A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of V mm3, has a 2 mm thick wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of the container is 10 mm, then the value of $\frac{V}{250\pi } $, is

4

The correct answer is option (3) : 4

Let the inner radius of the container be r mm and height be g mm. Then,

$V= \pi r^2 h $

Let M be the volume of material used is making the container.

Then,

$M= \pi ( r + 2)^2 (h + 2) - \pi r^2 h $

$⇒M = \pi (r + 2 )^2 \left(\frac{V}{\pi r^2 }+ 2\right) - V$

$⇒ M = V \left(\frac{4}{r}+\frac{4}{r^2}\right) + 2\pi ( r + 2)^2 $

$⇒\frac{dM}{dr}= V \left(-\frac{4}{r^2}-\frac{8}{r^3}\right) + 4\pi ( r + 2) $

It is given that M is minimum when r = 10 mm.

$∴\left(\frac{dM}{dr}\right) = 0 ⇒V \left(-\frac{4}{100}-\frac{8}{1000}\right) + 48 \pi = 0 $

$⇒V = 1000 \pi ⇒ \frac{V}{250\pi } = 4 $.