A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of `V m^3`, has a 2 mm 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 the material used to make the container is minimum when the inner radius of the container is `10 mm`. then the value of `V/(250 pi)` is

Answer

4

Explanation

Let the inner radius and inner length be ra and h respectively
`therefore v= pir^(2)h`
Let the volume of material be M
`rarr M=pi(r+2)^(2)2+pi(r+2)^(2)h-pir^(2)h`
`=2p[i(r+2)^(2)+4pih(r+1)]`
`=22pi(r+2)^(2)+2(r+1V)/(pir^(2))`
`rarr (dM)/(dr)=2pi 2(r+2)+(2V)/(pi)(-1)/(r^(2))-(2)/(3)`
Given volume is maximum when r=10
`therefore (dM)/(dr)=0` when r=10
`rarr 24+(2V)/(pi)(-10-2)/(10^(3))=0` ltMbrgt `rarr (24V)//(10^(3)pi)=24`
`rarr=V=10^(3)pi`
`rarr (V)/(250pi)=4`