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March 6, 2015

March 6, 2015

Posted by **Michael** on Tuesday, March 27, 2012 at 9:28pm.

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 two-by-two 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 three-by-three arrangement? What if the farmer subdivides into n2 pens in an n-by-n arrangement?

- Calculus -
**MathMate**, Wednesday, March 28, 2012 at 11:12amL=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=(L-nx)/(n+1)

A=xy

=x(L-nx)/(n+1)

For maximum area,

dA/dx=0

which gives

(L-2nx)/(n+1)=0

Solving for x,

x=L/2n

So total Area

A=xy

=x((L-nx)/(n+1)

=(L/2n)((L-n(L/2n))/(n+1)

=L^2/(4n(n+1))

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