A rectangular field is to be enclosed by a fence and divided into three lots by fences parallel to one of the sides. Find the dimensions of the largest field that can be enclosed with 800 feet of fencing.
Help me please!!!!!!!!!!! THANK YOU
2 * L + 4*w = 800
so
L = 400 - 2w
A = L * w
A = (400 -2w)w
A = -2 w^2 +400 w
dA/dw = 0 at max
= -2 w + 200
w = 100
L = 400 -200 = 200
Why did the scarecrow win an award? Because he was outstanding in his field!
In order to find the dimensions of the largest field, let's break down the problem step by step.
Let's assume that the length of the rectangle is L and the width is W. Since the field is divided into three lots by fences parallel to one side, there will be two fences dividing the length and two fences dividing the width.
Given that there is 800 feet of fencing, we can set up the equation:
2L + 4W = 800
In this equation, 2L represents the length of the fences on the sides, and 4W represents the length of the fences dividing the width.
Now, let's isolate L in terms of W:
2L = 800 - 4W
L = (800 - 4W) / 2
L = 400 - 2W
We can now substitute this value for L into the equation for the perimeter of the field:
P = 2L + 2W
P = 2(400 - 2W) + 2W
P = 800 - 4W + 2W
P = 800 - 2W
Since the fences parallel to the width have a length of W, we know that:
P(W) = 800 - 2W
To find the dimensions of the largest field, we need to maximize the area. The area of the field is given by:
A = L * W
A = (400 - 2W) * W
A = 400W - 2W^2
To find the maximum area, we can take the derivative of the area equation with respect to W and set it equal to zero:
dA/dW = 400 - 4W = 0
Solving for W, we get W = 100. Substituting this back into the perimeter equation, we find:
P(100) = 800 - 2(100) = 600
Therefore, the dimensions of the largest field that can be enclosed with 800 feet of fencing are: length = 400 - 2(100) = 200 feet and width = 100 feet.
I hope this helps!
To find the dimensions of the largest field that can be enclosed with 800 feet of fencing, we need to consider the perimeter of the field.
Let's assume that the length of the field is L, and the width of the field is W.
Since the field is rectangular, the perimeter is given by:
Perimeter = 2L + 3W
According to the problem, the total fencing available is 800 feet:
Perimeter = 800
Substituting the values, we have:
2L + 3W = 800
To find the dimensions for maximum area, we need to express one of the variables in terms of the other. Let's solve the equation for L:
2L = 800 - 3W
L = (800 - 3W)/2
Now, we can express the area of the field in terms of W:
Area = L * W
Area = ((800 - 3W)/2) * W
Area = (800W - 3W^2)/2
To find the maximum area, we can take the derivative of the area function with respect to W and set it equal to zero:
d(Area)/dW = 0
800 - 6W = 0
Solving for W:
6W = 800
W = 800/6
W = 133.33 feet (approx.)
Substituting this value back into our perimeter equation:
2L + 3(133.33) = 800
2L + 400 = 800
2L = 400
L = 200 feet
Therefore, the dimensions of the largest field that can be enclosed with 800 feet of fencing are approximately:
Length = 200 feet
Width = 133.33 feet
To solve this problem, we need to find the dimensions of the rectangular field that will maximize its area, given the constraint of using 800 feet of fencing.
Let's assume the length of the rectangular field is L and the width is W.
According to the problem, the field must be divided into three lots by fences parallel to one of the sides. This means that the field will have two parallel fences dividing it into three equal parts.
So, we can divide the field into three equal sections, where each section will have a width of W/3.
Now, let's calculate the total amount of fencing needed.
1. Two sides of the rectangular field will each have a length L.
2. The bottom side of each section will have a length W/3, and the top side will have a length W/3 as well.
3. The two fences dividing the field into three lots will each have a length L.
Adding up these lengths, we get the following equation:
2L + (2/3)W + (2/3)W + L = 800
Simplifying the equation, we have:
4L + (4/3)W = 800
Now, we need to express one variable in terms of the other to make it a single-variable equation and maximize the area.
Solving for L, we find:
L = (800 - (4/3)W) / 4
Now, let's express the total area, A, in terms of W:
A = LW
Substituting the value of L, we get:
A = [(800 - (4/3)W) / 4] * W
To maximize the area, we can take the derivative of A with respect to W, set it equal to zero, and solve for W.
dA/dW = [(800 - (4/3)W) / 4] - (4W/3) = 0
Now, we can solve this equation for W.
[(800 - (4/3)W) / 4] - (4W/3) = 0
[(800 - (4/3)W) - (16W/3)] / 4 = 0
[2400 - 4W - 16W] / 12 = 0
2400 - 20W = 0
20W = 2400
W = 2400 / 20
W = 120 feet
Now that we have the value of W, we can substitute it back into the equation for L:
L = (800 - (4/3)W) / 4
L = (800 - (4/3)(120)) / 4
L = (800 - 160) / 4
L = 640 / 4
L = 160 feet
Therefore, the dimensions of the largest field that can be enclosed with 800 feet of fencing are 160 feet by 120 feet.