Question

Find the volume of the largest cylinder that can be inscribed in a sphere of radius $rcm$.

• A. $\frac{\pi r^3}{3\sqrt{3}}$
• B. $\frac{4\pi r^{2}h}{3\sqrt{3}}$
• C. $4\pi r^3$
• D. $\frac{4\pi r^3}{3\sqrt{3}}$

Answer 1

Answer: (D)

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 $\frac{d^{2}V}{d{h}^2}=-\frac{3\pi h}{2}$
For maximum or minimum values of $V$, we must have
$\frac{dV}{dh}=0$
$\Rightarrow \pi r^2-\frac{3\pi h^2}{4}=0$
$\Rightarrow h^2=\frac{4r^2}{3}$
$\Rightarrow h=\frac{2}{\sqrt{3}}r$
$\left[\text{neglecting}h=\frac{-2}{\sqrt{3}}r\because \frac{d^{2}V}{d{h}^2}{|}_{h=\frac{-2}{\sqrt{3}}r}>0\right]$
Now, ${\left(\frac{d^{2}V}{d{h}^2}\right)}_{h=\frac{2r}{\sqrt{3}}}=-\sqrt{3}\pi r<0$.
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 $R=\sqrt{\frac{2}{3}}r.$
$\left[\text{neglecting -ve value of}R\right]$
$\therefore$ The maximum volume of the cylinder is given by
$V=\pi R^{2}h=\pi \left(\frac{2}{3}{r}^2\right)\left(\frac{2r}{\sqrt{3}}\right)=\frac{4\pi r^2}{3\sqrt{3}}$