Question 213554: A farmer wants to make a rectangular enclosure along the side of a barn and then divide the enclosure into two pens with a fence constructed at a right angle to the barn. If 300 yards of fencing are available, what are the dimensions of the largest section that can be enclosed?
Firstly, the wording of this particular problem has thrown both myself, family, and friends for a loop. I have drawn, erased, and redrawn several rough sketches of this barn-and-fence problem and still feel as though it is wrong. I am at a loss of going about how to solve it. Any assistance would be greatly appreciated!
Thank you in advance
-B
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A farmer wants to make a rectangular enclosure along the side of a barn and then divide the enclosure into two pens with a fence constructed at a right angle to the barn. If 300 yards of fencing are available, what are the dimensions of the largest section that can be enclosed?
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_________________Barn
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I think it looks something like this:
Then we can write an equation for the 300 yds of fencing
L + 3W = 300
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L = (300-3W)
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Area = L*W
Replace L with (300-3W)
A = (300-3W) * W
Arrange as a quadratic equation
A = -3W^2 + 300W
Find the axis of symmetry for Width that gives max area x = -b/(2a)
in this equation a=-3; b=300
W = -300/(2*-3)
W = -300/-6
W = +50 ft is the width for max area
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Find the length
L = 300 - 3(50)
L = 150 ft is the length
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Max area would be 150 * 50 = 7500 sq/ft
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