Show that the right circular cone of least curved surface

Let r and h be the radius and the height (altitude) of the cone respectively.

Then, the volume (V) of the cone is given as:

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

The surface area (S) of the cone is given by,

S = πrl (where l is the slant height)

$=\pi r \sqrt{r^{2}+h^{2}}$

$=\pi r \sqrt{r^{2}+\frac{9 V^{2}}{\pi^{2} r^{4}}}=\frac{\pi r \sqrt{\pi^{2} r^{6}+9 V^{2}}}{\pi r^{2}}$

$=\frac{1}{r} \sqrt{\pi^{2} r^{6}+9 V^{2}}$

$\therefore \frac{d S}{d r}=\frac{r \cdot \frac{6 \pi^{2} r^{5}}{2 \sqrt{\pi^{2} r^{6}+9 V^{2}}}-\sqrt{\pi^{2} r^{6}+9 V^{2}}}{r^{2}}$

$\quad=\frac{3 \pi^{2} r^{6}-\pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}}$

$\quad=\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}}$

$\quad=\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}}$

Now,$\frac{d S}{d r}=0 \Rightarrow 2 \pi^{2} r^{6}=9 V^{2} \Rightarrow r^{6}=\frac{9 V^{2}}{2 \pi^{2}}$

Thus, it can be easily verified that when $r^{6}=\frac{9 V^{2}}{2 \pi^{2}}, \frac{d^{2} S}{d r^{2}}>0$.

$\therefore$ By second derivative test, the surface area of the cone is the least when $r^{6}=\frac{9 V^{2}}{4}$

When $r^{6}=\frac{9 V^{2}}{2 \pi^{2}}, h=\frac{3 V}{\pi r^{2}}=\frac{3}{\pi r^{2}}\left(\frac{2 \pi^{2} r^{6}}{9}\right)^{\frac{1}{2}}=\frac{3}{\pi r^{2}} \cdot \frac{\sqrt{2} \pi r^{3}}{3}=\sqrt{2} r$.

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to $\sqrt{2}$ times the radius of the base.