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A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to

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Question

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. 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 the side of the required square be x,

Length of the box (l) = (45 - 2x)

And width of the box (b) = (24 - 2x)

Height of the box (h) = x

∴ Volume of the box, V = l × b × h

V = x (45 - 2x) · (24 - 2x)

= (4x3 - 138x2 + 1080x)           ...(1)

On differentiating equation (1) with respect to x,

`(dV)/dx =` 12x2 - 276x + 1080         ...(2)

For maximum value of V, `(dV)/dx = 0`

or 12x2 - 276x + 1080 = 0  or x2 - 23x + 90 = 0

or x2 - 18x - 5x + 90 = 0   or  x(x - 18) - 5 (x - 18) = 0

or (x - 18)(x - 5) = 0

`therefore` x = 5, 18

Differentiating equation (2) again with respect to x, `(d^2V)/dx^2` = 24x - 276

At, x = 5 `(d^2V)/dx^2` = 24 × 5 - 276 = (negative value)

∴ The value of V will be maximum at x = 5.

∴ The side of the square will be 5 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 18 | Page 233

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