A rectangular field is to be enclosed by a fence and divided into three equal rectangular parts by two other fences. find the maximum area that can be enclosed and separated in this way with 1200m of fencing.
let the width of the whole rectangle be x m (there will be 4 of these)
let the length be y m
then 4x + 2y = 1200
2x + y = 600
y = 600 - 2x
Area = xy
= x(600-2x)
= -2x^2 + 600x
Now, I don't know if you are studying Calculus.
If you do, then
d(Area)/dx = -4x + 600
= 0 for a max area
x = 150
then y = 600 - 2(150) = 300
and the max area is (150)(300) = 45000
If you don't know Calculus, complete the square on the above quadratic
you should end up with
Area = -2(x-150)^2 + 45000
45000
To find the maximum area that can be enclosed and separated in this way, we need to determine the dimensions of the rectangular field and the two dividing fences.
Let's assume the length of the rectangular field as "L" and the width as "W."
According to the problem, we have three equal rectangular parts, so the width of each part will be W/3.
We can represent the dimensions of the field and the two dividing fences as:
Field:
Length = L
Width = W
Dividing Fences:
Length = L
Width = W/3
To calculate the total fencing required, we add up the lengths of all the sides:
Field:
2(L + W)
Dividing Fences:
2(L + W/3) + 2(L + W/3)
Given that the total amount of fencing available is 1200m, we can equate this to the total fencing required:
2(L + W) + 2(L + W/3) + 2(L + W/3) = 1200
Simplifying the equation:
2L + 2W + (2L)/3 + (2W)/3 + (2W)/3 = 1200
Multiplying both sides of the equation by 3 to eliminate the fractions:
6L + 6W + 2L + 2W + 2W = 3600
Simplifying further:
8L + 10W = 3600
To find the maximum area, we need to maximize the product of the length and width. So, let's solve this equation to find the optimum values.
We'll use the method of substitution to simplify the equation:
From 8L + 10W = 3600, we get L = (3600 - 10W)/8.
The area A is given by A = LW.
Substituting the value of L, we have A = ((3600 - 10W)/8) * W.
To find the maximum area, we can differentiate the equation with respect to W and equate it to zero:
dA/dW = ((1800 - 5W)/4) = 0
Simplifying, we get:
1800 - 5W = 0
Solving for W, we find:
W = 360.
Substituting this value back into the equation L = (3600 - 10W)/8, we get:
L = (3600 - 10*360)/8 = 240.
Therefore, the maximum area that can be enclosed and separated in this way with 1200m of fencing is given by:
Area = Length * Width = 240 * 360 = 86,400 square meters.
To find the maximum area that can be enclosed and separated in this way, we need to optimize the dimensions of the rectangle.
Let's consider the rectangle's dimensions:
- Length: L
- Width: W
The question states that the rectangle is divided into three equal parts by two other fences. This means that the rectangle will have two additional fences inside, dividing it into three sections with equal widths.
To calculate the total fencing required, we need to consider:
- Four sides of the rectangle: 2L + 2W
- Two interior fences: 2W
Given that the total fencing available is 1200m, we can set up the equation:
2L + 2W + 2W = 1200
Simplifying the equation:
2L + 4W = 1200
L + 2W = 600
L = 600 - 2W
Next, we need to calculate the area of the rectangle, which is given by:
Area = Length x Width
Area = L x W
Area = (600 - 2W) x W
Now, we can find the maximum area by differentiating the equation of the area with respect to W and setting it to zero:
d(Area)/dW = (600 - 4W) = 0
Solving this equation gives us:
600 - 4W = 0
4W = 600
W = 150m
Substituting the value of W back into the equation for L:
L = 600 - 2W
L = 600 - 2(150)
L = 600 - 300
L = 300m
Therefore, the maximum area that can be enclosed and separated in this way is:
Area = Length x Width
Area = 300m x 150m
Area = 45000m²
So, the maximum enclosed area is 45000 square meters.