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

S is the total surface area of the closed right circular cylinder whose radius is r, height is h, and volume is V.
Then surface area S = 2 * pi * r* h + pi * r^2 * h

 ------ (i)
V = ½ [Sr – 2*pi*r^3]

Therefore, dV / dr = ½ [S – 6 * pi * r^2 ]

For extrema (maxima or minima) dV/dr=0

Therefore S – 6 * pi * r^2 = 0 
S – 6 * pi * r^2 = 0
Plugging this value of S in eq.(i),we get

h (at V-extrema)= 2r
For maxima, d2V / dr2 = – 6* pi*r = negative 

Therefore, V is maximum when height, h = 2r = diameter of the base.