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Q.
Find the volume of the largest cylinder that can be inscribed in a sphere of radius $r\, cm$.
Application of Derivatives
Solution:
Let $h$ be the height and $R$ be the radius of the base of the inscribed cylinder. Let $V$ be the volume of the cylinder. Then,
$V=\pi R^{2}h ...(i) $
From $ \Delta OCA $, we have
$r^{2} = \left(\frac{h}{2}\right)^{2} + R^{2} $
$\Rightarrow R^{2} = r^{2} - \frac{h^{2}}{4}$
$ \therefore V = \pi\left(r^{2} - \frac{h^{2}}{4}\right) h$
$ \Rightarrow V = \pi r^{2}h - \frac{\pi}{4} h^{3}$
$\Rightarrow \frac{dv}{dh} = \pi r^{2} - \frac{3\pi h^{2}}{4}$ and $ \frac{d^{2}V}{dh^{2}} = - \frac{3\pi h}{2}$
For maximum or minimum values of $V$, we must have
$\frac{dV}{dh} = 0 $
$\Rightarrow \pi r^{2} - \frac{3\pi h^{2}}{4} = 0$
$ \Rightarrow h^{2} = \frac{4r^{2}}{3}$
$ \Rightarrow h= \frac{2}{\sqrt{3}} r$
$ \qquad\left[ {\text{neglecting}}\,\, h=\frac{-2}{\sqrt{3} } r \because\frac{d^2V}{dh^{2}}\bigg|_{h=\frac{-2}{\sqrt{3}} r} >0\right]$
Now, $\left(\frac{d^{2}V}{dh^{2}}\right)_{h=\frac{2r}{\sqrt{3}} }= -\sqrt{3}\pi r < 0 $.
Thus, $V$ is maximum when $h =\frac{2r}{\sqrt{3}}$ .
Putting $h= \frac{2r}{\sqrt{3}}$ in $R^{2} = r^{2} -\frac{h^{2}}{4}$, we obtain $R = \sqrt{\frac{2}{3}}r. $
$\qquad\left[{\text{neglecting -ve value of}}\, R\right] $
$\therefore $ The maximum volume of the cylinder is given by
$ V= \pi R^{2}h = \pi \left(\frac{2}{3} r^{2}\right)\left(\frac{2r}{\sqrt{3}}\right) = \frac{4\pi r^{2}}{3\sqrt{3}}$
