Show that the height of the right circular cylinder of greatest volume which can be inscribed in a right circular cone of height h and radius r is one-third of the cone, and the greatest volume of the cylinder is $\frac{4}{9}$times the volume of the cone.

The below figure represents the cylinder inscribed in cone,

Schematic diagram
In a cone, radius= R, height=H
Semi vertical angle =θ
In a cylinder, radius= r, height=h
The expression of similar triangle,
$\begin{array}{l}\mathrm{△}\mathrm{ACE}~\mathrm{△}\mathrm{FCD}\\ \therefore \frac{\mathrm{CE}}{\mathrm{EA}}=\frac{\mathrm{CD}}{\mathrm{DF}}\end{array}$
Substitute the given values,
$\begin{array}{r}\therefore \frac{\mathrm{H}}{\mathrm{R}}=\frac{\mathrm{H}-\mathrm{h}}{\mathrm{r}}\\ \therefore \mathrm{h}=\mathrm{H}-\frac{\mathrm{H}}{\mathrm{R}}\mathrm{\gamma }\end{array}$
The expression of volume of cylinder,
$\mathrm{V}={\mathrm{\pi r}}^2\mathrm{h}$
Substitute the given value of height of cylinder,
$\begin{array}{l}\mathrm{V}={\mathrm{\pi r}}^2\left(\mathrm{H}-\frac{\mathrm{H}}{\mathrm{R}}\mathrm{r}\right)\\ \therefore \mathrm{V}={\mathrm{\pi r}}^2\mathrm{H}-{\mathrm{\pi r}}^3\frac{\mathrm{H}}{\mathrm{R}}\end{array}$
Differentiate the above expression with respect to radius r and equate it to zero,
$\begin{array}{l}\therefore \frac{\mathrm{d}}{\mathrm{dr}}\left(\pi r^{2}H-\pi r^3\frac{H}{R}\right)=0\\ \therefore 2\pi rH-3\pi r^2\frac{H}{R}=0\\ \therefore \left.r=\frac{2R}{3}\text{ (because radius }>0\right)\end{array}$
Second differentiate at the given value of radius is negative. So, it gives maxima.
Substitute the value of radius in the height of cylinder.
$\begin{array}{l}\therefore \mathrm{h}=\mathrm{H}-\frac{\mathrm{H}}{\mathrm{R}}\left(\frac{2\mathrm{R}}{3}\right)\\ \therefore \mathrm{h}=\frac{\mathrm{H}}{3}\end{array}$
Substitute the calculated value of radius and height in the expression of volume of cylinder.
$\begin{array}{l}\therefore {\mathrm{V}}_{max}=\mathrm{\pi }{\left(\frac{2\mathrm{R}}{3}\right)}^2\left(\frac{\mathrm{H}}{3}\right)\\ =\mathrm{\pi }\left(\frac{4{\mathrm{R}}^2}{9}\right)\left(\frac{\mathrm{H}}{3}\right)\\ =\frac{4}{9}\left(\frac{1}{3}{\mathrm{\pi R}}^2\mathrm{H}\right)\end{array}$
The expression for volume of cone,
${\mathrm{V}}_{\mathrm{c}}=\frac{1}{3}{\mathrm{\pi R}}^2\mathrm{H}$
Substitute the volume of cone in the expression of maximum volume of cylinder.
$\therefore {\mathrm{V}}_{max}=\frac{4}{9}{\mathrm{V}}_{\mathrm{c}}$
Therefore, the greatest volume of cylinder that can be inscribed in cone has 4/9 volume of cone. That have proved.