# Calculus

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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 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 -

I 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 -

Might 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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