Question

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is ${\tan }^{-1}\sqrt{2}$

Answer 1

The slant height is given by
$l=\sqrt{h^2+r^2}$
Now volume is
$V=\frac{\pi }{3}\left(r^2\right)h$
$=\frac{\pi }{3}(l^2-h^2)h$
$\frac{dV}{dh}=\frac{\pi }{3}[l^2-h^2-h(2h\left)\right]$
Now for critical points
$\frac{dV}{dh}=0$
Or
$l^2-h^2-2h^2=0$
$l^2=3h^2$
$l=\sqrt{3}h$
$\frac{h}{l}=\frac{1}{\sqrt{3}}=\cos \theta$ where $\theta$ is the semi vertical angle.
Hence
$\tan \theta =\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }$
$=\frac{\frac{\sqrt{2}}{\sqrt{3}}}{\frac{1}{\sqrt{3}}}$
$=\sqrt{2}$
Therefore
$\tan \theta =\sqrt{2}$
Or
$\theta ={\tan }^{-1}\left(\sqrt{2}\right)$