Problem 68 - College Algebra

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 maximum area that can be enclosed by 80 yards of fencing is 400 square yards, accomplished with dimensions 20 yards by 20 yards for the rectangle.

Step 1: Setting up the equation

Given 80 yards of fencing to enclose a rectangular region, let us denote the length of the rectangle as \(x\) and its width as \(y\). The total length of the fence, which is double the sum of the length and width, must equal 80. This can be represented as: \(2x + 2y = 80\), simplifying, we get \(x + y = 40\). This gives \(y = 40 - x\). Likewise, the area \(A\) of the rectangle can be represented as: \(A = xy\). Substituting \(y\) from the first equation to the second, we obtain an equation representing the area in terms of \(x\): \(A = x(40 - x)\).

Step 2: Finding the maximum area

To find the maximum area, we need to differentiate \(A\) with respect to \(x\) and find the value of \(x\) that maximizes \(A\). Differentiating, we get the derivative \(A' = 40 - 2x\). We then set the derivative equal to zero and solve for \(x\): \(40 - 2x = 0\). This gives that \(x = 20\). Using the first equation, we can find \(y = 40 - 20 = 20\). So, the dimensions that give the maximum area are 20 yards by 20 yards.

Step 3: Computing the maximum area

Substituting \(x = 20\) into the area equation: \(A = 20 * 20 = 400\) square yards. Therefore, the maximum area that can be enclosed by 80 yards of fencing is 400 square yards.