SOLUTION: A carpenter is building a rectangular room with a fixed perimeter of 324 ft. What dimentions would yield the maximum area? What is the maximum area? I understand the formula P=2

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Question 508439: A carpenter is building a rectangular room with a fixed perimeter of 324 ft. What dimentions would yield the maximum area? What is the maximum area? I understand the formula P=2l+2w will help solve this, but am unsure how to find the length and width. Any help you can give is greatly appreciated. Thanks.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A carpenter is building a rectangular room with a fixed perimeter of 324 ft. What dimentions would yield the maximum area? What is the maximum area?
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2(L + W) = 324
L+W = 162
L = (162-W)
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Area = LW
Area = (162-W)W
A = 162W-W^2
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You have a quadratic equation with a = -1, b = 162, c = -A
Maximum Area occurs when W = -b/(2a) = -162/(2*-1) = 81 ft.
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Solve for "L":
L + W = 162
L = 162-81 = 81 ft.
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Maximum Area = 81^2 = 6561 sq. ft.
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Cheers,
Stan H.
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