/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 You have 600 feet of fencing to ... [FREE SOLUTION] | 91Ó°ÊÓ

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You have 600 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?

Short Answer

Expert verified
The dimensions that will maximize the area are a length of 300 feet and a width of 150 feet, giving a maximum area of 45000 square feet.

Step by step solution

01

Define the variables

Let's assume the length of the plot (the side along the river) is \( L \) feet and the width (the side perpendicular to the river) is \( W \) feet. As we do not require a fence along the length, we only need fencing along the width and one length, i.e.,\( 2W + L = 600 \).
02

Express the area in terms of one variable

The area \( A \) of a rectangle is given by \( A = L \times W \). We can express \( L \) in terms of \( W \) from the fencing equation as \( L = 600 - 2W \). Substituting this into the area equation gives \( A = W \times (600 - 2W) = 600W - 2W^2 \).
03

Find the maximum area

To maximize the area, we differentiate \( A \) with respect to \( W \) and set it equal to 0, thus we have \( \frac{dA}{dW} = 600 - 4W = 0 \). Solving this for \( W \) yields \( W = 150 \) feet.
04

Find the length of the plot

Substituting \( W = 150 \) into the equation \( L = 600 - 2W \) to find the length, this gives us \( L = 600 - 2 \times 150 = 300 \) feet.
05

Calculate the maximum area

Substituting \( L = 300 \) and \( W = 150 \) into \( A = L \times W \) we find that the maximum area is \( A = 300 \times 150 = 45000 \) square feet.

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