Question

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2R}{\sqrt{3}}$. Also, find the maximum volume of the cylinder.

Answer 1

Let $R$ be the radius of the sphere.
Let $h$ be the height and $x$ be the diameter of the cylinder.
In $\Delta ABC$,
Using Pythagoras theorem,
$(CB{)}^2+(AB{)}^2=(AC{)}^2$
$\Rightarrow h^2+x^2=(2R{)}^2$
$\Rightarrow x^2=4R^2-h^2$

Volume of cylinder is given by,
$V=\pi (\text{radius}{)}^2\times (\text{height})$
$=\pi {\left(\frac{x}{2}\right)}^{2}h$
$=\pi \left(\frac{4R^2-h^2}{4}\right)h$
$=\frac{4\pi R^{2}h}{4}-\frac{\pi h^3}{4}$
$=\pi R^{2}h-\frac{\pi h^3}{4}$

Differentiate w.r.t. $x$,
$\Rightarrow \frac{dV}{dh}=\pi R^2-\frac{3\pi h^2}{4}$
Put $\frac{dV}{dh}=0\Rightarrow \pi R^2-\frac{3\pi h^2}{4}=0$
$\Rightarrow h^2-\frac{4R^2}{3}=0$
$\Rightarrow h=\frac{2R}{\sqrt{3}}$

Now,
$\Rightarrow \frac{d^{2}V}{d{h}^2}=0-\frac{6\pi h}{4}=-\frac{3\pi h}{4}<0$
$\therefore h=\frac{2R}{\sqrt{3}}$ is a point of maxima.
$\Rightarrow x^2=4R^2-h^2=4R^2-\frac{4R^2}{3}=\frac{8R^2}{3}$

$V_{max}=\frac{\pi }{4}{x}^{2}h$
$=\frac{\pi }{4}\left(\frac{8R^2}{3}\right)\left(\frac{2R}{\sqrt{3}}\right)$
$=\frac{4\pi R^3}{3\sqrt{3}}\text{cubic unit}$