Question

a farmer is planning a rectangular area for her chickens. the area of the rectangle will be 200 square feet. three sides of the rectangle will be formed by fencing, which cost 5$ per foot, the fourth side of the rectangle will be formed by a portion of the barn wall, which requires no fencing. in order to minimize the cost of fencing, how long should the fourth wall be?

Answer 1

Let

x---------> the length side of the rectangular area

y---------> the width side of the rectangular area

we know that

the area of the rectangle is equal to

`A=x*y\\ A= 200\ ft^(2) \\ x*y=200`

`y=(200)/(x)` -----> equation
`1`

The perimeter of the rectangle is equal to

`P=2x+2y`

but remember that the fourth side of the rectangle will be formed by a portion of the barn wall

so

`P=x+2y` -----> equation
`2`

To minimize the cost we must minimize the perimeter

Substitute the equation
`1` in the equation
`2`

`P=x+2*[(200)/(x) ]`

Using a graph tool

see the attached figure

The minimum of the graph is the point
`(20,40)`

that means for
`x=20\ ft`

the perimeter is a minimum and equal to
`40\ ft`

Find the value of y

`y=(200)/(x)`

`y=(200)/(20)`

`y=10\ ft`

The cost of fencing is equal to

`5*40= \$200`

therefore

the answer is

the length side of the the fourth wall will be
`20\ ft`