Question

A farmer wants to fence a rectangular pasture using a river as one side. The pasture must contain 180,000 square meters of space for the herd to frolic. What is the least amount of fencing required to enclose such a pasture?

Answer 1

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=$l\times b=xy$

$xy=180000$

$y=\frac{180000}{x}$

Fencing required=2x+y

$P=2x+\frac{180000}{x}$

Differentiate w.r.t x

$\frac{dP}{dx}=2-\frac{180000}{x^2}$

$\frac{dP}{dx}=0$

$2-\frac{180000}{x^2}=0$

$\frac{180000}{x^2}=2$

$x^2=\frac{180000}{2}=90000$

$x=\sqrt{900000}=300$m

It is always positive because length cannot be negative.

$y=\frac{180000}{300}=600$m

Again differentiate w.r.t x

$\frac{d^2P}{dx^2}=\frac{360000}{x^3}$

Substitute x=300

$\frac{d^2P}{dx^2}=\frac{360000}{27000000}=0.013>0$

The fence is minimum at x=300

Therefore, fencing required  to enclose such a pasture

$P=2x+y=2(300)+600=1200 m$