A square piece of tin of side 18cm is to be 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 square to be cut off so that the volume of the box is the maximum possible?

Solution:

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

Then, the length and the breadth of the box will be (18 - 2x) cm each and the height of the box be x cm.

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

V (x) = x (18 - 2x)²

Hence,

V' (x) = (18 - 2x)² - 4x (18 - 2x)

= (18 - 2x)[18 - 2x - 4x]

= (18 - 2x)(18 - 6x)

= 6 × 2 (9 - x)(3 - x)

= 12 (9 - x)(3 - x)

V" (x) = 12 (- 9 - x)(3 - x)

= - 12 (9 - x + 3 - x)

= - 12 (12 - 2x)

= - 24 (6 - x)

Now,

V' (x) = 0

Þ x = 9, x = 3

If, x = 9 then the length and the breadth will become 0.

Hence, x ≠ 9

When, x = 3

Then,

V" (3) = - 24 (6 - 3)

= - 72 < 0

By the second derivative test, x = 3 is the point of local maxima of V.

Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 17

A square piece of tin of side 18cm is to be 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 square to be cut off so that the volume of the box is the maximum possible?

Summary:

Square of side 3 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