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

Question 76383: 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 vertex form to find the maximum area.
Answer:
Show work in this space.

Answer by stanbon(75887) (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. Use the vertex form to find the maximum area.
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400=2l+2w
200=l+w
Area = lw
l=200-w
Substitute to get:
Area = (200-w)w
Area = 200w-w^2
This is a quadratec with a=-1, b=200
Maximum occurs at w=-b/2a = -200/-2 = 100
Then length = 200-width = 100 ft
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Using the vertex form on -w^2+200= Area
w^2-200w = -Area
Complete the square:
w^2-200w+100^2 = 100^2-Area
(w-100)^2 = 10000-Area
Vertex is (100,10000) meaning the width is 100 ft and
the Area is 10,000.
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Cheers,
Stan H.

Answer:

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