Alexis B.
asked • 03/05/21The back of George's property is a creek. George would like to enclose a rectangular area, using the creek as one side and fencing for the other three sides, to create a pasture. If there is 160 feet of fencing available, what is the maximum possible area of the pasture?
Raymond B. answered • 03/05/21
Math, microeconomics or criminal justice
2W + L = 160
L = 160-2W
A= LW = (160-2W)W =160W -2W^2
A' = 160-4W = 0
4W = 160
W = 40 feet
L = 160-2(40) = 80 feet
max Area = 40(80) = 3200 square feet
or in the x y coordinates, let Area =y,
y = -2x^2 +160x
0 = -2x^2 + 160x
2x^2 -160x = 0
x^2 -80x = 0
x^2 -80x + 40^2 = 40^2
(x-40)^2 = 40^2
x-40 = 40
x =80 feet for length
160-2W = 80
2W =80
W = 40 feet
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As this problem is listed as algebra 2 or precalc, it should be solved using completing the squares (defining the parabola, with the solution at the vertex).03/05/21