Problem 68 - Algebra and Trigonometry

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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 \(A = W(40 - W)\).

Step 2: 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, \(W = 20\).

Step 3: 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.

Step 4: 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.