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Question
Solve the following : An open box with a square base is to be made out of given quantity of sheet of area a2. Show that the maximum volume of the box is `a^3/(6sqrt(3)`.
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Solution
Let x be the side of square base and h be the height of the box.
Then x2 + 4xh = a2
∴ h = `(a^2 - x^2)/(4x)` ...(1)
Let V be the volume of the box.
Then V = x2h
∴ V = `x^2((a^2- x^2)/(4x))` ...[By (1)]
∴ V = `(1)/(4)(a^2x - x^3)` ...(2)
∴ `"dV"/dx = (1)/(4)"d"/"dx"(a^2x - x^3)`
= `(1)/(4)(a^2 xx 1 - 3x^2)`
= `(1)/(4)(a^2 - 3x^2)`
and
`(d^2V)/(dx^2) = (1)/(4).d/dx(a^2 - 3x^2)`
= `(1)/(4)(0 - 3 xx 2x)`
= `-(3)/(2)x`
Now, `"dV"/dx = 0 "gives" (1)/(4)(a^2 - 3x^2)` = 0
∴ a2 – 3x2 = 0
∴ 3x2 = a2
∴ x2 = `a^2/(3)`
∴ x = `a/sqrt(3)` ...[∵ x > 0]
and
`((d^2V)/dx^2)_("at" x = a/sqrt(3)`
= `-(3)/(2) xx a/sqrt(3)`
= `-sqrt(3)/(2) a < 0`
∴ V is maximum when x = `a/sqrt(3)`
From (2), maximum volume = `[1/4(a^2x - x^3)]_("at" x = a/sqrt3)`
= `(1)/(4)(a^2 xx a/sqrt(3) - a^3/(3sqrt(3)))`
= `(1)/(4)((2a^3)/(3sqrt(3)))`
= `a^3/(6sqrt(3)`
Hence, the maximum volume of the box is `a^3/(6sqrt(3)` cu. unit.
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