Question

A cylinder is inscribed in a sphere of radius $3$ units. Find the curved surface area of the cylinder which has the maximum volume.

• A. $6\sqrt{2}\pi$
• B. $12\sqrt{2}\pi$
• C. $8\sqrt{3}\pi$
• D. $6\sqrt{3}\pi$

Answer 1

The correct option is B $12\sqrt{2}\pi$

Let's assume that the height and radius of the cylinder is $l$ and $r$ respectively.
Then the volume of the cylinder
$V=\pi r^{2}l$...(1)
According to the question
from the above figure
$r^2+\frac{l^2}{4}=9$
From here we have
$r^2=9-\frac{l^2}{4}$...(2)
Putting the value of $r^2$ into equation (1), we get
$V=\pi (9-\frac{l^2}{4})l$
Now for maximum value of V

$\frac{dV}{dl}=0$
from here we have
$9\pi -\frac{3\pi l^2}{4}=0$
$l=\pm 2\sqrt{3}$
Now at $l=2\sqrt{3}$
$\frac{d^{2}V}{d{l}^2}=(-)ve$
So, at $l=2\sqrt{3}$ volume is maximum
From equation (2) we have
$r=\sqrt{6}$
So, the surface area of the cylinder
= $2\pi rl$ = $2\pi \sqrt{6}\times 2\sqrt{3}$
= $12\sqrt{2}\pi$