Question

A cylinder of radius $x$ and height $2h$ is to be inscribed in a sphere of radius $R$ centred at $O$ as shown in figure.


i. Evaluate the volume of cylinder in terms of $h$.
ii. The cylinder has maximum volume when $h=?$.

• A. i. $V=2\pi h(R^2-h^2)$
  ii. $h=\frac{R}{\sqrt{3}}$
• B. i. $V=2\pi h(R^2-h^3)$
  ii. $h=\frac{R}{\sqrt{3}}$
• C. i. $V=2\pi h(R^2-2h^2)$
  ii. $h=\frac{R}{\sqrt{5}}$
• D. i. $V=3\pi h(R^2-h^2)$
  ii. $h=\frac{2R}{\sqrt{3}}$

Answer 1

The correct option is A i. $V=2\pi h(R^2-h^2)$

ii. $h=\frac{R}{\sqrt{3}}$
Volume of cylinder $V=\pi x^{2}H$ (where, $H=2h$)

From Pythagoras theorem,
$x^2=R^2-h^2$
$V=2\pi h(R^2-h^2)$

For volume to be maximum $\frac{dV}{dh}=0\text{and}\frac{d^{2}V}{d{h}^2}<0$

$V=2\pi h(R^2-h^2)=2\pi h{R}^2-2\pi h^3$

$\frac{dV}{dh}=2\pi R^2-6\pi h^2=0$
$\therefore h=\frac{R}{\sqrt{3}}$

Minimum value is zero at $h=0$, so at $h=\frac{R}{\sqrt{3}}$ value should be maximum.
Let's confirm this,
$\frac{d^{2}V}{d{x}^2}=-12\pi <0$

Hence, at $h=\frac{R}{\sqrt{3}}$ volume of cylinder is maximum.