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

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> 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       Log On


   



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(75887) About Me  (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.