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 $\alpha$ is one-third that of the cone and the greatest volume of cylinder is $\frac{4}{27}\pi h^3{\tan }^2\alpha$.

Answer 1

Let $VAB$ be the cone of height $h$, semi vertical angle $\alpha$ and let $x$be the radius of the base of the cylinder $A^{\prime }{B}^{\prime }DC$ which is inscribed in the cone $VAB$.

Then $O{O}^{\prime }$ height of the cylinder $=VO-V{O}^{\prime }=(h-x\cot \alpha )\pi x^2$

Now differentiating with respect to $x$,

$\frac{dV}{dx}=2\pi xh-3\pi x^2\cot \alpha$

For maxima or minima $V,\frac{dV}{dx}=0$

$2\pi xh-3\pi x^2\cot \alpha =0$

$\Rightarrow 2\pi xh=3\pi x^2\cot \alpha$

$\Rightarrow x=\frac{3\pi x^2\cot \alpha }{2\pi h}$

$\Rightarrow 2hx=3\pi x^2\cot \alpha$

$\Rightarrow 2h=3\pi x\cot \alpha$

$\Rightarrow x=\frac{2h}{3\pi \cot \alpha }$

We know $\tan \alpha =\frac{1}{\cot \alpha }$

Thus $x=\frac{2h}{3}\tan \alpha$

Now $\frac{d^{2}V}{d{x}^2}=2\pi h-6\pi x\cot \alpha$

When $x=\frac{2h}{3}\tan \alpha$, we have

$\frac{d^{2}V}{d{x}^2}=\pi (2h-4h)=-2\pi h<0$

$\Rightarrow V$ is maximum when $x=\frac{2h}{3}\tan \alpha$

$\Rightarrow O{O}^{\prime }=h-x\cot \alpha$

$=h-\frac{2h}{3}$

$=\frac{h}{3}$

The maximum volume of cylinder is $V=\pi {\left(\frac{2h}{3}\tan \alpha \right)}^2(h-\frac{2h}{3})$

$\Rightarrow V=\frac{4}{27}\pi h^3{\tan }^2\alpha$