A farmer has a 100m chainlink to fence a rectangular plot.what is the greatest area he can enclose with the chainlink
a = x y
100 / 2 = x + y
so x =50-y
a = (50 -y)y = 50 y - y^2
or
y^2 - 50 y = -a
I do not know if you do calculus so will find vertex of parabola.
complete the square
y^2 - 50 y + 625 = -a + 625
(y-25)^2 = -1(a-625)
vertex at y = 25 , area = 625 so a square
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using calculus
a = -y^2 + 50 y
da/dy = 0 at min or max = -2 y +50
so 2 y = 50
y = 25 again
To maximize the area enclosed by the chainlink fence, the farmer should create a square plot since a square has the largest area for a given perimeter.
1. Start by dividing the 100m chainlink into four equal sides since the square has four equal sides.
2. Divide 100m by 4, which gives us 25m per side of the square.
3. The area of a square is calculated by squaring the length of one side. Hence, the area of the square plot would be 25m x 25m.
4. Multiply the values to find the maximum area: 25m x 25m = 625 square meters.
Therefore, the greatest area the farmer can enclose with the 100m chainlink fence is 625 square meters.
To find the greatest area that can be enclosed with a 100m chainlink, we need to determine the dimensions of the rectangular plot that would give us the largest area.
Let's assume the length of the rectangular plot is L meters and the width is W meters. To maximize the area, we can use the following equation for the perimeter of a rectangle:
Perimeter = 2L + 2W
Since we have a 100m chainlink, the perimeter of the rectangle should be equal to 100m:
2L + 2W = 100
Let's solve this equation for one of the variables, say W, in terms of L:
2W = 100 - 2L
W = 50 - L
Now we can express the area of the rectangle in terms of L:
Area = Length * Width = L * (50 - L) = 50L - L^2
To find the maximum area, we need to find the value of L that maximizes the Area. We can do this by finding the maximum point on the graph of the quadratic function Area = 50L - L^2.
In this case, we can use the vertex formula to find the x-coordinate of the vertex (which corresponds to L in our problem):
x-coordinate of vertex = -b / (2a)
In our case, a = -1 (since the coefficient of L^2 is -1) and b = 50. Plugging these values into the formula, we get:
x-coordinate of vertex = -50 / (2 * -1) = 25
So, the width of the rectangle that will give us the maximum area is W = 50 - L = 50 - 25 = 25 meters.
Thus, the length of the rectangle is also 25 meters, making it a square.
The greatest area the farmer can enclose with the 100m chainlink is obtained by squaring the side length:
Area = Length * Width = 25m * 25m = 625 square meters.
Therefore, the greatest area the farmer can enclose with the chainlink is 625 square meters.