Applied Math Gr.11
posted by Greg .
A farmer wants to enclose three sides of a rectangular pasture unsing 1000 yards of fencing. The fourth side does not require fencing because it borders a river.
What dimensions (length and width) should the farmer choose in order to enclose the greatest area?
a) Find at least five ordered pairs, comparing pasture length to its area. Remember: only 1000 yards of fencing available.
(Don't i need the maximum to determine this?)
b) Plot your five data points (length and area) on a grid with scales, and join them roughly. You should get the shape of a parabola.
c) Use your graping calculator to find the quadratic function that related to the data.
d) Graph resulting function by hand. State coordinateds of the point of the graps that yields max area. Find max area the farmer can fence in. What are the dimensions?
Can someone help me with this. I am so lost.
All Help is appreciated.

Applied Math Gr.11 
tchrwill
Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area?
Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P  2x) = Px  2x^2.
Taking the first derivitive and setting equal to zero, dA/dx = P  4x = 0, x becomes P/4.
With x = P/4, we end up with a rectangle with side ratio of 2:1.
.....The short side is P/4.The traditional calculus approach would be as follows.
.....The long side is (P  2(P/4)) = P/2.
Therefore, it can be unequivicably stated that of all possible rectangles with a given perimeter, one side being a given external boundry, the rectangle with side ratio of 2:1 encloses the maximum area.
The prime factorization os 1000 is 2^3(5^3).
Therefore, the total number of factors(divisors) is F = (3 + 1)(3 + 1) = 16, namely 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 10, 125, 250, 500 and 1000 enabling 8 possible sets of dimensions. 
Applied Math Gr.11 
Anonymous
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