The volume of the greatest cylinder which can be inscribed in a cone of height 30 cm and semi-vertical angle 30^o is

From geometry, wo have $\frac{\mathrm{r}}{30\tan {30}^{\circ }}=\frac{30-\mathrm{h}}{30}$or $\mathrm{h}=30-\sqrt{3}\mathrm{r}$
Now, the volume of cylinder, $\mathrm{V}={\mathrm{\pi r}}^2\mathrm{h}={\mathrm{\pi r}}^2(30-\sqrt{3}\mathrm{r})$
Now, let $\frac{\mathrm{dV}}{\mathrm{dr}}=0\text{ or }\mathrm{\pi }\left(60\mathrm{r}-3\sqrt{3}{\mathrm{r}}^2\right)=0\text{ or }\mathrm{r}=\frac{20}{\sqrt{3}}$
Hence, ${\mathrm{V}}_{max}=\mathrm{\pi }{\left(\frac{20}{\sqrt{3}}\right)}^2\left(30-\sqrt{3}\frac{20}{\sqrt{3}}\right)$
                    $=\mathrm{\pi }\frac{400}{3}\times 10=\frac{4000\mathrm{\pi }}{3}$