Question

Find the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $a$.

• A. $2a/3$
• B. $\frac{2a}{\sqrt{3}}$
• C. $a/3$
• D. $a/5$

Answer 1

Answer: (B)

Let $r$ be the radius of the base and $h$ be the height of the cylinder $ABCD$ which is inscribed in a sphere of radius $a$. It is obvious that for maximum volume the axis of the cylinder must be along the diameter of the sphere. Let $O$ be the centre of the sphere such that
$OL=x$. Then, $C{A}^2=O{L}^2+A{L}^2$
$\Rightarrow AL=\sqrt{a^2-x^2}$
Let $V$ be the volume of the cylinder. Then,
$V=\pi {\left(AL\right)}^2\times LM$
$\Rightarrow V=\pi {\left(AL\right)}^2\times 2\left(OL\right)$
$\Rightarrow V=2\pi \left(a^2-x^2\right)\times 2x$

$\Rightarrow V=2\pi \left(a^{2}x-x^3\right)$
$\Rightarrow \frac{dV}{dx}=2\pi \left(a^2-3x^3\right)$
and $\frac{d^{2}V}{d{x}^2}=-12\pi x$
For maximum or minimum, we must have
$\frac{dV}{dx}=0$
$\Rightarrow 2\pi \left(a^2-3x^3\right)=0$
$\Rightarrow x=\frac{a}{\sqrt{3}}$
$\left[\text{neglecting}x=\frac{-a}{\sqrt{3}}\because \frac{d^{2}V}{d{x}^2}{|}_{x=\frac{-a}{\sqrt{3}}}>0\right]$
Clearly, ${\left(\frac{d^{2}V}{d{x}^2}\right)}_{x=a/\sqrt{3}}=-12\pi \times \frac{a}{\sqrt{3}}<0$
$\therefore V$ is maximum when $x=\frac{a}{\sqrt{3}}$
Hence, height of the cylinder $LM=2x=\frac{2a}{\sqrt{3}}$.