A circular cylindrical container, open at the top, and having a capacity of 24pi cubic inches, is to be manufactured. If the cost of the material used for the bottom of the container is three times that used for the curved part and if there is no waste of material, find the dimensions which will minimize the cost

Answer: radius r = 2 inches

height h = 6 inches

Step-by-step explanation:

Given;

Volume V = 24πin3

Volume of a cylinder is given by

V = πr^2h

h = V/πr^2 ... 1

Where, h = height and r = radius of cylinder

For the surface area of the cylinder with open top. we have,

S = 2πrh + πr^2

For the cost of materials used, let k represent the cost of materials used for the body of the cylinder.

Then, for the bottom will be 3k

Total cost will be represented by C, which gives

C = 2πrhk + 3πr^2k ... 2

Substituting eqn 1 to 2, we have;

C = 2πrVk/πr^2 + 3πr^2k

C = 2Vk/r + 3πr^2k

The material cost is minimum at dC/dr = 0

dC/dr = - 2Vk/r^2 + 6πrk = 0

6πrk = 2Vk/r^2

r^3 = 2V/6π

r = (2*24π/6π) ^-3

r = (8) ^-3

r = 2

Substituting r = 2 into eqn1

h = 24π/π (2^2)

h = 24/4 = 6

h = 6