Answer:

Step-by-step explanation:
<u>Maximization With Derivatives</u>
Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.
We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.
The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.
The amount of fence needed to enclose the external and the internal divisions is

We know the total fencing is 1/2 miles long, thus

Solving for x

The total area of the pasture is

Substituting x


Differentiating with respect to y

Equate to 0

Solving for y

And also

Compute the second derivative

Since it's always negative, the point is a maximum
Thus, the maximum area is
