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

Let $r$ and $h$ be the radius and height respectively of the cone $ABC$

Volume, $V = \cfrac{1}{3}\pi {r^2}h \Rightarrow h = \cfrac{{3V}}{{\pi {r^2}}}$

Curved surface area, $S = \pi rl$
$= \pi r\left( {\sqrt {{h^2} + {r^2}} } \right) = \pi r\left( {\sqrt {{{\left( {\cfrac{{3V}}{{\pi {r^2}}}} \right)}^2} + {r^2}} } \right)$
${S^2} = {(\pi r)^2}\left[ {{{\left( {\cfrac{{3V}}{{\pi {r^2}}}} \right)}^2} + {r^2}} \right] = \cfrac{{{{(3V)}^2}}}{{{r^2}}} + {\pi ^2}{r^4}$

Let $Z = {S^2}$
$\Rightarrow Z = \cfrac{{{{(3V)}^2}}}{{{r^2}}} + {\pi ^2}{r^4}$ …(i)

Differentiating (i) w.r.t. $r$, we get $\cfrac{{dZ}}{{dr}} = \cfrac{{ - 2{{(3V)}^2}}}{{{r^3}}} + 4{\pi ^2}{r^3}$

For maximum/minimum surface area,
$\cfrac{{dZ}}{{dr}} = 0$

$\Rightarrow \cfrac{{ - 2{{(3V)}^2}}}{{{r^3}}} + 4{\pi ^2}{r^3} = 0 \Rightarrow - 2{(3V)^2} + 4{\pi ^2}{r^6} = 0$

$\Rightarrow - {(3V)^2} + 2{\pi ^2}{r^6} = 0 \Rightarrow 2{\pi ^2}{r^6} = {(3V)^2}$

$\Rightarrow {r^6} = \cfrac{{{{(3V)}^2}}}{{2{\pi ^2}}}$

$= \cfrac{{{{\left( {3\left( {\cfrac{1}{3}\pi {r^2}h} \right)} \right)}^2}}}{{2{\pi ^2}}} = \cfrac{{{r^4}{h^2}}}{2}$
$\Rightarrow {r^2} = \cfrac{{{h^2}}}{2} \Rightarrow 2{r^2} = {h^2}$

$\cfrac{{{d^2}Z}}{{d{r^2}}} = - 2{(3V)^2}\left[ {\cfrac{{ - 3}}{{{r^4}}}} \right] + 12{\pi ^2}{r^2}$

and ${\left( {\cfrac{{{d^2}Z}}{{d{r^2}}}} \right)_{{r^2} = \cfrac{{{h^2}}}{2}}} > 0$

Hence the surface area is minimum at $2{r^2} = {h^2} \Rightarrow h = \sqrt 2 r$