Problem 67 - College Algebra

Question

You have 50 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 enclosed area are both 12.5 yards for length and width, making the enclosure a square. The maximum area that can be enclosed is 156.25 square yards.

Step 1: Formula for Perimeter of a Rectangle

The total length of the fencing represents the perimeter of a rectangle which is given by \(2*(Length + Width)\). We have 50 yards of fencing so \(2*(Length + Width) = 50\). We can simplify this by dividing both sides with 2 to give \(Length + Width = 25\) yards.

Step 2: Express Length in terms of Width

We can express the Length in terms of Width from the above equation, by solving for Length. This gives us \(Length = 25 - Width\) yards.

Step 3: Area of a Rectangle in terms of Width

The area \(A\) of a rectangle is given by \(Length*Width\). Substituting the expression of Length from the previous step into this yields \(A = Width*(25 - Width)\) or equivalently \(A = 25*Width - Width^2\) square yards.

Step 4: Find Maximum Area using Calculus

To find the maximum area, we find the derivative of the area with respect to Width, set it to 0 and solve for Width. This gives us \(A' = 25 - 2*Width = 0\), solving for Width provides \(Width = 12.5\) yards.

Step 5: Determine Length and Maximum Area

Substitute the Width, 12.5 yards, back into perimeter equation to find the Length, which is also \(25 - 12.5 = 12.5\) yards. Substitute Width and Length into the area equation to find the maximum area which gives \(A = 12.5 * 12.5 = 156.25\) square yards.

Step 6: Verification of Maximum

To verify that 156.25 square yards is indeed the maximum area, we perform a second derivative test. The second derivative of the area equation is \(A'' = -2\), which is negative, confirming that the area is maximized.