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This problem is a classic optimization question where we need to maximize the area enclosed by a given amount of fencing.

Let's break it down:

### Key Information:
- The rancher has **2100 feet of fencing**.
- There are **6 corrals** arranged in two rows of three, as shown in the diagram.
- The total dimensions of the enclosure are \( x \times y \) feet, where \( x \) is the total length and \( y \) is the total width.

The fencing is used for:
1. **Two horizontal fences along the total width** (\( y \)).
2. **Four vertical fences along the total length** (\( x \)) that separate the corrals.

The constraints of the problem arise from the total available fencing and the need to maximize the enclosed area.

### Step 1: Express the total fencing constraint
From the layout, the total fencing used consists of:
- **2 long fences** along the width \( y \).
- **4 shorter fences** along the length \( x \).
  
Additionally, there are **2 extra fences** (running vertically) within the enclosure to separate the three corrals.

Thus, the total amount of fencing is given by the equation:
\[
2y + 4x + 2x = 2100
\]
Simplifying this equation:
\[
2y + 6x = 2100
\]
\[
y + 3x = 1050 \quad \text{(solving for y)} \quad y = 1050 - 3x
\]

### Step 2: Express the area
The total area of the enclosure is:
\[
A = x \times y
\]
Substituting \( y = 1050 - 3x \) into the area formula:
\[
A(x) = x \times (1050 - 3x)
\]
\[
A(x) = 1050x - 3x^2
\]

### Step 3: Maximize the area
To find the value of \( x \) that maximizes the area, we take the derivative of \( A(x) \) with respect to \( x \) and set it equal to zero:
\[
\frac{dA}{dx} = 1050 - 6x
\]
Setting this equal to zero to find the critical points:
\[
1050 - 6x = 0
\]
\[
6x = 1050
\]
\[
x = 175
\]

### Step 4: Find the corresponding \( y \)
Substitute \( x = 175 \) into the equation for \( y \):
\[
y = 1050 - 3(175)
\]
\[
y = 1050 - 525
\]
\[
y = 525
\]

### Step 5: Calculate the maximum area
Now that we have \( x = 175 \) feet and \( y = 525 \) feet, the maximum area is:
\[
A = 175 \times 525 = 91875 \text{ square feet}
\]

### Final Answer:
- The dimensions that maximize the area are \( x = 175 \) feet and \( y = 525 \) feet.
- The maximum enclosed area is **91,875 square feet**.

---

Do you need further explanation on any step, or would you like more details? Here are some related questions:

1. How do you solve optimization problems using calculus?
2. How does changing the layout affect the use of the fencing?
3. Can this method be generalized for more than six corrals?
4. What happens if there is less fencing available for the same number of corrals?
5. How do we verify that this critical point gives a maximum area?

**Tip:** For optimization problems, always check the critical points by using the first or second derivative test to confirm whether it is a maximum or minimum.

A rancher has 2100 feet of fencing to construct six corrals. Find the dimensions that maximize the enclosed area. What is the maximum area?

Learn how to maximize the area of six corrals using 2100 feet of fencing. This problem involves optimization techniques, algebraic expressions, and calculus derivatives. Suitable for students in Grades 10-12.

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