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

(A) Let $\theta$ be the semi-vertical angle of the cone.
It is clear that $\theta \in [0, \frac{\pi}{2}]$.
Let $r$,$h$,and $l$ be the radius,height,and the slant height of the cone respectively.
The slant height $l$ of the cone is given as constant.
Now,$r = l \sin \theta$ and $h = l \cos \theta$.
The volume $(V)$ of the cone is given by $V = \frac{1}{3} \pi r^2 h$.
Substituting the values of $r$ and $h$,we get:
$V = \frac{1}{3} \pi (l^2 \sin^2 \theta)(l \cos \theta) = \frac{1}{3} \pi l^3 \sin^2 \theta \cos \theta$.
Differentiating $V$ with respect to $\theta$:
$\frac{dV}{d\theta} = \frac{\pi l^3}{3} [\sin^2 \theta(-\sin \theta) + \cos \theta(2 \sin \theta \cos \theta)]$
$= \frac{\pi l^3}{3} [-\sin^3 \theta + 2 \sin \theta \cos^2 \theta]$.
For maximum or minimum volume,set $\frac{dV}{d\theta} = 0$:
$\sin^3 \theta = 2 \sin \theta \cos^2 \theta \Rightarrow \tan^2 \theta = 2 \Rightarrow \tan \theta = \sqrt{2} \Rightarrow \theta = \tan^{-1} \sqrt{2}$.
Now,check the second derivative:
$\frac{d^2V}{d\theta^2} = \frac{\pi l^3}{3} [-3 \sin^2 \theta \cos \theta + 2 \cos^3 \theta - 4 \sin^2 \theta \cos \theta] = \frac{\pi l^3}{3} [2 \cos^3 \theta - 7 \sin^2 \theta \cos \theta]$.
Substituting $\sin^2 \theta = 2 \cos^2 \theta$ at $\tan \theta = \sqrt{2}$:
$\frac{d^2V}{d\theta^2} = \frac{\pi l^3}{3} [2 \cos^3 \theta - 7(2 \cos^2 \theta) \cos \theta] = \frac{\pi l^3}{3} [2 \cos^3 \theta - 14 \cos^3 \theta] = -4 \pi l^3 \cos^3 \theta$.
Since $\theta \in [0, \frac{\pi}{2}]$,$\cos \theta > 0$,so $\frac{d^2V}{d\theta^2} < 0$.
Thus,the volume is maximum when $\theta = \tan^{-1} \sqrt{2}$.