Answer:
1200 m
Step-by-step explanation:
We are given that
Area of pasture,A=180000 square meter
We have to find the least amount of fencing required to enclose such a pasture.
Let length of pasture,=x
Width of pasture,b=y
Area of pasture=[tex]l\times b=xy[/tex]
[tex]xy=180000[/tex]
[tex]y=\frac{180000}{x}[/tex]
Fencing required=2x+y
[tex]P=2x+\frac{180000}{x}[/tex]
Differentiate w.r.t x
[tex]\frac{dP}{dx}=2-\frac{180000}{x^2}[/tex]
[tex]\frac{dP}{dx}=0[/tex]
[tex]2-\frac{180000}{x^2}=0[/tex]
[tex]\frac{180000}{x^2}=2[/tex]
[tex]x^2=\frac{180000}{2}=90000[/tex]
[tex]x=\sqrt{900000}=300[/tex]m
It is always positive because length cannot be negative.
[tex]y=\frac{180000}{300}=600[/tex]m
Again differentiate w.r.t x
[tex]\frac{d^2P}{dx^2}=\frac{360000}{x^3}[/tex]
Substitute x=300
[tex]\frac{d^2P}{dx^2}=\frac{360000}{27000000}=0.013>0[/tex]
The fence is minimum at x=300
Therefore, fencing required to enclose such a pasture
[tex]P=2x+y=2(300)+600=1200 m[/tex]
