A farmer wants to build a rectangular fence using the side of his barn as one side. He has 200 yards of fencing available. Find the maximum possible area he can enclose with the fence.
Let x be the length of the side parallel to the barn.
The other two side lengths are therefore 100 - x/2
Write the equatuion for Area as a function of x, and set the derivative dA/dx = 0. That will let you solve for x. It will turn out to be 100 yards. The area will then be 50*100 = 5000 sq yds
I know the answer is 5000 and that the equation to figure out the answer is in quadratic form. I am just not sure on how to get to the answer, so it would be great if someone could show the steps on how to get to the answer. It would be greatly appreciated.
I understand how you got the answer, but I need to start with 2W+L=200 and somehow turn that into a quadratic equation(ax^2+bx+c) to obtain the answer.
Let x = L
Then W = (200 - x)/2 = 100 - (x/2)
That is exactly what I did. It does not matter whether I call the length x or L. The algebra is the same.
The area is then A(x) = 100 x - x^2/2
If they don't want you to use calculus, then complete the square to find the x that gives maximum Area.
2 A(x) = 200 x -x^2
= -(x-100)^2 + 10,000
That clearly has a maximum whan x = 100, and that value corresponds to 2A = 10,000. or A = 5,000
1=(X-100)/((X+100)/2)) / ((2-1)/((2+3)/2)
Solve for X
I know X =200, but don't understand how to reach that anwser.
is this really algebra, im doing this now in freshman year algebra 2
To find the maximum possible area the farmer can enclose with the fence, we need to determine the dimensions of the rectangle that will yield the largest area.
Let's assume the length of the rectangle is L (in yards) and the width is W (in yards). Since one side of the fence is the side of the barn, the length of the fence used will be L + W + L = 2L + W.
According to the problem, the farmer has 200 yards of fencing available, so we can set up the equation:
2L + W = 200 (Equation 1)
We also know that the area of the rectangle is given by A = L * W.
To proceed, we need to express one of the variables (either L or W) in terms of the other variable. In this case, let's solve Equation 1 for W:
W = 200 - 2L (Equation 2)
Substituting Equation 2 into the formula for the area:
A = L * (200 - 2L)
A = 200L - 2L^2
Now we have the area of the rectangle, A, expressed in terms of the length, L. To determine the maximum possible area, we need to find the vertex of the parabolic function A = 200L - 2L^2.
The vertex of a parabola can be found using the formula: L = -b / (2a), where a and b are the coefficients of the quadratic equation.
In our equation, a = -2 and b = 200, so substituting these values into the formula:
L = -200 / (2 * -2)
L = -200 / -4
L = 50
The length of the rectangle that will yield the maximum area is 50 yards. Now substitute this value back into Equation 2 to find the corresponding width:
W = 200 - 2 * 50
W = 200 - 100
W = 100
The width of the rectangle is 100 yards.
Finally, calculate the maximum possible area using these values:
A = L * W
A = 50 * 100
A = 5000 square yards
Therefore, the maximum possible area the farmer can enclose with the fence is 5000 square yards.