Discussion Forum

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 2mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2mm 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 10mm, then the value of  $\frac{V}{250\pi}$  is

Answer:

Explanation:

Here, volume of cylindrical container,

       V=  $\pi r^{2}h$           .............(i)

 and let volume of the material used be T.

$\therefore  T=\pi \left[(r+2)^{2}-r^{2}\right]h+\pi(r+2)^{2}\times2$

 $T=\pi \left[(r+2)^{2}-r^{2}\right]\frac{V}{\pi r^{2}}+2\pi(r+2)^{2}$

                                    $\left[\because  V=\pi r^{2}h\Rightarrow h=\frac{V}{\pi r^{2}}\right]$

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

 On differentiating  w.r.t , we get

                  $\frac{dT}{dr}=2V.\left(\frac{r+2}{r}\right).\left(\frac{-2}{r^{2}}\right)+4\pi (r+2)$

 At,            r=10,    $\frac{dT}{dr}=0$

 Now,   $0=(r+2).4\left(\pi-\frac{V}{r^{3}}\right)$

$\Rightarrow$                 $\frac{V}{r^{3}}=\pi,$      where r=10

$\Rightarrow$         $\frac{V}{1000}=\pi, or \frac{V}{250\pi}=4$