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.
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
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
amazing, I found an area above of 11250. Math analysis is a gift from God.
I agree with Bob
Normal and <small>small</small> text
oops - thought I read a square was required - sorry and may God come to your assistance with math
To find the maximum possible area of the pasture, we need to determine the dimensions of the rectangle first.
Let's assume the length of the rectangle (parallel to the creek) is 'L' and the width (perpendicular to the creek) is 'W'.
The available fencing is used for three sides of the rectangle, which gives us:
2L + W = 300 (since two sides are equal in length)
Now, we need to express W in terms of L to form the area equation.
W = 300 - 2L
To find the maximum area, we differentiate the area equation with respect to L and set it equal to zero:
A = L * W
A = L * (300 - 2L)
A = 300L - 2L^2
dA/dL = 300 - 4L = 0
Solving the equation, we find:
300 - 4L = 0
4L = 300
L = 75
Substituting L = 75 back into the equation W = 300 - 2L, we obtain:
W = 300 - 2*75
W = 300 - 150
W = 150
So, the maximum possible area of the pasture is:
A = L * W
A = 75 * 150
A = 11250 square feet