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Chapter 10: Problem 62

You need to enclose a rectangular region with 200 feet of fencing. Experiment with different lengths and widths to determine the maximum area you can enclose. Which quadrilateral encloses the most area?

Short Answer

Expert verified
With 200 feet of fencing, the maximum area that can be enclosed is 2500 square feet. This is achieved when the fencing is arranged to form a square with sides of 50 feet each.

Step by step solution

01

Define the variables and the equations

Assign the dimensions of the rectangle as length \( L \) and width \( W \). The perimeter of the rectangle is fixed at 200 ft. So, the equation is \(2L + 2W = 200\). By manipulating this equation, we can express \(L\) as \(L = 100 - W\). To maximize the area \(A\) of the rectangle, we express \(A\) in terms of \(W\) as \(A = L * W = (100 - W) * W\).
02

Find the maximum area

To find the maximum area, we take the derivative of \(A\) with respect to \(W)\), i.e., \(A' = 100 - 2W\). Setting \(A' = 0\), we solve for \(W\) to get \(W = 50\). We substitute \(W = 50\) into the equation for \(L\) and we also get \(L = 50\). Hence the dimensions that maximize the area are when \(L = W = 50\)
03

Confirm the maximum

To make sure that these dimensions indeed maximize the area and not minimize it, we take the second derivative of \(A\) with respect to \(W)\), i.e., \(A'' = -2\). Since \(A'' < 0\), this confirms that \(A\) is at its maximum when \(W = 50\).

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