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April 16, 2014

April 16, 2014

Posted by **Michael** on Monday, March 26, 2012 at 4:31pm.

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 -
**bobpursley**, Monday, March 26, 2012 at 4:41pmI don't understand the two by two arrangement.

What you need to do is write an expression for the total amount of fence (outside+inside lengths)=900

it will be in terms of fractions of length, width.

Then you want to maximize area:

Area= Outsidelength*width.

Now in the fence length needed equation, solve for w. Put that in the area equation.

Take the derivative dA/dl, set to zero, and solve for l.

- Calculus -
**Steve**, Monday, March 26, 2012 at 4:46pmMight as well go for the general nxn case, then we can plug in what we want.

If there are n^2 small pens, with width x and length y, with length parallel to the long fence, then the total fence used is

n(n+1)x + n^2y = 900

y = (900-n(n+1)x)/n^2

A = n^2 xy

= n^2 x (900-n(n+1)x)/n^2

= x(900-n(n+1)x)

= 900x - n(n+1)x^2

dA/dx = 900 - 2n(n+1)x

max at x = 900/[2n(n+1)]

x = 450/[n(n+1)]

y = 450/n^2

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