Question

Show that the radius of a closed right circular cylinder of given surface area and maximum volume is equal to half of its height.

Answer 1

Let S be the given surface area of a closed right circular cylinder whose radius is r and height is h.

∴ S = 2πrh + 2πr²

⇒ `h = (S - 2πr^2)/(2πr)`   ...(i)

Now, volume of cylinder (V) = πr²h

∴ `V = πr^2 ((S - 2πr^2)/(2πr))`   ...[From equation (i)]

⇒ `V = (Sr - 2πr^3)/2`

∴ `(dV)/(dr) = (S - 6πr^2)/2`   [Differentiating w.r.t. r.] ...(ii)

And `(d^2V)/(dr^2) = (-12πr)/2` = – 6πr   ...[Again diff. w.r.t. r.] ...(iii)

For maximum and minimum value,

Put `(dV)/(dr) = 0`   ...[From equation (ii)]

⇒ `(S - 6πr^2)/2 = 0`

⇒ S = 6πr²

Putting the value of S in equation (i), we get

`h = (6πr^2 - 2πr^2)/(2πr)`

= `(4πr^2)/(2πr)`

= 2r

or `r = h/2`

And `[(d^2V)/(dr^2)]_(r = h/2) = -6π xx h/2`   ...[From equation (iii)]

= – 3πh < 0   ...[∵ h cannot be negative]

So, volume is maximum when `r = h/2`.

Hence Proved.