SOLUTION: A dog owner has 250 feet of fencing to enclose a rectangular run for his dogs. If he wants the maximum possible area, what should the length and width of the rectangle be?

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Question 205701: A dog owner has 250 feet of fencing to enclose a rectangular run for his dogs. If he wants the maximum possible area, what should the length and width of the rectangle be?

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
Perimeter of rectangle = 2* Length + 2* Width
.
P =250 = 2L +2W,,,,,dividing by 2
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125 = L+W,,,,,W=125-L
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Area = L*W = L*(125-L) = 125L -L^2
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making a table
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L,,,,,,,Area
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1,,,,,,,124
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60,,,,,,3900
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62,,,,,,3906
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62.5,,,,3906.25,,,,,max,,,,L=W=62.5
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63,,,,,,3906
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The max area of the rectangle occurs when L=W=62.5 ft,,,,
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This makes rectangle into a square
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