SOLUTION: 4) John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his patio (area of

Question 43553This question is from textbook College Algebra
: 4) John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his 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.
Show clearly the algebraic steps which prove your dimensions are the maximum area which can be obtained.
Answer:
Show work in this space.
This question is from textbook College Algebra

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his 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.
Perimeter =2l+2w=300 ft.
Dividing by 2 you get:
l+w=150
Area=(length)(width)
Let length be "l"
Then width = "150-l"
Area=(l)(150-l)
Area=150l-l^2
This is a quadratic equation where a=-1 and b=150.
The maximum area is where l=-b/2a=-150/(-2)=75
So the maximum area is as follows:
Area=lw=(75)(150-75)=75(75)=5625 sq. ft.
Cheers,
Stan H.
-b/2a=

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