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प्रश्न
An open box is to be made out of a piece of a square card board of sides 18 cms by cutting off equal squares from the comers and turning up the sides. Find the maximum volume of the box.
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उत्तर
Let the side of the square cut off from the corners be x cm.
Therefore, each side of the square box is (18 – 2x) cms and the height is x cms.
Let V be the volume of the box.
V = Area of the base × Height
V = (18 − 2x)2x
V = (324 − 72x + 4x2) x
∴ V = 4x3 − 72x2 + 324x
Differentiating w.r.t. x, we get
`(dV)/dx = 12x^2 - 144x + 324`
`therefore (d^2V)/dx^2 = 24x - 144`
For maximum volume, `(dV)/dx = 0`
∴ 12x2 − 144x + 324 = 0
∴ x2 − 12x + 27 = 0
∴ (x − 3) (x − 9) = 0
∴ x − 3 = 0 or x − 9 = 0
∴ x = 3 or x = 9
But x ≠ 9
∴ x = 3
For x = 3
`((d^2V)/dx^2)_(x = 3) = 24(3) - 144 = -72 < 0`
The volume of the box is maximum when x = 3.
∴ Maximum value of the box = (18 − 6)2 (3)
= 432 cc
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