Question

A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use? Note: The answer to this problem requires that you enter the correct units. (Enter your answer with a space before the units)

Answer 1

Answer:

6000 ft

Step-by-step explanation:

Let length of rectangular field=x

Breadth of rectangular field=y

Area of rectangular field=$1500000$ square ft

Area of rectangular field=$l\times b$

Area of rectangular field=$x\time y$

$1500000=xy$

$y=\frac{1500000}{x}$

Fencing used ,P(x)=$x+x+y+y+y=2x+3y$

Substitute the value of y

P(x)=$2x+3(\frac{1500000}{x})$

$P(x)=2x+\frac{4500000}{x}$

Differentiate w.r.t x

$P'(x)=2-\frac{4500000}{x^2}$

Using formula:$\frac{dx^n}{dx}=nx^{n-1}$

$P'(x)=0$

$2-\frac{4500000}{x^2}=0$

$\frac{4500000}{x^2}=2$

$x^2=\frac{4500000}{2}=2250000$

$x=\sqrt{2250000}=1500$

It is always positive because length is always positive.

Again differentiate w.r.t x

$P''(x)=\frac{9000000}{x^3}$

Substitute x=1500

$P''(1500)=\frac{9000000}{(1500)^3}>0$

Hence, fencing is minimum at x=1 500

Substitute x=1 500

$y=\frac{1500000}{1500}=1000$

Length of rectangular field=1500 ft

Breadth of rectangular field=1000 ft

Substitute the values

Shortest length of fence used=$2(1500)+3(1000)=6000 ft$

Hence, the shortest length of fence that the rancher can used=6000 ft