Question

If $\theta$ is the semi vertical angle of a cone of maximum volume and given slant height, then $\tan \theta$ is given by

• A. 2
• B. 1
• C. $ \sqrt{2} $
• D. $ \sqrt{3} $

Answer 1

Answer: (C)

Volume of cone, $V=\frac{\pi}{3} r^{2} h$

$\Rightarrow V=\frac{\pi}{3} r^{2} \sqrt{l^{2}-r^{2}}$
On differentiating w.r.t., $r$ we get
$\frac{d V}{d r}=\frac{\pi}{3}\left[2 r \sqrt{l^{2}-r^{2}}+\frac{r^{2}}{2 \sqrt{l^{2}-r^{2}}}(-2 r)\right]$
Put $\frac{d V}{d r}=0$
$\Rightarrow 2 r\left(\sqrt{l^{2}-r^{2}}\right)-\frac{r^{3}}{\sqrt{l^{2}-r^{2}}}=0$
$\Rightarrow \left.r 2\left(l^{2}-r^{2}\right)-r^{3}\right]=0$
$\Rightarrow 2 l^{2}-3 r^{2}=0$
$\Rightarrow r=\pm \sqrt{\frac{2}{3}}$
$\therefore $ At $r=\sqrt{\frac{2}{3}}, \frac{d^{2} V}{d r^{2}}<0,$ maxima
$\therefore h=\sqrt{l^{2}-\frac{2}{3} l^{2}}=\frac{l}{\sqrt{3}}$
In $\Delta A B C, \tan \theta =\frac{r}{h}$
$=\frac{\sqrt{\frac{2}{3}}}{\frac{l}{\sqrt{3}}}=\sqrt{2}$