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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.
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Solution
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.
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