Answer:
Length of rectangular field=45 ft
Width of rectangular field=30 ft
Step-by-step explanation:
We are given that
Area of rectangular field,A=1350 square feet
Let
Length of rectangular field=x
Width of rectangular field=y
a fence out parallel to y
Fence used,P=2x+2y+y=2x+3y
Area of rectangular field,[tex]A=l\time b=xy[/tex]
[tex]1350=xy[/tex]
[tex]y=\frac{1350}{x}[/tex]
[tex]P(x)=2x+3(\frac{1350}{x})=2x+\frac{4050}{x}[/tex]
Differentiate w.r.t x
[tex]P'(x)=2-\frac{4050}{x^2}[/tex]
[tex]P'(x)=0[/tex]
[tex]2-\frac{4050}{x^2}=0[/tex]
[tex]2=\frac{4050}{x^2}[/tex]
[tex]x^2=\frac{4050}{2}=2025[/tex]
[tex]x=\sqrt{2025}=45[/tex]
It is positive because length cannot be negative.
Again, differentiate w.r.t x
[tex]P''(x)=\frac{8100}{x^3}=\frac{8100}{(45)^3}=0.089>0[/tex]
Fencing needed is minimum at x=45
Substitute the value
[tex]y=\frac{1350}{x}=\frac{1350}{45}=30[/tex]
Hence, Length of rectangular field=45 ft
Width of rectangular field=30 ft
The lengths of the sides of the rectangular field so as to minimize the amount of fencing needed is 45 feet by 30 feet
Let x represent the length of the fence and y represent the width of the fence.
Since the area is 1350, hence:
Area = length * width = x * y
1350 = xy
y = 1350/x
A fence divide the field in half and is parallel to one of the sides of the rectangle. Hence:
Amount of fencing needed (P) = x + x + y + y + y = 2x + 3y
Amount of fencing needed (P) = 2x + 3(1350/x) = 2x + 4050/x
To minimize the amount of fencing needed, dP/dx = 0, hence:
dP/dx = 2 - 4050/x²
2 - 4050/x² = 0
2 = 4050/x²
2x² = 4050
x² = 2025
x = 45 feet
y = 1350/x = 1350/45 = 30 feet
Hence, the lengths of the sides of the rectangular field so as to minimize the amount of fencing needed is 45 feet by 30 feet
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