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

Let $\theta$ be the semi vertical angle of the cone, where $l$ be the given slant height, then radius of base is $\left( r \right) = l\sin \theta$,
height$\left( h \right) = l\cos \theta$
( is right angled triangle)

and volume of cone$= \cfrac{1}{3}\pi {r^2}h$

$\Rightarrow V = \cfrac{1}{3}\pi {\left( {l\sin \theta } \right)^2}l\cos \theta = \cfrac{1}{3}\pi {l^3}{\sin ^2}\theta \cos \theta$

$\Rightarrow \cfrac{{dV}}{{d8}} = \cfrac{1}{3}\pi {l^3}\left\{ {\left( {{{\sin }^2}\theta } \right)\left( { - \sin \theta } \right) + \cos \theta \times 2\sin \theta cos\theta } \right\}$

$= \cfrac{1}{3}\pi {l^3}\sin \theta \left[ { - {{\sin }^2}\theta + 2\left( {1 - {{\sin }^2}\theta } \right)} \right]$
$= \cfrac{1}{3}\pi {l^3}\sin \theta {\cos ^2}\theta \left[ {2{{\sec }^2}\theta - 3{{\tan }^2}\theta } \right]$

$= \cfrac{1}{3}\pi {l^3}\sin \theta {\cos ^2}\theta \left[ {2 - {{\tan }^2}\theta } \right]$
[For maximum or minimum volume, ]

$\cfrac{{dV}}{{d\theta }} = 0$
$\Rightarrow \cfrac{1}{3}\pi {l^3}\sin \theta {\cos ^2}\theta \left( {2 - {{\tan }^2}\theta } \right) = 0 \Rightarrow {\tan ^2}\theta = 2$

$\Rightarrow \tan \theta = \left( {\sqrt 2 } \right)$
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {\sqrt 2 } \right)$

$\cfrac{{{d^2}V}}{{d{\theta ^2}}} = \cfrac{1}{3}\pi {l^3}{\cos ^3}\theta \left( {2 - 7{{\tan }^2}\theta } \right)$

$\Rightarrow {\left( {\cfrac{{{d^2}V}}{{d{\theta ^2}}}} \right)_{\tan \theta = \sqrt 2 }}$

$= \cfrac{1}{3}\pi {l^3}{\left( {\cfrac{1}{{\sqrt 3 }}} \right)^3}\left( {2 - 7 \times 2} \right) = - \cfrac{{4\pi {l^3}}}{{3\sqrt 3 }} < 0$

Thus, $V$ is maximum, when $\tan \theta = \sqrt 2$ or $\theta = {\tan ^{ - 1}}\sqrt 2$

Thus, the semi-vertical angle of the cone is ${\tan ^{ - 1}}\sqrt 2$.