Maximum Volume of Cone With Fixed Slant Height

Given a cone of fixed slant height
$l$
, what are the radius of the base and the height of the cone that give the maximum volume?

$l=\sqrt{r^2+h^2} \rightarrow r^2=l^2-h^2$

The volume of a cone is given by
$V= \frac{ \pi}{3}r^2h=\frac{\pi}{3}(l^2-h^2)h=\frac{\pi}{3} (l^2h-h^3)$

$\frac{dV}{dh}= \frac{\pi}{3}(l^2-3h^2)$

The volume is maximum when
$\frac{dV}{dh}=0 \rightarrow l^2-3h^2=0 \rightarrow h=\frac{l}{\sqrt{3}}$
.
Then
$r=\sqrt{l^2-h^2}=\sqrt{l^2-( \frac{l}{\sqrt{3}})^2}= h \sqrt{\frac{2}{3}}$
.
The maximum volume is
$V= \frac{\pi}{3} r^2h= \frac{\pi}{3} \frac{2l^2}{3} \frac{l}{\sqrt{3}}= \frac{2 \pi l^3 \sqrt{3}}{27}$