone be $h$ and the semi-vertical angle be $\alpha$. Let the radius of the base of the cone be $r = h \tan \alpha$.
Let the radius and height of the inscribed cylinder be $R$ and $H$ respectively.
By similar triangles,$\frac{H}{r-R} = \frac{h}{r}$.
Substituting $r = h \tan \alpha$,we get $\frac{H}{h \tan \alpha - R} = \frac{h}{h \tan \alpha} = \frac{1}{\tan \alpha}$.
Thus,$H = \frac{h \tan \alpha - R}{\tan \alpha} = h - \frac{R}{\tan \alpha}$.
The volume of the cylinder is $V = \pi R^{2} H = \pi R^{2} (h - \frac{R}{\tan \alpha}) = \pi h R^{2} - \frac{\pi R^{3}}{\tan \alpha}$.
To maximize $V$,we find $\frac{dV}{dR} = 2 \pi h R - \frac{3 \pi R^{2}}{\tan \alpha}$.
Setting $\frac{dV}{dR} = 0$,we get $2 \pi h R = \frac{3 \pi R^{2}}{\tan \alpha}$,which gives $R = \frac{2}{3} h \tan \alpha$.
Then $H = h - \frac{\frac{2}{3} h \tan \alpha}{\tan \alpha} = h - \frac{2}{3} h = \frac{1}{3} h$.
Since $\frac{d^{2}V}{dR^{2}} = 2 \pi h - \frac{6 \pi R}{\tan \alpha} = 2 \pi h - 4 \pi h = -2 \pi h < 0$,the volume is maximum at $H = \frac{1}{3} h$.
The maximum volume is $V = \pi (\frac{2}{3} h \tan \alpha)^{2} (\frac{1}{3} h) = \pi (\frac{4}{9} h^{2} \tan^{2} \alpha) (\frac{1}{3} h) = \frac{4}{27} \pi h^{3} \tan^{2} \alpha$.