Problem 70 - 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. Four 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 of the playground that give the maximum enclosed area are: \(x\) feet by \(y\) feet. The maximum area is \(max A\) square feet. Both the values of \(x\), \(y\) and \(max A\) depend on the solutions of the mentioned equations.

Step 1: Translate the problem into mathematical form

Use the given conditions to express the problem as a function of area and perimeter. If we label the length of the rectangular playground as \(x\) (in feet) and the width as \(y\) (in feet), and the third division also has length \(x\) (because the additional fence is parallel to one side), then the total fence length equals \(400\) feet, hence the perimeter of the playground is given as \(2x + 3y = 400\). The total enclosed area is expressed as \(A = x*y\).

Step 2: Express width \(y\) as a function of length \(x\)

Solving for \(y\) in terms of \(x\) from the perimeter equation we get \(y = (400 - 2x) / 3\). Substitute this in the equation for \(A\) to yield area completely in terms of \(x\), namely \(A = x * ((400 - 2x) / 3)\).

Step 3: Calculate the derivative of \(A\)

To find the maximum enclosed area, we need to take the derivative of the area function with respect to \(x\), i.e., \(A' = (400x - 2x^2) / 3 - x\).

Step 4: Find critical points

Set \(A'\) equal to zero and solve for \(x\) to find the critical points. Solve \(A' = 0\), which gives two solutions denoted by \(x_1\) and \(x_2\). These are the points where the maximum or minimum of the enclosed area can occur. Discard any negative solution since \(x\) (length) can't be negative. The remaining value is the length \(x\) that maximizes the area.

Step 5: Evaluate the maximum area

Substitute the optimal \(x\)-value into the area function \(A\) to get the maximum area. If you substitute this \(x\)-value into the equation \(y = (400 - 2x) / 3\), you will obtain the corresponding width \(y\) which together with \(x\) give the dimensions that maximize the area.