Question 71236: 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. Use the vertex formula to find the maximum area.
Answer:
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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 are-1a 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. Use the vertex formula to find the maximum area.
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Perimeter = 2(l+w)=300 ft
l+w=150
l=150-w
Area = lw
Area = (150-w)w
A = -w^2 + 150w
This is a quadratic with a= -1; b= 150
The vertex is at w = -b/2a = -150/-2 = 75 ft. (this is the width of the rectangle)
l = 150-w = 150 - 75 = 75 ft (this is the length of the rectangle)
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Cheers,
Stan H.
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