The volume of a cylindrical tin can with a top and bottom is to be 54π cubic inches. If a minimum amount if tin is to be used to construct the can, what must be the radius, in inches, of the can?

Answer 1

volume = V = 54 = height times pi times radius squared = hpi(r^2)

h = 54/pi(r^2)

surface area = A = 2pi(r^2) + h(2pi(r)) = 2pi(r^2) + [(54/pi(r^2)](2pi(r))

A = 2pi(r^2) + 108/r

take the derivative and set equal to zero

A' = 4pi(r) -108/r^2 = 0

multiply by r^2

4pi(r^3) - 108 = 0

r^3 = 108/4pi = 27/pi

r = cube root of 27/pi = 3/pi^(1/3)

radius = about 4.39 inches minimizes the surface area