Let radius and height of the cylinder be R and H respectively. Volume of the cylinder V = πR 2 H———————— (1) AE = AC – EC = h – H Therefore, tan α = DE/AE = R/(h –H) -> R = (h – H)tan α ——————– (2) Putting R in (1) we get V = π(h – H) 2 tan 2 α.H Or, V = πtan2 α(h – H) 2———— (3) Differentiating w.r.t H, we get dV/dH = πtan2 α[(h – H) 2 .1 + H.2(h –H)( –1 )] = πtan 2 α(h – H)[h – H – 2H] = πtan 2 α(h – H)(h – 3H) ——————(4) For maxima and minima, dV/dH = 0 Therefore, πtan 2 α(h – H)(h – 3H) =0 -> H = h or H = h/3. As, H = h is not possible, hence, H = h/3. Now, d 2 V/dH2 = πtan 2 α[(h – H )(– 3)+ (h – 3H)(– 1)] =d 2 V/dH 2 = πtan2 α[6H – 4h] =(d 2 V/dH 2 )atH=h/3 = πtan 2 α(6.h/3 –4h) = πtan 2 α(2h – 4h) = πtan 2 α(– 2h) < 0 Therefore, volume is maximum at H =h/3. Putting H in (3), we get Maximum volume = πtan 2 α(h –h/3) 2 .h/3 = πtan 2 α.4h 2 /9.h/3 = 4/27πh^3 tan^2 α. {Ans}