Question

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

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

Answer 1

Answer: (C)

$\sqrt{2}$

From the figure:
Let $OB=l$, $OA=l\cos \theta$ and $AB=l\sin \theta$ where, $0\le \theta \le \pi /2$.
Then $V=\frac{\pi }{3}{\left(AB\right)}^2\left(OA\right)=\frac{\pi }{3}{l}^3{\sin }^2\theta \cos \theta$
$\Rightarrow \frac{dV}{d\theta }=\frac{\pi }{3}{l}^3\sin \theta (3{\cos }^2\theta -1)$
So from $dV/d\theta =0$, we get $\sin \theta =0$ or $\cos \theta =1/\sqrt{3}$.
Also, $V\left(0\right)=0$, $V\left(/pi/2\right)=0$ and $V\left({\cos }^{-1}\frac{1}{\sqrt{3}}\right)=\frac{2\pi l^3}{9\sqrt{3}}$
Hence $V$ is maximum when $\cos \theta =1/\sqrt{3}$ i.e., $\tan \theta =\sqrt{2}$.