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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.
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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