Arectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. four hundredfour hundred and eightyand eighty feet of fencing is used. find the dimensions of the playground that maximize the total enclosed area. what is the maximum area?

h = 120 ft;    w = 80 ft

A = 9600 ft^2

Explanation:

Let h and w be the dimensions of the playground. The area is given by:

A = h*w   (eq1)

The total amount of fence used is:

L = 2*h + 2*w + w    (eq2)    (an extra distance w beacuse of the division)

Solving for w:

w = L - 2/3*h = 480 - 2/3*h   (eq3)  Replacing this into the area eq:

We derive this and equal zero to find its maximum:

  Solving for h:

h = 120 ft.  Replacing this into eq3:

w = 80ft

Therefore the maximum area is:

A = 9600 ft^2

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