A farmer is fencing a rectangular area for his farm using the straight portion of a river as one side of the rectangle. If the farmer has 2400 ft of fence, find the dimension of the rectangle that gives the maximum area for the farm
So we need 2 widths and 1 length
let the width be x
and the length be y
2x + y = 2400
y = 2400-2x
Area = A = xy
= x(2400-2x)
= 2400x - 2x^2
dA/dx = 2400 - 4x = 0 for a max of A
4x = 2400
x = 600
then y = 2400-1200 = 1200
field is 600 ft by 1200 ft
To find the dimensions that give the maximum area for the farm, we need to analyze the problem and formulate an equation.
Let's assume the length of the rectangle is L and the width is W. We know that the straight portion of the river acts as one side of the rectangle, so the length L will be the portion of the fence parallel to the river.
The remaining two sides of the rectangle will have a combined length of 2400 ft - L.
Since the area of a rectangle is given by the formula A = length × width, we can express the area A as:
A = L × (2400 ft - L)
Now, we need to determine the value of L that maximizes the area A. To do so, we can graph the equation and identify the vertex, which represents the maximum point.
We can start by expanding the equation:
A = 2400L - L^2
Now, let's rewrite it in the standard quadratic form:
A = -L^2 + 2400L
To find the value of L that maximizes A, we can recall that the vertex of a quadratic equation in the form Ax^2 + Bx + C is given by the x-coordinate x = -B/(2A).
Therefore, substituting the values A = -1, B = 2400, and C = 0, we find:
L = -2400/(2*(-1))
L = -2400/(-2)
L = 1200
So the length of the rectangle that gives the maximum area is 1200 ft.
To find the width, W, we subtract the length from the total length of the fence:
W = 2400 - L
W = 2400 - 1200
W = 1200
Therefore, the dimensions of the rectangle that maximize the area are a length of 1200 ft and a width of 1200 ft.