The volume of the greatest cylinder which can be inscribed in a cone of height $\text{30 cm}$ and semi – vertical angle ${30}^0$ is 

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