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Chapter 11: Problem 66

You have 80 yards of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

Short Answer

Expert verified
The dimensions of the rectangle that maximize the enclosed area are 20 x 20 yards, and the maximum area is 400 square yards.

Step by step solution

01

Define Variables

Let's denote the sides of the rectangle as \( x \) and \( y \). Then, the perimeter of the rectangle is \( 2x + 2y \) and the area is \( A = x \cdot y \). As given, we have 80 yards of fencing, making \( 2x+2y = 80 \) or \( x + y = 40 \).
02

Express One Variable in Terms of the Other

From the perimeter equation \( x + y = 40 \), we can express \( y \) in terms of \( x \) as \( y = 40 - x \).
03

Substitute and Simplify

Now replace \( y \) in the area equation \( A = x \cdot y \) with \( 40 - x \). After this replacement, \( A = x \cdot (40 - x) = 40x - x^2 \).
04

Find the Maximum Area

To find the maximum, we differentiate the area function with respect to \( x \) and set it equal to zero. \( \frac{dA}{dx} = 40 - 2x = 0 \), implies \( x = 20 \). Substituting \( x = 20 \) back into \( y = 40 - x \) will give \( y = 20 \) also.
05

Compute the Maximum Area

Substitute \( x = 20 \) and \( y = 20 \) into the area formula \( A = x \cdot y = 20 \cdot 20 = 400 \) yard square.

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