Acylindrical can is to have volume 1500 cubic centimeters. determine the radius and the height which will minimize the amount of material to be used. note that the surface area of a closed cylinder is s=2rh+2r2 and the volume of a cylindrical can is v=r2h.

Assuming R and H:
So volume is pir^2 * H = 1500 and H = 1500/(pir^2)  while surface area is A= 2pir*H + 2pir^2 
A = 2pir(r+h)= 2piR^2 + 2pir*1500/(pir^2)= 2piR^2 + 3000/r
For A to take minimum, get the derivative 4pir - 3000/R^2 and let it be 0 
4pir^3 - 3000 = 0 
r = cbrt(3000/(4pi)) ≈ 6.20
h = 1500/(pi(6.20)^2) ≈ 12.42