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

The Correct Option is C

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} $ .