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
[tex] A=x*y\\ A= 200\ ft^{2} \\ x*y=200 [/tex]
[tex] y=\frac{200}{x} [/tex] -----> equation [tex] 1 [/tex]
The perimeter of the rectangle is equal to
[tex] P=2x+2y [/tex]
but remember that the fourth side of the rectangle will be formed by a portion of the barn wall
so
[tex] P=x+2y [/tex] -----> equation [tex] 2 [/tex]
To minimize the cost we must minimize the perimeter
Substitute the equation [tex] 1 [/tex] in the equation [tex] 2 [/tex]
[tex] P=x+2*[\frac{200}{x} ] [/tex]
Using a graph tool
see the attached figure
The minimum of the graph is the point [tex] (20,40) [/tex]
that means for [tex] x=20\ ft [/tex]
the perimeter is a minimum and equal to [tex] 40\ ft [/tex]
Find the value of y
[tex] y=\frac{200}{x} [/tex]
[tex] y=\frac{200}{20} [/tex]
[tex] y=10\ ft [/tex]
The cost of fencing is equal to
[tex] 5*40= \$200 [/tex]
therefore
the answer is
the length side of the the fourth wall will be [tex] 20\ ft [/tex]
