Question

Show that the height of the cylinder of maximum volume that can be inscribed in a cone of height his $\frac{h}{3}$

Answer 1

Let radius $CD$ of inscribed cylinder be $x$ and height $OC$ be $H$ and $q$ be the semi-vertical angle of cone.

$\therefore ,OC=OB-BC$

$\Rightarrow H=h-x\cot \theta$

Now, volume of cylinder

$V=\pi x^2(h-x\cot \theta )$

$V=\pi (x^{2}h-x^3\cot \theta )$

Form maximum or minimum value,

$\frac{dV}{dx}=0\Rightarrow \pi (2xh-3x^2\cot \theta )=0$

$\Rightarrow \pi x(2h-3x\cot \theta )=0$

$\Rightarrow 2h-3x\cot \theta =0$ as $x=0$ is not possible.

$\therefore x=\frac{2h}{3}\tan \theta$

Now, $\frac{d^{2}V}{d{x}^2}=\pi (2h-6x\cot \theta )$

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

${\left[\frac{d^{2}V}{d{x}^2}\right]}_{x=\frac{2h\tan \theta }{3}}=2\pi h-6\pi \times \frac{2h}{3}\tan \theta \cot \theta$

$2\pi h-4\pi h=-2\pi h<0$

Hence, volume will be maximum when $x=\frac{2h}{3}\tan \theta$

$\therefore$ height of cylinder

$H=h-x\cot \theta$

$=h-\frac{2h}{3}\tan \theta \cot \theta$

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

Thus, the height of the cylinder is $\frac{1}{3}$ of height of cone.