Question

The volume of the largest cylinder that can be inscribed in a sphere of radius ${}^{\prime }{r}^{\prime }cm$ is (in cubic units)

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

Answer 1

The correct option is A $\frac{4\pi r^3}{3\sqrt{3}}$

In $\Delta$ OAB

$r^2=R^2+x^2$ ...... $\left(i\right)$

Volume of cylinder $=\pi R^{2}h$

$\Rightarrow V=\pi R^2\times 2x$

$\Rightarrow V=2\pi R^{2}x$

$\Rightarrow V=2\pi \left(r^2-x^2\right)x$ ....... [from $\left(i\right)$]

$\Rightarrow V=2\pi \left(r^{2}x-x^3\right)$ ..... $\left(ii\right)$

$\Rightarrow \frac{dV}{dx}=2\pi \left(r^2-3x^2\right)$

and $\frac{d^{2}V}{d{x}^2}=2\pi \left(-6x\right)$

for critical points

$\frac{dV}{dx}=0$

$\Rightarrow 2\pi \left(r^2-3x^2\right)=0$

$\Rightarrow r^2=3x^2$

$\Rightarrow x=\frac{r}{\sqrt{3}}$

Now, ${\left(\frac{d^{2}V}{d{x}^2}\right)}_{x=\frac{r}{\sqrt{3}}}=2\pi \left(-6\times \frac{r}{\sqrt{3}}\right)<0$

So, volme of cylinder is maximum when $x=\frac{r}{\sqrt{3}}$

Therefore maximum volume of cylinder

$=2\pi \left(r^{2}x-x^3\right)$ ....... [from $\left(ii\right)$]

$=2\pi \left(r^2\times \frac{r}{\sqrt{3}}-\frac{r^3}{3\sqrt{3}}\right)$

$=2\pi \left(\frac{3r^3-r^3}{3\sqrt{3}}\right)$

$=\frac{4\pi r^3}{3\sqrt{3}}$