Problem 69 - College Algebra

Question

A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Six hundred feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area. What is the maximum area?

Step-by-Step Solution

Answer

The dimensions that maximize the area are 100 feet by 200 feet, and the maximum area is 20000 square feet.

Step 1: Understanding the Question

Let the length of the rectangle be \( x \) (in feet), and its width be \( y \) (in feet). The rectangle is also divided into two equal rectangles by a fence parallel to the sides of length \( x \). So, the total length of the fencing used is \( 2y \) (for the opposite sides) + \( 3x \) (for the three divisions), which is given as 600 feet.

Step 2: Setting up the equations

From the information given, we can formulate two equations. The first is the equation for the perimeter: \( 2y + 3x = 600 \). The second equation is for the area of the rectangle. Since a rectangle's area is calculated as it's length times its width, the area \( A \) is given by \( A = xy \). From the perimeter equation, we can express \( y \) in terms of \( x \): \( y = (600 - 3x) / 2 \).

Step 3: Substituting and Differentiating the Area Equation

We substitute \( y \) from step two into the area equation to give the area in terms of only one variable, \( x \). This gives \( A = x(600 - 3x) / 2 \). To find the value of \( x \) that maximizes the area, we differentiate \( A \) with respect to \( x \), set the derivative equal to zero, and solve for \( x \) to obtain the critical points.

Step 4: Solving for x

Setting the derivative of the area equal to zero gives \( x = 100 \) or \( x = 200 \). But since the length of the rectangle cannot exceed half the total length of the fence, only \( x = 100 \) is physically meaningful.

Step 5: Finding the Other Dimension and Maximum Area

Substitute \( x = 100 \) into the equation \( y = (600 - 3x) / 2 \) to find \( y = 200 \). The maximum area is therefore \( A = 100 * 200 = 20000 \) square feet.