a rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the?

Question

playground. 360 feet of fencing is used. find the dimensions of the playground that maximize the total enclosed area.

Answer 1

L = length of one side
W = width of one side
Perimeter = 2L + 3W = 360
Area = LW
L = 180 - 1.5W
Maximize Area
Area = (180 - 1.5W)W
Area = 180W - 1.5W^2
Area' = 180 - 3W
W = 60
L = 180 - 1.5*60 = 90

60*90 maximizes the enclosed area

Answer 2

There are 5 fences; the top, bottom, left, right, and middle (see diagram below). Let's assume the middle fence is parallel to the left and right. The top fence and bottom fence are both of length 'y'. The left, right, and middle fences are all of length 'x'.
.__
|_|_|<-- x
^-- y
The total fence used (or parameter) would be:
3*x + 2*y = 360ft
Solving for y --> y=180ft - (3/2)*x
The area enclosed is length*width = x*y. We want to maximize this area, so we'll take the derivative of the formula we find for the area and set it equal to zero.
Area = x*y = x*(180ft - (3/2)*x) = 180*x - (3/2)*x^2
derivative(Area) = d/dx(180*x - (3/2)*x^2) = 180 - 3*x = 0
3*x = 180
x = 60ft
y = 180 - (3/2)*60 = 180 - 90 = 90 ft
x*y=60*90=5400ft^2

As a simple check, if x=50ft then y=180-75=105ft, then x*y = 105*50=5250ft^2. Since this is LESS than 5400ft^2, we're still in good shape.
If x=70ft then y=180-105=75ft, then x*y = 70*75 = 5250ft^2. Since this is LESS than 5400ft^2, we are very confident with our answer!