Respuesta :
The required dimensions are 500 meters and 250 meter for the fencing to be minimum.
Uses for maxima and minima:
If a rectangle has dimensions x and y, then its area is equal to x and y, and its perimeter is equal to 2(x + y). The edge of a figure has anything to do with fencing. Use the idea of maxima and minima by combining all the variables into a single variable.
According to the given data:
Let the dimensions of the rectangular pasture be x and y.
Equate the area of the rectangle to 125,000 and solve for y.
Where:
xy = 125000
y = 125000/x
If one side of the rectangle is left out, figure out how much fencing or perimeter is needed.
P = 2(x + y) - x
= x + 2y.
Set y = 125000/x into obtained perimeter.
P = x + 2(125000/x).
= x + (2,50,000/x)
Differentiate P with respect to x.
dP/dx = 1 - (2,50,000/x²)
Equate the obtained derivative to 0 and solve for x (can not be negative).
1 - (2,50,000/x²) = 0
(2,50,000/x²) = 1
x² = 2,50,000
x = 500
Calculate the second derivative of P and set x = 500.
d²P/dx² = 1 - (2,50,000/x³)
= (2,50,000/x³)>0
Calculate y by setting x=500 into y = 125000/x .
y = 125000/x
y = 125000/500
y = 250
The required dimensions are 500 meters and 250 meter for the fencing to be minimum.
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