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A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the

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Question

A square piece of tin of side 18 cm is to made into a box without a top  by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Sum
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Solution

Let x cm be the length of each side of the square which is to be cut off from each corner of the square tin sheet of side 18 cm.

Let V be the volume of the open box formed by folding up the flaps, then

V = x (18  - 2x) (18 - 2x) = 4x (9 - x)2

= 4 (x3 - 18x2 + 81x)

Differentiate w.r.t.x., we get

`(dV)/dx = 4(3x^2 - 36x + 81) = 12 (x^2 - 12x + 27)`

For maximum / minimum volume

`(dV)/dx = 0`

⇒ 12 (x2 - 12x + 27) = 0

⇒ 12 (x - 3) (x - 9) = 0

⇒ x = 3, 9 but 0 < x < 9

⇒ x = 3

`((d^2V)/dx^2) = 12 (2x - 12) = 24 (x - 6)`

and `((d^2V)/dx^2)_(x=3) = 24 (3 - 6) = -72 <0`

⇒ V has a maximum at x = 3

Hence, the volume of the box is at its maximum when the side of the square to be cut off is 3 cm.

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Chapter 6: Application of Derivatives - Exercise 6.5 [Page 233]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 6 Application of Derivatives
Exercise 6.5 | Q 17 | Page 233

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