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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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Question

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.

Sum
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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 the height is x cms.


Let V be the volume of the box.

V = Area of the base × Height

V = (18 − 2x)2

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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Chapter 2.2: Applications of Derivatives - Long Answers III

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SCERT Maharashtra Mathematics and Statistics (Arts and Science) [English] 12 Standard HSC
Chapter 2.2 Applications of Derivatives
Long Answers III | Q 8

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