Question

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

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

Answer 1

Answer: (C)

$\sqrt{2}$

Answer 2

In $\Delta OBC,$ $\sin \theta =\frac{BC}{OC}$
$\Rightarrow$ $r=l\sin \theta$ .. (i) and $\cos \theta =\frac{OB}{OC}$
$\Rightarrow$ $h=l\cos \theta$ ...(ii) Now, $V=\frac{1}{3}\pi {{r}^{2}}h$
$=\frac{1}{3}\pi {{l}^{2}}{{\sin }^{2}}\theta .l\cos \theta$
$\Rightarrow$ $V=\frac{\pi {{l}^{3}}}{3}{{\sin }^{2}}\theta \cos \theta$
$\therefore$ $\frac{dV}{d\theta }=\frac{\pi {{l}^{3}}}{3}[2\sin \theta co{{s}^{2}}\theta +{{\sin }^{2}}\theta -(\sin \theta )]$
$=\frac{\pi {{l}^{3}}}{3}\sin \theta (2co{{s}^{2}}\theta -{{\sin }^{2}}\theta )$ and $\frac{{{d}^{2}}V}{d{{\theta }^{2}}}=\frac{\pi {{l}^{3}}}{3}\cos \theta (2co{{s}^{2}}\theta -{{\sin }^{2}}\theta )$ $+\frac{\pi {{l}^{3}}}{3}\sin \theta (2co{{s}^{2}}\theta -{{\sin }^{2}}\theta )=0$ On putting $\frac{dV}{d\theta }=0,$ for maxima or minima
$\therefore$ $\frac{\pi {{l}^{3}}}{3}\sin \theta (2{{\cos }^{2}}\theta -{{\sin }^{2}}\theta )=0$
$\Rightarrow$ $\tan \theta =\sqrt{2}$ At $\theta ={{\tan }^{-1}}\sqrt{2}$ $\frac{{{d}^{2}}V}{d{{\theta }^{2}}}=-ve<0$
$\therefore$ Volume is maximum at $\tan \theta =\sqrt{2}$ .