A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 100m^2. what are the dimensions of each pen that minimize the amount of fence that must be used?

Well, to minimize the amount of fence used, we need to find the dimensions that are both rectangular and adjacent. So, let's put on our thinking caps and start solving this mathematical puzzle, shall we?

Since each pen has an area of 100m², we can express the length as L and the width as W. Therefore, we have:

L × W = 100

Now, if we make four adjacent pens, we need to consider the fence used between them. To minimize this, we want the pens to share as much fence as possible. With adjacent pens, we can merge the fences in between.

Let's consider the fence between the first two pens. If we label the width of the pen L and the length of the shared fence x, then the width of the second pen is still W, and the length of the shared fence for the other two sides is also x. We can simplify the equation as:

2L + x = 4W

Since we want to minimize the fence used, we need to look for the smallest values. In this case, let's set L equal to W, making the pens squares:

2W + x = 4W

Solving this, we find:

x = 2W

Substituting this into our original area equation:

W × W = 100

W² = 100

Now, to find the dimensions of each pen, we can take the square root of both sides:

W = √100

W = 10

And since L is equal to W, each pen will have dimensions of 10m by 10m.

Voila! We found the magical solution! Using square pens with dimensions of 10m by 10m will minimize the amount of fence used. Plus, it'd make for some happy little farm animals living in their cozy squares!

I hope that puts a smile on your face!

To minimize the amount of fence that must be used, we need to find the dimensions of each pen.

Let's represent the width of each pen as x meters and the length as y meters. Since each pen has an area of 100m^2, we can express this as:

Area = Length * Width
100 = x * y

We want to minimize the amount of fence used, which is the sum of the perimeters of the four pens. The perimeter of a rectangle is found by adding the lengths of all its sides, so the perimeter P of a single pen is:

P = 2x + 2y

Since there are four identical pens, the total fence length, F, is:

F = 4P

Substituting the value of P, we have:

F = 4(2x + 2y)
F = 8x + 8y

To minimize F, we can differentiate F with respect to one variable (x or y) and set the derivative equal to zero. Let's differentiate F with respect to x:

dF/dx = 8

Setting this equal to zero, we have:

8 = 0

This equation has no solutions, which indicates that the minimum value occurs at the boundary of the feasible region. Since x represents the width, there are no constraints on its value, so we can set x = 0 and find the corresponding value of y.

When x = 0, the area equation becomes:

100 = 0 * y
100 = 0

This equation has no solutions, which means that setting the width to zero is not feasible. Therefore, we cannot minimize the amount of fence used.

In conclusion, there are no dimensions of each pen that minimize the amount of fence that must be used given the conditions.

The equation for the perimeter of the adjacent pens is incorrect. They are being built against the barn meaning that the farmer does not use fencing against the barn. The correct equation to model that situation would thus be 5x+4y.

Now, if the barn is used as one wall of the pen, meaning only 3 sides have to be fenced, then

p = 5x+4y = 5x + 4(100/x) = 5x + 400/x
p' = 5 - 400/x^2

p' = 0 at x = 4√5
y = 25/√5

If each pen has width x and length y (against the barn),

the area is 4xy, and the fence used is 5x+8y

So, each pen has area 100.

p = 5x + 8(100/x) = 5x + 800/x
dp/dx = 5 - 800/x^2

dp/dx = 0 when x = √160 = 4√10

x = 4√10
y = 25/√10