Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal to the diameter of its base.

Let r and h be radius and height of given cylinder of surface area S. If V be the volume of cylinder then V = πr²h

S = 2πr² + 2πrh ⇒ h = (S - 2πr²)/2πr

V = πr²[(S - 2πr²)/2πr] = rS/2 - πr³/2

dV/dr = 1/2(S - 3πr²)

For maximum or minimum value of V, dV/dr = 0

1/2(S - 3πr²) = 0 ⇒ S - 3πr² = 0 ⇒ r² = S/3π ⇒ r = √(S/3π)

d²V/dr² = -3πr

d²V/dr²|r=√(S/3π) = -3π√(S/3π) = -ve

Hence for r = √(S/3π), volume V is maximum.

h = (S - 2π(S/3π))/2π√(S/3π) = (S/3)/2π√(S/3π) = S/6π√(S/3π) = √(S/3π)

h = (S - 2πr²)/2πr = (S - 2π(S/3π))/2π√(S/3π) = S/(6π√(S/3π)) = √(S/3π) / 2π√(S/3π) = 1/2√(3π/S) = 2r

Therefore, for maximum volume height of cylinder is equal to diameter of its base.