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The length of fence will be equal to the length of the perimeter of the rectangular field.
For a given given area the perimeter of a rectangle is minimum when the rectangle is a square.
Therefor the fence for the rectangular field with area of 30 cm^2 will be for a square field with this area.
The measure of each side of this field will be:
Side = 30^(1/2)
The perimeter of the field will be 4 time the side. Thus:
Perimeter = 4*Side = 4*30^(1/2)
The area of the rectangle = 30 cm^2.
To have fence in least amount, we should have the least length of the fence.
Since there is a stone wall on one side, we have only 3 sides of the rectangle.
So let x be one side then the adjacent side must be 30/x.
So the length of fence = 2x+30/x which shoild be minimum.
L(x) = 2x+30/x
L'(x) = 2-30/x^2
L'(c) = 0 gives: 2-30/x^2 = 0.
2x^2 -30 = 0
x^2 = 15
x = sqrt15.
L"(x) = (2-30/x^2)' = -(-1/2)30/x^3 = 15/(x^3) > for x = sqrt15.
Therefore L(x) is least when x = sqrt15 , as L'(sqrt15) = 0 and L"(sqrt15) >0.
Therefore L(sqrt15) = 2sqrt15+30/sqrt15 = 2sqrt15 +2sqrt15 = 4sqrt15 is the minimum length of the fence.
Let us assume the length of the field is L. Now the area of a rectangular field is given as length * width.
Here area = length * width
=> 30 = L * W
=> W = 30 / L
Now the amount of fencing required is equal to 2*W + L as one of the sides has a stonewall and I assume that it is equal to the longer side so that we need to use as little of the fence as possible.
Now Fencing : F = 2*W + L
= 2* 30 / L + L
= 60 / L + L
Now differentiating this with respect to L we get
dF / dL = 60* (-1) / L^2 + 1
Equating this to 0
-60 / L^2 +1 = 0
=> L^2 = 60
=> L = sqrt 60
Therefore the length is sqrt 60. Width is 30 / sqrt 60 and the length of fencing required is sqrt 60 + 60 / sqrt 60 = 2* sqrt 60
or 15.49 cm.
The required result is 15.49 cm.
If we need to determine a maximum or minimum amount of something, we have to create a function that depends on the amount.
After creating the function, we'll differentiate it. Then, we'll calculate the roots of the first derivative. These roots, if they exist, represent the extremes of a function.
In our case, we need to determine the minimum amount to be fenced.
We'll choose as amount one dimension of the rectangle. We'll choose the length and we'll note it as x.
We'll find the other dimension of the rectangle, namely the width, using the formula of area.
A = l*w
We'll substitute area by 30.
30 = x*w
We'll divide by x:
w = 30/x
Since the perimeter of the rectangle is calculated with respect to x, we'll create the function that depends on x, to determine the minimum amount to be fenced.
The formula of the perimeter of a rectangle is;
P = 2(l+w)
We'll create the function:
P(x) = 2(x + 30/x)
Now, we'll differentiate
P'(X) = (2x + 60/x)'
P'(X) = 2 - 60/x^2
We'll calculate the solution of P'(X).
P'(X) = 0
2 - 60/x^2 = 0
60/x^2 = 2
2x^2 - 60 = 0
We'll divide by 2:
x^2 - 30 = 0
x^2 = 30
x1 = +sqrt30
x2 = -sqrt30 is rejected because a length cannot be negative.
P(sqrt30) = 2sqrt30+ 60/sqrt30
P(sqrt30) = (60 + 60)/sqrt30
P(sqrt30) = 120/sqrt30
P(sqrt30) = 120sqrt30/30
P(sqrt30) = 4sqrt30
P(sqrt30) = 4sqrt30 cm is the least amount to be fenced.
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