Question

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and semi vertical angle $α$ is one-third that of the cone and the greatest volume of cylinder is $\frac{4}{27}πh^3 \,tan^2α$

Answer 1

The given right circular cone of fixed height $(h)$ and semi-vertical angle $(α)$ can be drawn as:
Here, a cylinder of radius $R$ and height $H$ is inscribed in the cone.
Then, $∴GAO=α, OG=r,OA=h,OE=R$, and $CE=H.$
We have,$r=h\,tanα$
Now, since $∆AOG$ is similar to $∆CEG$, we have:
$\frac{AO}{OG}=\frac{CE}{EG}$
$⇒\frac{h}{r}=\frac{H}{r-R}\,\,\,\, [EG=OG-OE]$
$⇒H=\frac{h}{r}(r-R)$
$=\frac{h}{h\,tanα}(h\,tanα-R)$
$=\frac{1}{tanα}(h\,tanα-R)$
Now, the volume $(V)$ of the cylinder is given by,
$V=πR^2H=\frac{πR^2}{tanα}(h\,tanα-R)$
$=πR^2h-\frac{πR^3}{tanα}$
$∴\frac{dV}{dR}=2πRh-\frac{3πR^2}{tanα}$
Now $\frac{dV}{dR}=0$
$⇒2πRh=\frac{3πR^2}{tanα}$
$⇒2h\,tanα=3R$
$⇒R=\frac{2h}{3}tanα$
Now,$\frac{d^2V}{dR^2}=2πh-\frac{6πR}{tanα}$
And for $R=\frac{2h}{3tanα}$,we have
$\frac{d^2V}{dR^2}=2πh-\frac{6π}{tanα}(\frac{2h}{3}tanα)$
$=2πh-4πh=-2πh<0$
∴By second derivative test, the volume of the cylinder is the greatest when
$R=\frac{2h}{3}tanα$
When $R=\frac{2h}{3}tanα,H=\frac{1}{tanα}(htanα-\frac{2h}{3}tanα)$
$=\frac{1}{tanα}(\frac{h\,tanα}{3})=\frac{h}{3}$
Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.
Now, the maximum volume of the cylinder can be obtained as:
$π(\frac{2h}{3}tanα)^2(\frac{h}{3})=π(\frac{4h^2}{9}tan^2α)(\frac{h}{3})$
$=\frac{4}{27}πh^3tan^2α$
Hence, the given result is proved