Question

An open cylindrical tank whose base is a circle is to be constructed of metal sheet so as to contain a volume of `pia^3`cu cm of water. Find the dimensions so that the quantity of the metal sheet required is minimum.

Answer 1

Let x be the radius of the base, h be the height, V be the volume and S be the total surface area of the cylindrical tank.

Then V = `pia^3`                              ...(Given)

∴ `pix^2h = pia^3`

∴ h = `a^3/x^2`                             ...(1)
Now, S = `2pixh + pix^2`

= `2pix(a^3/x^2) + pix^2`            ...[By (1)]

= `(2pia^3)/x + pix^2`

∴ `"dS"/dx = d/dx((2pia^3)/ x + pix^2)`

= 2πa³ (– 1)x^–2 + π x 2x

= `(-2pia^3)/x^2 + 2pix`
and
`(d^2S)/(dx^2) = d/dx((-2pia^3)/x^2 + 2pix)`

= – 2πa³(– 2)x^–3 + 2π x 1

= `(4pia^3)/x^3 + 2pi`

Now, `"dS"/dx = 0  "gives" (-2pia^3)/x^2 + 2pix` = 0

∴ – 2πa³ + 2πx³ = 0
∴ 2πx³ = 2πa³
∴ x = a
and
`((d^2S)/dx^2)_("at"   x = a)`

= `(4pia^3)/a^3 + 2pi`
= 6π > 0
∴ by the second derivative test,
S is minimum when x = a
When x = a, from (1)

h = `a^3/a^2` = a

Hence, the quantity of metal sheets is minimum when radius = height = a cm.