Problem 69 - Algebra and Trigonometry

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 provide the maximum area are 150 feet by 100 feet and the maximum area is 15,000 square feet.

Step 1: Define the Variables

Let's assume the width of the field be \(x\) feet and the length be \(y\) feet. Since the playground is divided into two by another fence which is parallel to one of its sides, there will be three \(y\) and two \(x\) in total.

Step 2: Set Up the Equation

According to the problem, 600 feet of fencing is used. So, the total length of the fence is equal to \(3y + 2x = 600\). We can rearrange the equation to express \(y\) in terms of \(x\) and substitute into the formula for the area of a rectangle, which is \(A = x * y\). Now, our equation will be \(A = x * (200 - 2/3 * x)\)

Step 3: Find the Derivative

To find the maximum area, we need to find the derivative of the area function. The derivative of the area \(A\), is \(A' = 200 - \frac{4}{3}x\).

Step 4: Solve for Critical Points

Setting the derivative equal to zero gives the critical points. Therefore, \(200 - \frac{4}{3}x = 0\). Solving this equation will give us \(x=150\). Substituting \(x=150\) into \(3y + 2x = 600\) gives us \(y = 100\). So, the dimensions that maximize the area are a length of 150 feet and a width of 100 feet.

Step 5: Find the Maximum Area

Substitute \(x = 150\) and \(y = 100\) into the area formula \(A = x * y\). So, \(A = 150 * 100 = 15000\). The maximum total enclosed area is 15000 square feet.