SOLUTION: 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
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Question 70670: 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 dimesions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation. Use the vertex formula to find the maximum area. Answer by ankor@dixie-net.com(22740) (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. Use the vertex formula to find the maximum area.
:
Generally max area is obtained in the form of a square but here is the algebra
:
2L + 2W = 300
Divide by 2:
L + W = 150
L = (150-W)
:
Area = L*W
Substitute (150-W) for L and you have a quadratic equation:
A = W(150-W)
A = -W^2 + 150W
:
Find the axis of symmetry of this equation: x = -b/(2a), in this equation:
a = -1, b = 150
:
W = -150/(2*-1)
W = -150/-2
W = 75
:
L = 150 - 75 = 75, so it is a square:
:
Find the vertex, substitute 75 for W in A = -W^2 + 150x
A = -(75^2) + 150(75)
A = -5625 + 11250
A = +5625 sq ft is the max area
:
If you graph, it you can see it clearly:
:
You can check it using 75*75 = 5625 sq ft
