SOLUTION: Hi, could you please help me solve this question. A rectangular area is enclosed by a fence and divided by another section of fence parallel to two of its side. The 500m of fence u

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Question 852390: Hi, could you please help me solve this question. A rectangular area is enclosed by a fence and divided by another section of fence parallel to two of its side. The 500m of fence used to enclose a maximum area. What are the dimensions of the fence? (leave answers in fraction form) Thank you!

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangular area is enclosed by a fence and divided by another section of fence parallel to two of its side.
The 500m of fence used to enclose a maximum area.
What are the dimensions of the fence? (leave answers in fraction form)
:
Because of the divided area, the perimeter will be:
2L + 3W = 500
therefore
2L = -3W + 500
L = -3%2F2W + 500%2F2
L = -1.5W + 250
:
Area = L*W
Replace L with (-1.5W+250)
A = (-1.5W+250)*W
A = -1.5W^2 + 250W
A quadratic equation, the max area occurs on the axis of symmetry x=-b/(2a)
In this equation, x=w, a=-1.5, b=250
w = %28-250%29%2F%282%2A-1.5%29
W = %28-250%29%2F%28-3%29
W = +831%2F3 meters is with width that gives max area
Find the Length
L = -1.5w+250
L = -1.5(83.33) + 250
L = -125 + 250
L = 125 meters is length for max area
Then
125 * 831%2F3 = 104162%2F3 sq meters is the max area