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### Problem:
A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals. We are tasked with:
1. Writing the area, \( A \), of the corral as a function of \( x \).
2. Finding the values of \( x \) and \( y \) that will maximize the area.

#### (a) Write the area \( A \) of the corral as a function of \( x \):

From the problem, we know that:
- The perimeter of the corral is constrained by the 200 feet of fencing.
- The total perimeter includes four vertical segments of length \( y \) and three horizontal segments of length \( x \).

The perimeter constraint is:
\[
4y + 3x = 200 \quad \text{(since there are 4 vertical segments of height \( y \), and 3 horizontal segments of width \( x \))}
\]
We can solve for \( y \):
\[
y = \frac{200 - 3x}{4}
\]

Now, the area of the corral, \( A \), is the product of length and width:
\[
A = x \cdot y = x \cdot \frac{200 - 3x}{4}
\]

Simplifying this:
\[
A(x) = \frac{x(200 - 3x)}{4}
\]
\[
A(x) = \frac{200x - 3x^2}{4}
\]
Thus, the area function is:
\[
A(x) = 50x - \frac{3x^2}{4}
\]

#### (b) Find the values of \( x \) and \( y \) that will produce the maximum area:

To maximize the area, we take the derivative of \( A(x) \) with respect to \( x \) and set it equal to zero to find the critical points.

1. Derivative of \( A(x) \):
   \[
   A'(x) = 50 - \frac{3}{2}x
   \]
2. Set the derivative equal to zero:
   \[
   50 - \frac{3}{2}x = 0
   \]
   Solving for \( x \):
   \[
   \frac{3}{2}x = 50
   \]
   \[
   x = \frac{50 \times 2}{3} = \frac{100}{3} \approx 33.33 \, \text{feet}
   \]

3. Now, substitute \( x = 33.33 \) back into the expression for \( y \):
   \[
   y = \frac{200 - 3(33.33)}{4} = \frac{200 - 100}{4} = \frac{100}{4} = 25 \, \text{feet}
   \]

So, the dimensions that maximize the area are approximately:
- \( x \approx 33.33 \, \text{feet} \)
- \( y = 25 \, \text{feet} \)

These values will give the maximum possible area.

### Do you have any questions or want more details?

Here are five follow-up questions for practice:
1. How would the area function change if there were three vertical segments and four horizontal segments?
2. What would happen to the optimal dimensions if the total fencing were increased to 300 feet?
3. How do you find the maximum area without using calculus?
4. How does the shape of the corral affect the maximum area that can be enclosed?
5. What is the importance of the derivative in finding maximum and minimum values?

**Tip**: Whenever you have a perimeter constraint problem, always express one variable in terms of the other using the given constraints to simplify the area equation.

A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals. The problem asks to: (a) Write the area of the corral as a function of x. (b) Find the values of x and y that will produce the maximum area.

In this problem, we maximize the area of two adjacent rectangular corrals with 200 feet of fencing. Using quadratic functions and calculus, we derive the area function and find the optimal dimensions for maximum area. Learn how to apply the quadratic formula and derivatives to solve this optimization problem.

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