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Q.
If $\theta$ is the semi-vertical angle of a cone of maximum volume and given slant height, then $\tan \theta$ is given by
ManipalManipal 2016
Solution:
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
$\frac{d V }{d r}=\frac{\pi}{3}\left[2 r \sqrt{l^{2}-r^{2}}+\frac{r^{2}}{2 \sqrt{1^{2}-r^{2}}}(-2 r)\right]$
Put $\frac{d V }{d r}=0$, 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 $\Delta ABC , \tan \theta=\frac{r}{h}=\frac{l \sqrt{\frac{2}{3}}}{l / 3}=\sqrt{2}$
