a farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides. What is the largest area that can be enclose?
3 w + 2 L = 10,000
so
w = (10,000 -2 L)/3
A = w L = (10,000/3)L -(2/3)L^2
that is a parabola (quadratic)
we could find the vertex or we could use calculus
first find the vertex
A is 0 when L = 0
A is 0 when (2/3)L =10,000/3
or when L = 5000
so vertex is halfway between
L = 2500
then
w = (10,000 - 5,000) /3 = 5,000/3 = 1667
Now with calculus
dA /dL = w dL/dL + l dw/dL = w + L dw/dL
when dA/dL = 0 we have an extreme
w = -L dw/dL
but w = 10,000/3 - (2/3) L
so
dw/dL = -2/3
so
w = (2/3) L
so
2 w = 10,000/3
w = 5000/3
then L = 10,000/4 = 2,500
which we knew from the parabola anyway
a farmers wants fence 250 h/a of land,
how much meters are in there?
To find the largest area that can be enclosed, we need to determine the dimensions of the rectangular field.
Let's assume the length of the rectangular field is "L" and the width is "W". We want to divide this field into two plots with a fence parallel to one of the sides, which means the fence will be of length "L".
Now let's calculate the perimeter of the rectangular field. Since we have three sides and two parallel sides, the perimeter P is given by:
P = 2L + W
Given that the farmer has 10,000 meters of fencing, we have:
2L + W = 10000
From this equation, we can express W in terms of L as:
W = 10000 - 2L
Now, let's calculate the area A of the rectangular field:
A = L * W
Substituting the expression for W:
A = L * (10000 - 2L)
To find the largest possible area, we need to find the maximum value of A. We can do this by finding the value of L that maximizes the area.
To find the maximum, we take the derivative of A with respect to L and set it equal to 0:
dA/dL = 10000 - 4L = 0
4L = 10000
L = 10000/4
L = 2500 meters
Now, let's substitute this value of L back into the expression for W:
W = 10000 - 2L
W = 10000 - 2(2500)
W = 10000 - 5000
W = 5000 meters
Therefore, the maximum area that can be enclosed is:
A = L * W
A = 2500 * 5000
A = 12,500,000 square meters
So, the largest area that can be enclosed is 12,500,000 square meters.
To find the largest possible area that can be enclosed, we need to determine the dimensions of the rectangular field that maximize the area. Let's break down the problem step by step:
1. Let's assume the length of the rectangular field is L (in meters) and the width is W (in meters). We want to maximize the area A.
2. The perimeter of the rectangular field will be the sum of all the sides, which is given by:
Perimeter = 2L + 2W
3. According to the problem, the farmer has 10,000 meters of fencing available to enclose the field. Therefore, we can write the equation for the perimeter as:
2L + 2W = 10,000
4. We need to find the relationship between the length (L) and the width (W) to maximize the area. Since we want to divide the field into two plots with a fence parallel to one of the sides, we can assume that one side of the rectangular field will be divided in half, resulting in two equal lengths: L/2 and L/2.
5. With this assumption, we can rewrite the equation for the perimeter using the given information:
2(L/2) + 2W = 10,000
L + 2W = 10,000
6. Now, we can isolate L in terms of W:
L = 10,000 - 2W
7. To maximize the area, we use the formula A = L * W. Substitute the value of L from step 6:
A = (10,000 - 2W) * W
8. Expand and rearrange the equation to find the maximum area:
A = 10,000W - 2W^2
9. Since we are trying to maximize the area, we need to find the maximum point of this quadratic equation. One way to do this is by finding the vertex of the quadratic. The x-value of the vertex represents the value of W that maximizes the area A.
10. The x-coordinate of the vertex of a quadratic equation in the form of A = ax^2 + bx + c is given by:
x = -b / 2a
11. Comparing the quadratic equation in step 8 with the general form of a quadratic equation, we have:
a = -2, b = 10,000, c = 0
12. Substitute the values into the formula for the x-coordinate of the vertex:
W(max) = -10,000 / (2 * -2)
W(max) = 10,000 / 4
W(max) = 2,500
13. Now that we have the value of W that maximizes the area, we can substitute it back into the equation for A to find the maximum area:
A = (10,000 - 2(2,500)) * 2,500
A = (10,000 - 5,000) * 2,500
A = 5,000 * 2,500
A = 12,500,000
Therefore, the largest area that can be enclosed is 12,500,000 square meters.