A farmer has 36 feet of fence to build a pigpen. He is going to use one of the sides of his barn as a side to the rectangular enclosure. Determine a function A that represents the total area of the enclosed region. What is the maximum area that can be enclosed?

A = L * w

Perimeter = L + 2 w = 36
so
L = 36-2w

A = (36-2w)w

-2 w^2 + 36 w = A

w^2 - 18 w = -A/2

w^2 - 18 w + 81 = -A/2 + 81

(w-9)^2 = -(1/2)(A-162)

vertex at
w = 9
area = 162 ft^2
====================
check
if w = 9
L = 36 - 18 = 18
area = 162, sure enough

Let's assume the length of the side of the pigpen parallel to the barn is denoted by "x" feet.

Since one side of the pigpen is the barn, there are only three sides that need fencing. The remaining two sides must have the same length to maximize the area of the enclosed region.

Therefore, the other two sides of the pigpen will each have a length of (36 - 2x)/2 = (18 - x) feet.

The area of the enclosed region can be calculated by multiplying the lengths of these two sides:

A(x) = x * (18 - x)

To find the maximum area that can be enclosed, we need to determine where the derivative of A(x) is equal to zero.

Let's differentiate A(x) with respect to x:

A'(x) = 18 - 2x

Setting A'(x) equal to zero:

18 - 2x = 0

Solving for x:

2x = 18
x = 9

To verify that this is a maximum, we can take the second derivative of A(x). If the second derivative is negative at x = 9, it confirms that it is a maximum.

Let's calculate the second derivative of A(x):

A''(x) = -2

Since the second derivative is negative, the value of x = 9 corresponds to a maximum area.

Substituting x = 9 into the equation for A(x):

A(9) = 9 * (18 - 9)
A(9) = 9 * 9
A(9) = 81

Therefore, the maximum area that can be enclosed is 81 square feet.

To determine a function A that represents the total area of the enclosed region, we will need to break down the problem and express the area in terms of a single variable. Let's assume the length of the pigpen parallel to the barn is x feet.

Since there are two sides parallel to the barn and the remaining two sides are perpendicular, the dimensions of the rectangular enclosure are x and (36 - 2x) feet.

The area A of a rectangle is given by the product of its length and width:
A = x * (36 - 2x) = 36x - 2x^2

To find the maximum area that can be enclosed, we need to determine the value of x that maximizes the area. This can be done by finding the vertex of the quadratic function A = -2x^2 + 36x.

The vertex of a quadratic function in the form A = ax^2 + bx + c is given by x = -b / (2a).

In this case, a = -2 and b = 36, so the x-coordinate of the vertex is:
x = -36 / (2 * -2) = -36 / -4 = 9.

To make sure we found the maximum, we can also check the endpoint values. Since we cannot have a negative length for the pen, the possible values for x are between 0 and 18 feet.

If we substitute x = 0 into A = -2x^2 + 36x, we get A = 0, and if we substitute x = 18, we also get A = 0. This means that the maximum area is achieved when x = 9 feet.

Substituting x = 9 into A = 36x - 2x^2, we get:
A = 36 * 9 - 2 * 9^2 = 324 - 162 = 162 square feet.

Therefore, the maximum area that can be enclosed is 162 square feet.