Question

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

• A. $4000\frac{\pi }{3}$ cubic cm
• B. $400\frac{\pi }{3}$ cubic cm
• C. $4000\frac{\pi }{\sqrt{3}}$ cubic cm
• D. None of these

Answer 1

The correct option is A $4000\frac{\pi }{3}$ cubic cm

From figure, we have
$\tan {30}^{\circ }=\frac{r}{30-h}$
$\Rightarrow h=30-\sqrt{3}r$
The volume of cylinder, $V=\pi r^{2}h=\pi r^2(30-\sqrt{3}r)$
Now, $\frac{dV}{dr}=0$
$\Rightarrow \pi (60r-3\sqrt{3}{r}^2)=0$
$\Rightarrow r=\frac{20}{\sqrt{3}}$
$\frac{d^{2}V}{d{r}^2}=\pi (60-6\sqrt{3}r)\Rightarrow {\left(\frac{d^{2}V}{d{r}^2}\right)}_{r=\frac{20}{\sqrt{3}}}=-60\pi <0$
Hence, $V_{max}=\pi {\left(\frac{20}{\sqrt{3}}\right)}^2(30-\sqrt{3}\times \frac{20}{\sqrt{3}})=\pi \frac{400}{3}\times 10$
$\Rightarrow V_{max}=\frac{4000\pi }{3}$