Question

Solve the following : Show that a closed right circular cylinder of given surface area has maximum volume if its height equals the diameter of its base.

Answer 1

Let r be the radius of the base, h be the height and V be the volume of the closed right circular cylinder, whose surface area is a² sq. units (which is given).

∴ 2πrh + 2πr² = a²

∴ 2πr(h + r) = a²

∴ h = `a^2/(2πr) - r`                   ...(1)

Now, V = πr²h

= `πr^2(a^2/(2πr) - r)`         ...[By (1)]

= `(1)/(2)a^2r - πr^3`

∴ `(dV)/(dr) = d/(dr)(1/2a^2r - πr^3)`

= `(1)/(2)a^2 xx 1 - π xx 3r^2`

= `a^2/(2) - 3πr^2`

and

`(d^2V)/(dr^2) = d/(dr)(a^2/2 - 3πr^2)`

= 0 – 3π × 2r

= – 6πr

For maximum volume,

`(dV)/(dr)` = 0

∴ `a^2/(2) - 3πr^2` = 0

∴ 3πr² = `a^2/(2)`

∴ r² = `a^2/(6π)`

∴ r = `a/sqrt(6π)`           ...[∵ r > 0]

and

`((d^2V)/(dx^2))_(at  r = a/sqrt(6π)`

= `- 6π(a/sqrt(6π)) < 0`

∴ V is maximum when r = `a/sqrt(6π)`

When r = `a/sqrt(6π)`, then from (1),

h = `a^2/(2π xx a/sqrt(6π)) - a/sqrt(6π)`

=  `sqrt(6πa)/(2π) - a/sqrt(6π)`

= `(6πa - 2πa)/(2πsqrt(6π)`

= `(4πa)/(2πsqrt(6π)`

= `(2a)/sqrt(6π)`

∴ h = 2r

Hence, the volume of the cylinder is maximum if its height is equal to the diameter of the base.