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This problem is an optimization problem in which we need to find the dimensions that maximize the enclosed area, given a fixed amount of fencing. Let's walk through it step by step.

### Step 1: Understand the Layout
You mentioned there is a figure, but since I can't view it, I'll assume the common configuration of the corral system:

- We have **6 corrals** arranged in a rectangular grid (either side by side or in a combination of rows and columns).
- If the corrals are laid out side by side, they will share internal fences.
- Let:
  - \( L \) be the total length of the corrals.
  - \( W \) be the width of each corral.

Since there are 6 corrals, and they are likely divided by internal fencing, the total amount of fencing used includes the perimeter and the internal dividers.

### Step 2: Fencing Constraints
The total fencing available is 1000 feet, which is used to fence the outer boundary and the internal dividers.

- The fencing for the length includes 2 external lengths and 5 internal lengths (because of the internal dividers between the 6 corrals).
  - Fencing for the length: \( 2L + 5W \).
- The fencing for the width includes just 2 external widths.
  - Fencing for the width: \( 2W \).

Thus, the total fencing constraint is:

\[
2L + 5W + 2W = 1000
\]
Simplifying:
\[
2L + 7W = 1000 \quad \text{(Equation 1)}
\]

### Step 3: Maximize the Area
The area \( A \) of the 6 corrals is given by the product of the length and width:
\[
A = L \times W
\]

### Step 4: Solve for \( L \)
From Equation 1, solve for \( L \) in terms of \( W \):
\[
L = \frac{1000 - 7W}{2}
\]

### Step 5: Express the Area Function
Substitute this expression for \( L \) into the area formula:
\[
A(W) = \left( \frac{1000 - 7W}{2} \right) \times W
\]
\[
A(W) = \frac{1000W - 7W^2}{2}
\]
\[
A(W) = 500W - \frac{7}{2}W^2
\]

### Step 6: Maximize the Area
To find the value of \( W \) that maximizes the area, take the derivative of \( A(W) \) with respect to \( W \), and set it equal to zero.

\[
\frac{dA}{dW} = 500 - 7W = 0
\]
\[
7W = 500
\]
\[
W = \frac{500}{7} \approx 71.43 \text{ feet}
\]

### Step 7: Calculate \( L \)
Now substitute \( W = 71.43 \) feet back into the expression for \( L \):

\[
L = \frac{1000 - 7(71.43)}{2} = \frac{1000 - 500}{2} = \frac{500}{2} = 250 \text{ feet}
\]

### Step 8: Maximum Area
The maximum area is:

\[
A = L \times W = 250 \times 71.43 \approx 17857.14 \text{ square feet}
\]

### Final Answer:
- **Dimensions**: \( L = 250 \) feet, \( W = 71.43 \) feet
- **Maximum Area**: Approximately **17,857.14 square feet**.

Would you like any additional details or explanations?

### Further questions:
1. Can the shape of the corrals be adjusted for better efficiency?
2. How would the solution change if there were only 4 corrals?
3. How would the internal fencing affect the outcome if the corral arrangement changed?
4. Can you think of other methods to optimize the area?
5. How does the second derivative test confirm this is a maximum?

### Tip:
In optimization problems, always start by expressing the constraints and objective function, then use calculus to find the maximum or minimum.

A rancher has 1000 feet of fencing to construct six corrals. Find the dimensions that maximize the enclosed area. What is the maximum area?

Explore how to determine the dimensions that maximize the enclosed area for six corrals using 1000 feet of fencing. This solution involves optimization and calculus, providing a detailed breakdown of constraints and maximizing area.

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