A can needs to hold 500 cm^3 of soup. To minimize costs the cans radius&height should be as small as possible?

Question

i need to know how to SOLVE the problem, not the answer necessarily

Volume of cylinder = (pi)r^2h
Surface area = 2(pi)r^2 = 2(pi)rh

Answer 1

The volume is 500 cm^3. Solve for r in terms of h. Then substitute the solution for r in the surface area equation. Then take the derivative of the surface area with respect to h. Find the value of h so dS/dh = 0. This gives the value at which the surface area reaches an extreme value - the minimum surface area.

Answer 2

S, surface area of the can = 2 pi r h + 2 pi r r = 2 pi r (r + h)

(assuming closed can)

volume = 500 = pi. r^2 h or h = 500 / (pi r^2)

so S = 2 pi r (r + 500/ pi. r^2) = 2 pi r^2 + 1000/r

differentiating S wrt r, we get dS/dr = 4 pi r - 1000/r^2

equating it to 0, we get 4 pi r^3=1000 or r^3 = 250/pi

we can find r

d^2S/dx^2 = 4pi+2000/r^3 >0 for the r value; it implies that S is minimum at the solved r value