Farmer Ed has
9,000 meters of fencing, and wants to enclose a rectangular plot that borders on a river. If Farmer Ed does not fence the side along the river, what is the largest area that can be enclosed?
x = length
y = width
x + 2 y = 9000 or x = 9000 - 2 y
A = x y
so
A = (9000-2y)y
2 y^2 - 9000 y = - A
y^2 - 4500 y = -A/2
y^2 - 4500 y +(4500/2)^2 = -A/2 + 5,062,500
(y-2250)^2 = -(1/2)(A-10,125,000)
so
width = 2250
length = 4500
area = 10,125,000 m^2
Well, Farmer Ed, it seems you have quite the predicament with that river! But don't worry, I'm here to help with a little bit of humor.
If we consider the side along the river as one of the shorter sides of the rectangle, we can imagine the other three sides forming a perfect square. That way, we can make the most out of our fence!
Now, since the total amount of fencing available is 9,000 meters, let's say each side of the square-shaped part requires x meters of fencing. That means the other side of the rectangle (also equal in length to the side of the square) will need 2x meters of fencing.
To find the maximum area, we'll use the formula A = length × width. The length is the side of the square (x meters), and the width is the sum of one side of the square (x meters) and the other side of the rectangle (2x meters).
A = x × (x + 2x)
A = x × 3x
A = 3x^2
Now, since we have 9,000 meters of fencing in total, we can set up the equation:
4x + 2x = 9,000
6x = 9,000
x = 1,500 meters
So the side of the square-shaped part should be 1,500 meters, and the other side of the rectangle should be 3,000 meters (2 × 1,500).
Now, with these dimensions, we can calculate the maximum area:
A = 3x^2
A = 3 × 1,500^2
A = 6,750,000 square meters
So Farmer Ed, the largest area you can enclose with 9,000 meters of fencing while leaving out the side along the river is approximately 6,750,000 square meters. Time to get creative and put that land to good use!
To find the largest area that can be enclosed, we need to determine the dimensions of the rectangular plot.
Let's assume the length of the rectangular plot is L and the width is W.
Given that Farmer Ed has 9,000 meters of fencing, we can calculate the perimeter of the rectangular plot:
Perimeter = 2L + W
Since we are not fencing the side along the river, the perimeter will be:
Perimeter = L + 2W
We also know that the perimeter is equal to 9,000 meters:
L + 2W = 9,000
Now, we need to express one of the variables in terms of the other so that we can maximize the area.
Rearranging the equation, we get:
L = 9,000 - 2W
The area of a rectangle is given by:
Area = Length x Width
Substituting the expression for L into the area formula, we get:
Area = (9,000 - 2W) x W
Now, we need to find the value of W that maximizes the area. We can do this by finding the maximum of the quadratic equation.
To find the maximum of the quadratic equation, we can calculate the vertex using the formula:
W = -b / (2a)
In this case, a = -2 and b = 9,000.
W = -9,000 / (2*(-2))
W = -9,000 / (-4)
W = 2,250
We can now find the value of L by substituting W back into the equation:
L = 9,000 - 2(2,250)
L = 9,000 - 4,500
L = 4,500
Therefore, the width is 2,250 meters and the length is 4,500 meters.
Finally, we can calculate the maximum area by multiplying the width and length:
Area = 2,250 x 4,500
Area = 10,125,000 square meters
So, the largest area that can be enclosed is 10,125,000 square meters.
To find the largest area that can be enclosed, we need to determine the dimensions of the rectangular plot. Let's assume the length of the plot is 'L' and the width is 'W'. Since the side along the river does not need to be fenced, the perimeter of the plot will be the sum of the lengths of the three remaining sides:
Perimeter = 2L + W
According to the problem, the total amount of fencing available is 9,000 meters. So we can write the equation:
2L + W = 9,000
Now, we need to express one variable in terms of the other and then substitute it back into the area formula.
By rearranging the equation, we can solve for W:
W = 9,000 - 2L
The area of the rectangular plot is calculated by multiplying the length and width:
Area = L * W
Substituting the expression for W:
Area = L * (9,000 - 2L) = 9,000L - 2L^2
To find the maximum area, we can take the derivative of the area equation with respect to L and set it equal to zero:
d(Area)/dL = 9,000 - 4L = 0
Solving this equation gives us:
4L = 9,000
L = 9,000/4
L = 2,250
Substituting the value of L back into the equation for W:
W = 9,000 - 2(2,250)
W = 4,500
Therefore, the dimensions of the rectangular plot that will maximize the area are L = 2,250 meters and W = 4,500 meters.
To find the largest enclosed area, we substitute these values back into the area equation:
Area = L * W = 2,250 * 4,500 = 10,125,000 square meters
Hence, the largest area that can be enclosed given 9,000 meters of fencing is 10,125,000 square meters.