a rectangular plot of farmland will be bounded on one side by a river and on the other sides by a single strand electric fence .
with 800m of wire at your disposal,what is the largest are you can enclose, and what are its dimensions ?
Hint: largest area would be a square.
To find the largest area you can enclose with 800m of wire, you need to determine the dimensions of the rectangular plot that would minimize the perimeter while maintaining its area.
Let's denote the length of the rectangular plot as L and the width as W.
Given that one side of the plot is bounded by a river, we have:
Perimeter = L + 2W = 800m ...(1)
To maximize the area, we need to express the area of the rectangle in terms of a single variable. The area (A) of a rectangle is given by:
Area = Length * Width
Since we want to express the area in terms of a single variable, we can solve equation (1) for L:
L = 800m - 2W
Now, substitute this value of L into the area equation:
Area = (800m - 2W) * W
= 800W - 2W^2
To find the maximum area, we need to differentiate the equation and set it equal to zero:
dA/dW = 800 - 4W = 0
Solving for W, we get:
4W = 800
W = 200m
Substitute this value of W into equation (1) to find L:
L = 800m - 2(200m)
L = 400m
So, the dimensions of the rectangular plot that would enclose the largest area with 800m of wire are 400m by 200m. The maximum area is:
Area = Length * Width
= 400m * 200m
= 80,000 square meters.
To find the largest area you can enclose with 800m of wire, you need to determine the dimensions of the rectangle that will maximize the area.
Let's say the length of the rectangle is L and the width is W. Since the river serves as one side of the plot, the total length of the electric fence will be L + 2W.
From the problem statement, it is given that the total amount of wire available is 800m. So we can set up the equation:
L + 2W = 800
Now, we need an equation to represent the area of the rectangle. The area of a rectangle is given by A = length × width, which in this case is A = LW.
To solve for the dimensions of the rectangle that maximize the area, we can express one variable in terms of the other. Let's solve the equation L + 2W = 800 for L:
L = 800 - 2W
Substituting this value of L into the equation for the area, we get:
A = (800 - 2W) × W
Expanding this equation, we have:
A = 800W - 2W^2
To maximize the area A, we differentiate the equation with respect to W and set it equal to zero:
dA/dW = 800 - 4W = 0
Solving this equation for W, we find:
800 - 4W = 0
4W = 800
W = 200
Now, we can substitute this value of W back into the equation L + 2W = 800 to find L:
L + 2(200) = 800
L + 400 = 800
L = 400
So the dimensions of the rectangle that maximizes the area are L = 400m and W = 200m.
To calculate the largest area, substitute the values of L and W back into the area equation:
A = 400 × 200
A = 80,000 square meters
Therefore, the largest area you can enclose is 80,000 square meters, and the dimensions of the rectangle are 400m by 200m.