Show that the right circular cone of least curved surface area and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.

(N/A) Let $r$ and $h$ be the radius and the height (altitude) of the cone respectively.
The volume $(V)$ of the cone is given by $V = \frac{1}{3} \pi r^{2} h$,which implies $h = \frac{3V}{\pi r^{2}}$.
The curved surface area $(S)$ of the cone is given by $S = \pi r l = \pi r \sqrt{r^{2} + h^{2}}$.
Substituting $h$,we get $S = \pi r \sqrt{r^{2} + \frac{9V^{2}}{\pi^{2} r^{4}}} = \frac{1}{r} \sqrt{\pi^{2} r^{6} + 9V^{2}}$.
To minimize $S$,we minimize $S^{2} = \frac{\pi^{2} r^{6} + 9V^{2}}{r^{2}} = \pi^{2} r^{4} + 9V^{2} r^{-2}$.
Let $f(r) = S^{2}$. Then $f'(r) = 4 \pi^{2} r^{3} - 18V^{2} r^{-3}$.
Setting $f'(r) = 0$,we get $4 \pi^{2} r^{3} = \frac{18V^{2}}{r^{3}}$,so $r^{6} = \frac{18V^{2}}{4 \pi^{2}} = \frac{9V^{2}}{2 \pi^{2}}$.
Substituting $V = \frac{1}{3} \pi r^{2} h$ into $r^{6} = \frac{9V^{2}}{2 \pi^{2}}$,we get $r^{6} = \frac{9}{2 \pi^{2}} \cdot (\frac{1}{9} \pi^{2} r^{4} h^{2}) = \frac{1}{2} r^{4} h^{2}$.
Thus,$r^{2} = \frac{1}{2} h^{2}$,which implies $h^{2} = 2r^{2}$,or $h = \sqrt{2} r$.
Since $f''(r) = 12 \pi^{2} r^{2} + 54V^{2} r^{-4} > 0$ for $r > 0$,the surface area is indeed minimized at $h = \sqrt{2} r$.