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
$\frac{dV}{dr}=\frac{\pi }{3}\left[2r\sqrt{l^2-r^2}+\frac{r^2}{2\sqrt{1^2-r^2}}(-2r)\right]$
Put $\frac{dV}{dr}=0$, we get
$r=\pm l\sqrt{\frac{2}{3}}$
At $r=l\sqrt{\frac{2}{3}},\frac{d^{2}V}{d{r}^2}<0,$ (maximum)
$\therefore h=\sqrt{l^2-\frac{2}{3}{l}^2}-\frac{l}{\sqrt{3}}$
In $\Delta ABC,\tan \theta =\frac{r}{h}=\frac{l\sqrt{\frac{2}{3}}}{l/3}=\sqrt{2}$