A rectangular sheet of tin 45cm by 24cm 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?

Solution:

Let the side of the square to be cut off be x cm.

Then, the height of the box is x cm , the length is (45 - 2x) cm and the breadth is (24 - 2x) cm.

Therefore, the volume V (x) of the box is given by,

V (x) = x (45 - 2x)(24 - 2x)

= x (1080 - 90x - 48x + 4x²)

= 4x³ - 138x² + 1080x

Hence,

V' (x) = 12x² - 276x + 1080

= 12 (x² - 23x + 90)

= 12 (x -18)( x - 5)

V" (x) = 24x - 276

= 12 (2x - 23)

Now,

V' (x) = 0

⇒ x = 18, x = 5

It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet.

Thus, x cannot be equal to 18.

When, x = 5

Then,

V" (5) = 12 (2 (5) - 23)

= 12 (10 - 23)

= 12 (- 13)

= - 156 < 0

By the second derivative test, x = 5 is the point of local maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 18

A rectangular sheet of tin 45cm by 24cm 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?

Summary:

Square of side 5 cm to be cut off so that the volume of the box is the maximum possible. By the second derivative test, x = 3 is the point of local maxima of V