Calculus
posted by Michael .
The Regional Farm Bureau (RFB) is preparing a brochure that offers advice about constructing pens for small farm animals, and they want us to be their consultants. They need us to carefully analyze the following situations and provide a detailed report. Then they will use our information to help them write the brochure.
In the example they wish to describe, it is assumed that the farmer has 900 feet of fencing with which to erect a rectangular pen alongside a long, existing fence (so the existing fence forms one side of the pen). Suppose the pen is to be subdivided into four parts in a twobytwo arrangement by including interior fences parallel to the outside boundaries. Then what dimensions make for the largest combined area? What if the farmer subdivides into nine pens in a threebythree arrangement? What if the farmer subdivides into n2 pens in an nbyn arrangement?

L=length of fence available = 900 ft
For an nxn arrangement, with total dimensions x (parallel to existing fence) and y (perpendicular).
total area, A =xy
Total length of fence
=(n+1)y+nx=L
y=(Lnx)/(n+1)
A=xy
=x(Lnx)/(n+1)
For maximum area,
dA/dx=0
which gives
(L2nx)/(n+1)=0
Solving for x,
x=L/2n
So total Area
A=xy
=x((Lnx)/(n+1)
=(L/2n)((Ln(L/2n))/(n+1)
=L^2/(4n(n+1))