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 of

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> 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 of      Log On


   

Question 36521: 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 by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
So thats:
2l+2w=400
l=200-w (subsituttion)
Area:
lw=A
Subsitute:
(200-w)(w)=A
200w-w^2=A
Differentate:
2w=200
w=100
Hence, the width and length are 100 ft each and the area is 100^2 ft squared.
Paul.