SOLUTION: Having trouble understanding this! 4) 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

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Having trouble understanding this! 4) 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       Log On


   

Question 35797: Having trouble understanding this!


4) 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.
Answer:
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Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
the maximum area within a rectangle is a rectangle in the form of a square

thus l=w and 2l+2w=400 2l+2l=400 or 4l=100 & 2w+2w=400 or 4w=400 or w=100
thus a square of 100 by 100 will provide the maximum area for a fence of 400
feet of lumber.
thus the area within this square of 100 by 100 feet is 100*100=10000 sq feet.