a cylindrical vat must hold 5m^3, the vat must be wider than it is tall, but no more than 3 m in diameter. What dimensions will use the least amount of material?

v = πr^2h = 5, so

h = 5/(πr^2)

a = 2πr^2 + 2πrh
= 2πr^2 + 10/r

da/dr = 4πr - 10/r^2
da/dr=0 when 4πr^3=10
r = ∛(5/(2π))
h = 5/(πr^2) = 5/(π(5/(2π))^(2/3)) = ∛(20/π)

since 3 > 2r > h, it works.