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

(N/A) 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,$l^{2}=r^{2}+h^{2} \quad \dots (1)$
$S=\pi r l+\pi r^{2}$
$S-\pi r^{2}=\pi r l$
$\Rightarrow l=\frac{S-\pi r^{2}}{\pi r}$
$V=\frac{1}{3} \pi r^{2} h$
$\Rightarrow V=\frac{1}{3} \pi r^{2} \sqrt{l^{2}-r^{2}} \quad (\text{by } (1))$
$\Rightarrow V^{2}=\frac{1}{9} \pi^{2} r^{4}\left(l^{2}-r^{2}\right)$
$\Rightarrow V^{2}=\frac{1}{9} \pi^{2} r^{4}\left[\left(\frac{S-\pi r^{2}}{\pi r}\right)^{2}-r^{2}\right]$
$\Rightarrow V^{2}=\frac{1}{9} \pi^{2} r^{4}\left[\frac{\left(S-\pi r^{2}\right)^{2}-\pi^{2} r^{4}}{\pi^{2} r^{2}}\right]$
$\Rightarrow V^{2}=\frac{1}{9} r^{2}\left[\left(S-\pi r^{2}\right)^{2}-\pi^{2} r^{4}\right]$
$\Rightarrow V^{2}=\frac{1}{9} r^{2}\left[S^{2}-2 \pi S r^{2}+\pi^{2} r^{4}-\pi^{2} r^{4}\right]$
$\Rightarrow V^{2}=\frac{1}{9}\left(r^{2} S^{2}-2 \pi S r^{4}\right)$
$2 V \frac{d V}{d r}=\frac{S^{2}}{9} (2 r)-\frac{2 \pi S}{9} (4 r^{3})$
$2 V \frac{d V}{d r}=\frac{2 r S}{9}\left(S-4 \pi r^{2}\right)$
For maximum volume,$\frac{d V}{d r}=0$
$\Rightarrow \frac{2 r S}{9}\left(S-4 \pi r^{2}\right)=0$
Since $r \neq 0$,we have $S=4 \pi r^{2}$
$\Rightarrow r^{2}=\frac{S}{4 \pi}$
Substituting $S=\pi r l+\pi r^{2}$,we get $4 \pi r^{2}=\pi r l+\pi r^{2}$
$\Rightarrow 3 \pi r^{2}=\pi r l$
$\Rightarrow l=3 r$
Let $\alpha$ be the semi-vertical angle.
$\sin \alpha=\frac{r}{l}=\frac{r}{3 r}=\frac{1}{3}$
$\Rightarrow \alpha=\sin ^{-1}\left(\frac{1}{3}\right)$