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Q.
Find the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $a$.
Application of Derivatives
Solution:
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, $CA^2 = OL^2 + AL^2$
$\Rightarrow AL = \sqrt{a^{2}-x^{2}}$
Let $V$ be the volume of the cylinder. Then,
$V = \pi\left(AL\right)^{2}\times LM$
$\Rightarrow V = \pi \left(AL\right)^{2}\times 2\left(OL\right)$
$\Rightarrow V = 2\pi \left(a^{2}-x^{2}\right)\times2x$
$\Rightarrow V = 2\pi \left(a^{2}x-x^{3}\right)$
$\Rightarrow \frac{dV}{dx} = 2\pi \left(a^{2}-3x^{3}\right)$
and $\frac{d^{2}V}{dx^{2}} = -12\,\pi x$
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}}$
$\left[\text{neglecting}\, x = \frac{-a}{\sqrt{3}} \because \frac{d^{2}V}{dx^{2}}\bigg|_{x = \frac{-a}{\sqrt{3}}} > 0\right]$
Clearly, $\left(\frac{d^{2}V}{dx^{2}}\right)_{x = a /\sqrt{3}} =-12\pi\times \frac{a}{\sqrt{3}} < 0$
$\therefore V$ is maximum when $x = \frac{a}{\sqrt{3}}$
Hence, height of the cylinder $LM = 2x = \frac{2a}{\sqrt{3}}$.
