Show that the semi vertical angle of a right circular cone of given surface area and maximum volume is ${\sin }^{-1}\frac{1}{\sqrt{3}}$

Let $r,h,l$ be the radius , height and slant height of the right circular cone respectively.

Let $S$ be the given surface area of the cone.

We have, $l^2=r^2+h^2$ ....(1)

$S=\pi rl+\pi r^2$

$S-\pi r^2=\pi rl$

$\Rightarrow l=\frac{S-\pi r^2}{\pi r}$ .....(2)

$V=\frac{1}{3}\pi r^{2}h$

$\Rightarrow V=\frac{1}{3}\pi r^2\sqrt{l^2-r^2}$ (by (1))

$\Rightarrow V^2=\frac{1}{9}{\pi }^2{r}^4(l^2-r^2)$

$\Rightarrow V^2=\frac{1}{9}{\pi }^2{r}^4\left[\right(\frac{S-\pi r^2}{\pi r}{)}^2-r^2]$

$\Rightarrow V^2=\frac{1}{9}{\pi }^2{r}^4\left[\frac{(S-\pi r^2{)}^2-{\pi }^2{r}^4}{{\pi }^2{r}^2}\right]$

$\Rightarrow V^2=\frac{1}{9}{r}^2\left[\right(S-\pi r^2{)}^2-{\pi }^2{r}^4]$

$\Rightarrow V^2=\frac{1}{9}{r}^2[S^2-2\pi S{r}^2+{\pi }^2{r}^4-{\pi }^2{r}^4]$

$\Rightarrow V^2=\frac{1}{9}(r^2{S}^2-2\pi S{r}^4)$

$2V\frac{dV}{dr}=\frac{S^2}{9}2r-\frac{2\pi S}{9}4r^3$

$2V\frac{dV}{dr}=\frac{2rS}{9}(S-4\pi r^2)$

For maximum volume, $\frac{dV}{dr}=0$

$\Rightarrow \frac{2rS}{9}(S-4\pi r^2)=0$

$\Rightarrow r=0$ or $S-4\pi r^2=0$

Since, $r$ cannot be 0

$\Rightarrow S=4\pi r^2$

$\Rightarrow r^2=\frac{S}{4\pi }$

$\Rightarrow r^2=\frac{\pi rl+\pi r^2}{4\pi }$

$\Rightarrow 4\pi r^2=\pi rl+\pi r^2$

$\Rightarrow 3\pi r^2=\pi rl$

$\Rightarrow l=3r$

Let $\alpha$ be the semi-vertical angle .

$\sin \alpha =\frac{r}{l}$

$\sin \alpha =\frac{r}{3r}$

$\Rightarrow \alpha ={\sin }^{-1}\left(\frac{1}{3}\right)$