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?

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=(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))

To help the Regional Farm Bureau analyze the situation and determine the dimensions that make for the largest combined area, we need to understand the problem and break it down step by step.

In the given example, the farmer has 900 feet of fencing to construct a rectangular pen alongside an existing fence. The task is to subdivide this pen into four parts in a two-by-two arrangement by including interior fences parallel to the outside boundaries. We want to find the dimensions that result in the largest combined area.

Let's start by visualizing the rectangular pen and understanding its dimensions. We have one existing side, which is formed by the long fence, and three sides that need to be constructed using the available fencing. Let's denote the length of the existing side as x and the length of the other two sides as y.

Since the pen is to be divided into four parts, there will be three interior fences dividing the pen into smaller sections. These fences are parallel to the outside boundaries and will divide the pen into equal dimensions.

To find the dimensions that result in the largest combined area, we need to maximize the total area of the four smaller pens within the larger rectangular pen.

Let's calculate the area of one of the smaller pens. The length of the smaller pen will be x/2, and the width will be y/2. Therefore, the area of one smaller pen will be (x/2) * (y/2) = xy/4.

Since there are four equal-sized pens within the larger rectangular pen, the total combined area of the four smaller pens will be 4 * (xy/4) = xy.

To maximize the combined area, we need to find the values of x and y that will maximize the expression xy.

To do this, we can use calculus by taking the derivative of the expression xy with respect to x and setting it equal to zero. Then, solve for x.

d(xy)/dx = (dy/dx) * x + (dx/dx) * y = y + 0 = y

Setting y equal to zero gives us 0y, which does not provide any meaningful solutions. Therefore, y cannot be zero.

Next, let's take the derivative of the expression xy with respect to y and set it equal to zero. Then, solve for y.

d(xy)/dy = (dy/dy) * x + (dx/dy) * y = x + 0 = x

Setting x equal to zero gives us 0x, which also does not provide any meaningful solutions. Therefore, x cannot be zero.

From the above calculations, we can conclude that neither x nor y can be zero to maximize the combined area.

Now, let's consider the constraints of the problem. The farmer has 900 feet of fencing, which means the total length of fencing used will be:

2x (existing fence) + 2y + y (interior fences) + y (interior fences) = 4x + 4y

Since the total length of the fencing is 900 feet, we can write the equation:

4x + 4y = 900

Simplifying the equation gives us:

x + y = 225

We can now express y in terms of x:

y = 225 - x

Substituting this into the expression xy, we have:

xy = x(225 - x) = 225x - x^2

To find the dimensions that maximize the combined area, we need to find the maximum point of the expression 225x - x^2. We can do this by finding the vertex of the quadratic function.

The vertex of a quadratic function in the form f(x) = ax^2 + bx + c can be found at x = -b/2a.

In our case, the function is f(x) = -x^2 + 225x, so the vertex can be found at x = -225/(2*(-1)) = 112.5.

Since the dimensions need to be positive, we can round x to the nearest whole number, resulting in x = 113. Then, we can find y by substituting x into y = 225 - x, giving us y = 225 - 113 = 112.

Therefore, the dimensions that make for the largest combined area in a two-by-two arrangement are approximately x = 113 feet and y = 112 feet.

To analyze the situations where the farmer subdivides into nine pens in a three-by-three arrangement or n^2 pens in an n-by-n arrangement, the same process can be applied. However, the expressions and equations will change accordingly.

For a three-by-three arrangement, the dimensions of each smaller pen will be (x/3) * (y/3). The total combined area of the nine pens can be calculated as 9 * (xy/9) = xy.

For an n-by-n arrangement, the dimensions of each smaller pen will be (x/n) * (y/n). The total combined area of the n^2 pens can be calculated as n^2 * (xy/n^2) = xy.

By following similar steps as outlined above, we can find the dimensions that result in the largest combined area for each respective arrangement.