## Maximum Volume of Cylinder in a Cone

Given a cone of fixed height
$H$
$R$
, what are the radius of the base and the height of the cylinder that give the maximum volume?

From the diagram,
$\frac{H-h}{}= \frac{r}{R} \rightarrow r= \frac{R(H-h)}{H}$
.
The volume of a cylinder is given by
$V= \pi r^2h= \pi (\frac{R(H-h)}{H})^2h=\frac{\pi R^2}{H^2}(H^2h-2Hh^2+h^3)$

$\frac{dV}{dh}= \frac{\pi R^2}{H^2}(H^2-4Hh+3h^2)$

The volume is maximum when
$\frac{dV}{dh}=0 \rightarrow H^2-4Hh+3h^2=0 \rightarrow (H-3h)(H-h)=0$
.
Then
$h=0 \rightarrow V=0$
or
$h= \frac{H}{3}$
and
$r=R \frac{H-h}{H}=R \frac{H-H/3}{H}= \frac{2}{3}R$
.
The maximum volume is
$V= \pi r^2h= \pi (\frac{2R}{3})^2 \frac{H}{3}= \frac{4 \pi R^2H}{27}$
.