Question

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 solid 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 V/250 π is

Answer 1

Let h be the height of the cylinder and r be the inner radius.

Then volume, V = πr²h

 

Let V1 be volume of the material.

Given that 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

 

V1 = π(r+2)²h – πr²h + π(r+2)²2

= πh[(r+2)² – r²] + π(r+2)²2

= πh[r²+4r + 4 – r²] + π(r+2)²2

= πh(4r + 4) + π(r+2)²2

= 4πrh + 4πh + 2π(r+2)²

= (4V/r) + (4V/r²) + 2π(r+2)²

 

For minimum material required, dV/dr = 0

=> (-4V/r²) – (8V/r³) + 4π(r+2) = 0

Put r = 10

=> (-4V/100) – (8V/1000) + 48π = 0

=> (-40V – 8V)/1000 = -48π

=> -48V/1000 = -48π

=> V/π = 1000

Divide both sides by 250

=> V/250 π = 4

 

 

Value of V/250 π = 4