# SOLUTION: Can anyone help me with this doozy? Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). She wants to

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 Click here to see ALL problems on Quadratic Equations Question 85337: Can anyone help me with this doozy? Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). She wants to maximize the area of her patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation. Use the Standard Form to find the maximum area.Answer by stanbon(60778)   (Show Source): You can put this solution on YOUR website!Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). She wants to maximize the area of her patio (area of a rectangle is length times width). What should the dimensions of the patio be? ------------ Perimeter = 2(length)+2(width)=400 ft. Length + width = 200 ft Let length be "x" The width = 200-x ---------------------- Area = length*width Area = x(200-x) Area = -x^2+200x ------------------ The maxiumum area occurs when x=-b/2a = -200/(2(-1))= 100 -------- So length = 100 ft width = 200-x = 100 ft. ================ Cheers, Stan H.