Answered: Heavy Duty Boxes would like to produce…

Name: Heavy Duty Boxes would like to produce a box at a lower cost for their
Uploaded: 2024-03-20T07:07:26.000Z
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Description: Heavy Duty Boxes would like to produce a box at a lower cost for their client. The box will have a volume of 108 cubic centimeters and needs to be twice as long as the width. The Customer has recommended the heaviest material and that all dimensions

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Heavy Duty Boxes would like to produce a box at a lower cost for their client. The box will have a volume of 108 cubic centimeters and needs to be twice as long as the width. The Customer has recommended the heaviest material and that all dimensions be within the nearest 10,000^th of a centimeter. The cost for the heaviest material is $0.15 per cm². Take into consideration that Heavy Duty Boxes marks up their pricing 150%.

Calculate the cost per box requested by the customer.

Expert Solution

Step 1 Set-up variables for unknown and relations between them

Let the width of box be $x cm$ , so, the length of the box will be $2 x cm$ . Also, let the height of box be $h cm$ .

The situation can be shown as follows:

Now, $Volume = Length \times Width \times Height$

$\Rightarrow 108 {cm}^{3} = 2 x \cdot x \cdot h {cm}^{3} {⇒2x}^{2} ·h=108 \Rightarrow x^{2} \cdot h = 54 \Rightarrow h = \frac{54}{x^{2}} \dots \dots \dots \dots \dots \dots \dots \dots (1)$

Step 2 Constructing a function of one variable

The area of metal required is surface area of box. If the surface area is minimum, then, the cost will be minimum.

$Surface Area, S = 2 (l \cdot b + b \cdot h + l \cdot h), where l, b, h are length, width and height, respectively.$

$\Rightarrow S = 2 (2 x \cdot x + x \cdot h + 2 x \cdot h) \Rightarrow S = 4 x^{2} + 2 x \cdot h + 4 x \cdot h \Rightarrow S = 4 x^{2} + 6 x \cdot h Substituting value of h from (1): \Rightarrow S (x) = 4 x^{2} + 6 x \cdot \frac{54}{x^{2}} . . . . . . . . . . . here S depends only on variable x \Rightarrow S (x) = 4 x^{2} + \frac{324}{x}$

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