Question

A rancher has 30,000 linear feet of fencing and wants to enclose a rectangular field and then divide it into four equal pastures with three internal fences parallel to one of the rectangular sides. what is the maximum area of each pasture?

Answer 1

Let

x--------> the length side of the rectangle

y-------> the width side of the rectangle

see the attached figure N 1 to better understand the problem

we know that

the perimeter of the rectangle is equal to

`P=2x+5y\\ P=30,000\ ft`

so

`2x+5y=30,000\\ 5y=30,000-2x`

`y=6,000-0.40x` --------> equation 1

the area of the rectangle is equal to

`A=x*y` -------> equation 2

substitute equation 1 in equation 2

`A=x*[6,000-0.40x]`

`A=-0.40x^(2) +6,000x`

Using a graph tool

see the attached figure

The vertex of the function is the point with the maximum area

the vertex of the function is the point (7,500, 22,500,000)

that means that

for x=7,500 ft

the maximum area is 22,500,000 ft^2

Find the value of y

`y=6,000-0.40*7,500`

`y=3,000\ ft`

the dimensions of the rectangle are 7,500 ft * 3,000 ft

the maximum area of each pasture is

(22,500,000 ft^2)/4=5,625,000 ft^2

therefore

the answer is

the maximum area of each pasture is 5,625,000 ft^2