Respuesta :
The length of the fencing around the park is 500 m.
The needed width of the fencing is 250 m.
Step-by-step explanation:
Let us assume the length of the field = x meters
Let us assume the breadth if the field = y meters
Now, let us assume the one length and twice the width.
So, the length of the total fencing = x + y + y = (x + 2 y) meters
Also, area of the field = Length x Breadth
= (x meters )( y meters) =x y sq m = 1,25,000 sq m
⇒ x y = 1,25,000
or, [tex]x = \frac{125000}{y}[/tex]
Putting this value in (x + 2 y), we get:
P = (x + 2 y) = [tex](\frac{125000}{y} ) + 2y[/tex]
Now, [tex]P'(y) = (-\frac{125000}{y^2} ) + 2 = 0[/tex] for minimizing the fencing perimeter.
On solving , we get:
[tex]125000 = 2y^2\\or, y^2 = 62,500\\\implies y = 250[/tex]
So, the width of the perimeter = y = 250 meters
[tex]x = \frac{125000}{y} = \frac{125000}{250} = 500\\\implies x = 500 m[/tex]
Hence, the needed length of the fencing is 500 m.
and the needed width of the fencing is 250 m.
The dimensions (in meters) that will use the least amount of fencing is 500m ; 250m
Let the sides represent L and W
Where:
L=Length
W=Width
First step
LW = 125,000 square meters
W = 125,000/L
Second step is to determine the perimeter
Perimeter= 2L + 2W
Perimeter= 2L + 125,000/L
Third step is for us to set the derivative to zero and then determine L
Perimeter = 2 - 250,000/L^2 = 0
2L^2 = 500,000
L^2=500,000/2
L^2 = 250,000
L = √(250,0000
L = 500 m
Fourth step is to determine W
W= 125,000/500
W= 250 m
Inconclusion The dimensions (in meters) that will use the least amount of fencing is 500m ; 250m
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