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Chapter 3: Problem 68

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 that maximize the area are both 20 yards, and the maximum area is 400 square yards.

Step by step solution

01

Formulate the area equation

Let the length and width of the rectangle be \(L\) and \(W\) respectively. The fencing available equals the perimeter of the rectangle. So, \(2L + 2W = 80\). Simplifying that, \(L + W = 40\). Express \(L\) in terms of \(W\) by rearranging the equation, we get \(L = 40 - W\). The area of a rectangle is given by \(A = L \times W\). Substitute \(L\) in terms of \(W\) into the area equation to give \(A = W(40 - W)\).
02

Find the critical point

Take the derivative of \(A\) with respect to \(W\). \(A'(W) = 40 - 2W\). Set this to zero and solve for \(W\), \(40 - 2W = 0\). Hence, \(W = 20\).
03

Verify the maximum area

To verify that \(W = 20\) indeed gives the maximum area, take the second derivative of \(A\) and substitute \(W = 20\). \(A''(W) = -2\). Since this is a constant negative value, \(A(W)\) is concave down and \(W = 20\) yields the maximum value.
04

Find the dimensions and the maximum area

Substitute \(W = 20\) into \(L = 40 - W\) to determine \(L\). Hence, \(L = 20\). The dimensions of the rectangle that maximize the area are both 20 yards. Substitute \(W = 20\) and \(L = 20\) into the area equation \(A = W(40 - W)\) to give \(A = 20(40 - 20)\). Hence, the maximum area is 400 square yards.

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