Question

show that the height of the right circular cylinder of maximum volume that
can be inscribed in a given right circular cone of height h is h/3 .​

Answer 1

Answer:

Let height of the cylinder be H and base radius be R.

Volume of cylinder V = πR2H ---------------------------------- (1)

ΔABC ~ ΔADE => AC/AE = BC/DE => h/(h – H) = r/R

Therefore, R = r[(h – H)/h] = r[1 – H/h] --------------------------- (2)

Putting R in (1), we get

V = πr2(1 – H/h)2H -------------------------------------- (3)

Differentiating w.r.t.H, we get

dV/dH = πr2[(1 – H/h)2.1 + H.2(1 – H/h)(–1/h)]

= πr2(1 – H/h)[1 – H/h – 2H/h]

dV/dH = πr2(1 – H/h)(1 – 3H/h)

For maximum or minimum, dV/dH = 0

Or, πr2(1 – H/h)(1 – 3H/h) = 0 => H = h, H = h/3.

But H = h is not possible. Hence, H = h/3.

Now, d2V/dH2 = πr2[(1 – H/h)(–3/h) + (1 – 3H/h)(–1/h)]

= (πr2)/h[0 – 3 + 3H/h – 1 + 3H/h]

= (πr2)/h[6H/h – 4]

Therefore, [d2V/dH2]atH=h/3 = πr2/h[6/h.h/3 – 4] = πr2/h(–2) < 0.

Therefore, volume is maximum at H = h/3. [Proved.]