Minimizing cost of a can/ Applied optimization questions

Question

A company wants to manufacture cylindrical aluminum cans with a volume of 1000 cubic centimeters. (These cans would have both top and bottom.) The material for the top and bottom of the can costs 5 cents per square centimeter, and the material for the sides costs 3 cents per square centimeters. Find the exact value of the radius of the base that would minimize the cost of constructing this can

Answer 1

They want the volume to be 1000 cc's and Volume for a cylinder is defined as the area of the base times the height so V = πr²h meaning 1000 = πr²h. That means h = 1000/(πr²)

The cost of the can is the surface are times the cost/area. The surface area = 2πr² + 2πrh. Cost (in cents) = 5(2πr²) + 3(2πrh) but since h = 1000/πr² we can say C(r) = 5(2πr²) + 3(2πr(1000/(πr²)). Simplifying, we get:

C(r) = 10πr² + 6000/r

To minimize cost, take the derivative and set it equal to zero.

C'(r) = 20πr - 6000/r²

20πr - 6000/r² = 0

20πr = 6000/r²

r³ = 300/π

r = cuberoot(300/π)