Question

2) A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 400 feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area. What is the maximum area?

Answer 1

Let the length of the playground be x

Width = y

Let another fence be parallel to the width of the playground.

Perimeter = 400

=> 2x + 3y = 400

=> 3y = 400 - 2x

=> y = (400 / 3) - (2/3)x

Area = Length * Width

=> A(x) = xy

=> A(x) = x((400 / 3) - (2/3)x)

=>

We know that for a quadratic function , maximum occurs at x = -b / 2a if a < 0

In this case, a = - 2 / 3 and b = 400 / 3

Since, a < 0, therefore, area is maximum when

x = - b / 2a

=> x = - (400 / 3) / 2(-2 / 3)

=> x = 100

Substituting the value of x we get

y = (400 / 3) - (2/3)(100) = 200 / 3

Therefore,

Length of the playground = 100 feet

Width of the playground = 200/3 feet = 66.67 feet

Area of the playground = Length * Width = 100 * (200 / 3) = 6666.67 square feet