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प्रश्न
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.
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उत्तर
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.
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