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 460 feet of fencing available, what is the maximum possible area of the pasture?
To find the maximum possible area of the pasture, we need to determine the dimensions of the rectangular area that can be enclosed using the creek as one side and the available fencing for the other three sides.
Let's call the length of the rectangular area L and the width W.
Since the creek is one side of the pasture, we can use the available fencing for the other three sides, which gives us:
2L + W = 460 feet
Now, we need to express one of the variables in terms of the other variable. We can choose to express W in terms of L, or L in terms of W. Let's express W in terms of L.
W = 460 - 2L
Now, we can substitute this expression for W in the formula for the area of a rectangle.
Area = L * W
Area = L * (460 - 2L)
To find the maximum possible area, we need to find the value of L that maximizes the area. We can do this by finding the vertex of the parabola given by the area equation.
The area equation is quadratic in nature, so it will have a vertex. The x-coordinate of the vertex represents the value of L that maximizes the area.
To find the x-coordinate of the vertex, we can use the formula: x = -b / (2a), where a is the coefficient of the squared term, and b is the coefficient of the linear term.
In our equation for the area, the coefficient of the squared term is -2, and the coefficient of the linear term is 460.
x = -(-2) / (2 * -2) = 1
Now, substitute this value of L back into the formula for W.
W = 460 - 2(1) = 458
So, the dimensions of the rectangular area that maximize the pasture area are L = 1 and W = 458. To find the maximum area, we can calculate:
Area = L * W = 1 * 458 = 458 square feet.
Therefore, the maximum possible area of the pasture is 458 square feet.
Let the side parallel to the creek be y ft
let the other two sides be x ft each
so you have y + 2x = 460
y = 460-2x
area = xy = x(460-2x)
= -2x^2 + 460x
this can be represented by a downwards pointing parabola of the form y = ax^2+bx+c
the x of such a parabola is -b/(2a)
= -460/-4 = 115
if x = 115, then y = 460-2(115) = 230
max area = xy = 115(230) = 26450 ft^2
or , by Calculus
d(area)/dx = -4x + 460 = 0 for a max of area
4x = 460
x = 115 , just as before, continue....