Problem 70 - 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. 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 that maximize the total enclosed area of the playground are 100 ft by 67 ft. The maximum area is 6700 square feet.

Step 1: Define the Variables and the Equations

Let's denote the length of the playground as \(x\) (in feet) and its width as \(y\) (in feet). Considering that there is a fence dividing the playground parallel to its width, we have two pieces of fences along the width. So, the total fencing of 400 feet comprises two lengths and three widths, hence the equation for the perimeter is: \(2x+3y=400\). The total area (A) of the playground is given by \(A = x \cdot y\).

Step 2: Express Area in Terms of One Variable

We can express \(y\) in terms of \(x\) from the perimeter equation: \(y = (400-2x)/3\). We substitute this value in the area equation. So, the area function becomes \(A(x) = x \cdot (400 - 2x)/3\).

Step 3: Take The Derivative of the Area Function

To find the maximum area, we should take the derivative of the area function \(A'(x) = \frac{(400 - 2x)/3 - (2x)}{3}\).

Step 4: Set the Derivative Equal to Zero and Solve

Set \(A'(x) = 0\) and solve for \(x\). We get \(x = 100\). Validate this result by looking at the second derivative. The second derivative \(A''(x) = -4/3\) is negative, hence \(x = 100\) gives a maximum area.

Step 5: Find the Corresponding Value of y

Substitute \(x = 100\) in the perimeter equation to find \(y\). We get \(y = (400 - 2*100)/3 = 67\). So, the maximum area is achieved for the playground dimensions 100 ft by 67 ft.

Step 6: Compute the Maximum Area

Substitute \(x = 100\) and \(y = 67\) into the area function, \(A = 100*67 = 6700\) square feet.