a rectangular field is to be enclosed by a fence. two fences, parallel to one side of the field, divided the field into 3 rectangular fields. if 2400 m of fence is available find the dimensions of the field giving the maximum area.
To find the dimensions of the field that give the maximum area, we need to understand the given problem and solve it step by step.
Let's assume the dimensions of the rectangular field as follows:
Length of the field = L
Width of the field = W
According to the problem, two parallel fences divide the field into three rectangular fields. This means there will be two vertical fences parallel to the width of the field.
Therefore, we can visualize the field as follows:
___________________
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|_________|_________|
To find the length of the fence required, we need to calculate the total length of the fence used to enclose the field.
Given that 2400 m of fence is available, we can calculate it using the formula:
Total fence length = Length of outer fence + 2 × Length of inner fence + 2 × Width of the field
As per the given problem, we have:
Length of the outer fence = L
Length of the inner fences (both) = W
Width of the field = W
Therefore, the equation becomes:
2400 = L + 2W + 2W
Simplifying the equation:
2400 = L + 4W
Now, let's solve for L in terms of W:
L = 2400 - 4W
The area of the rectangular field is given by the formula:
Area = Length × Width
Substituting the value of L from above into the area equation, we have:
Area = (2400 - 4W) × W
To find the dimensions of the field that give the maximum area, we need to differentiate the area equation with respect to W, and set it equal to zero.
d(Area)/dW = 0
Differentiating the equation:
d(Area)/dW = 2400 - 8W = 0
Simplifying the equation:
8W = 2400
Solving for W:
W = 2400 / 8
W = 300
Substituting the value of W back into the equation L = 2400 - 4W, we get:
L = 2400 - 4(300)
L = 2400 - 1200
L = 1200
Therefore, the dimensions of the field that give the maximum area are:
Length = 1200 m
Width = 300 m
as always, divide the fence equally among lengths and widths.
You have 2 lengths and 4 widths.
So, allocating 1200m to each, the dimensions are
1200/2 x 1200/4 = 600 x 300
Or, you can go through the math, using
2x + 4y = 2400
a = xy
and finding the maximum value for a