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Chapter 7: Problem 40

A box, constructed from a rectangular metal sheet, is \(21 \mathrm{~cm}\) by $16 \mathrm{~cm}\( by cutting equal squares of sides \)\mathrm{x}$ from the corners of the sheet and then tuming up the projected portions. The value of \(\mathrm{x}\) so that volume of the box is maximum is (A) \(\underline{1}\) (B) 2 (C) 3 (D) 4

Short Answer

Expert verified
A) 1 cm B) 3 cm C) 5 cm D) 7 cm Answer: A) 1 cm

Step by step solution

01

Find the expression for the volume of the box

Let x be the length of the side of the square cut from each corner. After cutting the squares and folding up the sides, the dimensions of the box will be (21-2x) cm in length, (16-2x) cm in width, and x cm in height. The volume of the box, V, can be expressed as: V(x) = (21-2x)(16-2x)x.
02

Find the first derivative of the volume expression

Differentiate the expression V(x) with respect to x: \[V'(x) = \frac{d}{dx}[(21-2x)(16-2x)x]\] To find the derivative, we will first use the product rule, which states that if u(x) and v(x) are differentiable functions of x, then the derivative of their product is: \[(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\] We'll use the product rule twice: first for u(x) = (21-2x)(16-2x) and v(x) = x, and then for u(x) = (21-2x) and v(x) = (16-2x). Applying the product rule, we find: \[V'(x) = [(21-2x)(16-2x)]'x + (21-2x)(16-2x)\] Now, we differentiate (21-2x) and (16-2x) and apply the product rule again: \[V'(x) = [(-2)(16-2x) + (21-2x)(-2)]x + (21-2x)(16-2x)\] \[V'(x) = (-32+4x -42+4x)x + (21-2x)(16-2x)\] Simplify the expression: \[V'(x) = (8x-74)x\]
03

Find the critical points

To find the critical points, set the first derivative equal to zero and solve for x: \[V'(x) = 0\] \[(8x-74)x = 0\] From this equation, we have two possible solutions for x: 1) x=0, which represents the original sheet before any squares are cut, so we can discard this solution as it doesn't form a box. 2) x=(74/8), which simplifies to x = 9.25. However, since the width of the original sheet is 16 cm, cutting a 9.25 cm square from each corner is not possible. Since none of the critical points are valid, the maximum volume must occur at one of the endpoints of the possible values of x.
04

Check the endpoints

The smallest value of x that we can have is x=1 cm, and the largest value we can have is x=7 cm (since 2x cannot exceed the width 16cm or the length 21cm). By checking the volume at these endpoints, we have: \[V(1) = (19)(14)(1) = 266\] \[V(7) = (7)(2)(7) = 98\] Since V(1) has a higher volume than V(7), the maximum volume will occur when x=1.
05

Provide the answer

The value of x for which the volume of the box is maximum is: x = 1 cm (option A).

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