Answer:
6000 ft
Step-by-step explanation:
Let length of rectangular field=x
Breadth of rectangular field=y
Area of rectangular field=[tex]1500000[/tex] square ft
Area of rectangular field=[tex]l\times b[/tex]
Area of rectangular field=[tex]x\time y[/tex]
[tex]1500000=xy[/tex]
[tex]y=\frac{1500000}{x}[/tex]
Fencing used ,P(x)=[tex]x+x+y+y+y=2x+3y[/tex]
Substitute the value of y
P(x)=[tex]2x+3(\frac{1500000}{x})[/tex]
[tex]P(x)=2x+\frac{4500000}{x}[/tex]
Differentiate w.r.t x
[tex]P'(x)=2-\frac{4500000}{x^2}[/tex]
Using formula:[tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]
[tex]P'(x)=0[/tex]
[tex]2-\frac{4500000}{x^2}=0[/tex]
[tex]\frac{4500000}{x^2}=2[/tex]
[tex]x^2=\frac{4500000}{2}=2250000[/tex]
[tex]x=\sqrt{2250000}=1500[/tex]
It is always positive because length is always positive.
Again differentiate w.r.t x
[tex]P''(x)=\frac{9000000}{x^3}[/tex]
Substitute x=1500
[tex]P''(1500)=\frac{9000000}{(1500)^3}>0[/tex]
Hence, fencing is minimum at x=1 500
Substitute x=1 500
[tex]y=\frac{1500000}{1500}=1000[/tex]
Length of rectangular field=1500 ft
Breadth of rectangular field=1000 ft
Substitute the values
Shortest length of fence used=[tex]2(1500)+3(1000)=6000 ft[/tex]
Hence, the shortest length of fence that the rancher can used=6000 ft
