question_answer

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)         \[\sqrt{3}\]

E)  \[\sqrt{3}+\sqrt{2}\]

Correct Answer: C

Solution :

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\[\tan \theta =\sqrt{2}\].