300=2w+L
L=300-2w
area=Lw=w(300-2w)
Notice this is a parabola. It's two zeroes are at w=0, and w=150
Because of symettry, the max must be halfway between the zeroes, so at max area, w=75
Area=75(150) max
The 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 300 feet of fencing available, what is the maximum possible area of the pasture.
6 answers
since the creek forms the fourth side of the pasture, fencing is required only for the other three sides
the max area is 10,000 sq ft, since each of the three fenced sides has length 100 ft
100 ft x 100 ft = 10,000 ft^2
the max area is 10,000 sq ft, since each of the three fenced sides has length 100 ft
100 ft x 100 ft = 10,000 ft^2
amazing, I found an area above of 11250. Math analysis is a gift from God.
I agree with Bob
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oops - thought I read a square was required - sorry and may God come to your assistance with math