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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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#### Solution

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 height is x cms.

Let V be volume of box.

∴ V = (18 – 2x)^{2}x

∴ V = 4(9 – x^{2}) · x

= 4(81 – 18x + x^{2})x

V = 4(x^{3} – 18x^{2} + 81x)

∴ `(dv)/(dx)` = 4(3x^{2} – 36x + 81)

= 12(x^{2} – 12x + 27)

= 12(x – 3)(x – 9)

Put `(dv)/(dx)` = 0,

∴ 12(x – 3)(x – 9) = 0

∴ x = 3 or x = 9

Since x = 9 is not possible,

∴ x = 3

Now `(d^2v)/(dx^2)` = 12(2x – 12)

= 24(x – 6)

\[\therefore \left. \frac{d^2V}{dx^2}\right]_{x=3}\] = 24(– 3)

= – 72 < 0

∴ V is maximum at x = 3

∴ If height of box is 3 cm, volume is maximum.

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