# SOLUTION: 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

Algebra ->  Algebra  -> Functions -> SOLUTION: 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      Log On

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 Question 37590: 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. Is this how I do it. P = 2l + 2w? Answer by stanbon(57347)   (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, and show how the maximum area of the patio is calculated from the algebraic equation. Yes, use P=2l+2w 400=2l+2w l+w=200 l=200-w Now, Area=lw A=(200-w)(w)=200w-w^2 Then, w^2-200w+A=0 This is a quadratic with a=1,b=-200, c=A The maximum is at -b/2a = -(-200)/2=100 So the dimensions are 100' by 100' Cheers, Stan H.