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

$A=x*y\\ A= 200\ ft^{2} \\ x*y=200$

$y=\frac{200}{x}$ -----> 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 $2$

$P=x+2*[\frac{200}{x} ]$

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

$y=\frac{200}{x}$

$y=\frac{200}{20}$

$y=10\ ft$

The cost of fencing is equal to

$5*40= \ 200$

therefore

the answer is

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