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
$\frac{dV}{dr}=\frac{\pi }{3}\left[2r\sqrt{l^2-r^2}+\frac{r^2}{2\sqrt{l^2-r^2}}(-2r)\right]$
Put $\frac{dV}{dr}=0$
$\Rightarrow 2r\left(\sqrt{l^2-r^2}\right)-\frac{r^3}{\sqrt{l^2-r^2}}=0$
$\Rightarrow r2\left(l^2-r^2\right)-r^3]=0$
$\Rightarrow 2l^2-3r^2=0$
$\Rightarrow r=\pm \sqrt{\frac{2}{3}}$
$\therefore$ 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 $\Delta ABC,\tan \theta =\frac{r}{h}$
$=\frac{\sqrt{\frac{2}{3}}}{\frac{l}{\sqrt{3}}}=\sqrt{2}$