When making a cylindrical can with a volume of 500ml , what dimensions will minimize the material to produce each can

πr^2 h = 500, so h = 500/(πr^2)

A = 2πr^2 + 2πrh = 2π(r^2 + 500/(πr))
dA/dr = 2π(2r - 500/(πr^2)) = (4πr^3 - 1000)/πr^2
dA/dr = 0 when r = 5∛(2/π), so h = 10∛(2π)