# 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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 Click here to see ALL problems on Rectangles 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(57347)   (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? ------------------- 2(L + W) = 324 L+W = 162 L = (162-W) --------------- Area = LW Area = (162-W)W A = 162W-W^2 ---- You have a quadratic equation with a = -1, b = 162, c = -A Maximum Area occurs when W = -b/(2a) = -162/(2*-1) = 81 ft. -- Solve for "L": L + W = 162 L = 162-81 = 81 ft. ----- Maximum Area = 81^2 = 6561 sq. ft. ===================================== Cheers, Stan H. =====================================