Calculus
posted by Adeola .
A cylindrical metal can is made to hold 500ml of soup. Determine the dimensions of the can that will minimize the amount of metal required(assume that the top and sides of the can are made from metal of the same thickness

half a liter
liter is 10*10*10 = 1,000 cm^3
a = 2(pi r^2) + 2 pi r h
v = pi r^2 h =500
so
h = 500/(pi r^2)
a = 2 pi r^2 + 1000/r
da/dr = 0 = 4 pi r  1000/r^2
pi r = 500/r^2
pi r^3 = 500
r^3 = 500/pi
r = 5.4 cm
h = 5.4 cm