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We can solve this problem by maximizing the area of each enclosure, given that the total length of the fence is 1,000 feet and we are dealing with two identical rectangular enclosures.

Let:
- \( w \) be the width (the shorter side) of one enclosure.
- \( l \) be the length (the longer side) of one enclosure.

### Step 1: Perimeter equation
The total perimeter includes the outer perimeter and the fence between the two enclosures. The total fencing used is \( 1,000 \) feet, so the perimeter equation is:
\[
2l + 3w = 1000
\]
This accounts for two lengths and three widths (since the middle dividing line adds one extra width).

### Step 2: Solve for \( l \) in terms of \( w \)
From the perimeter equation:
\[
2l = 1000 - 3w
\]
\[
l = \frac{1000 - 3w}{2}
\]

### Step 3: Area equation
The area of each enclosure is \( A = l \times w \). Substituting the expression for \( l \):
\[
A = w \times \frac{1000 - 3w}{2}
\]
\[
A = \frac{w(1000 - 3w)}{2}
\]
\[
A = \frac{1000w - 3w^2}{2}
\]
This is the function we need to maximize.

### Step 4: Maximize the area
To maximize the area, take the derivative of \( A \) with respect to \( w \), and set it equal to zero:
\[
\frac{dA}{dw} = \frac{1000 - 6w}{2} = 0
\]
Solving for \( w \):
\[
1000 - 6w = 0
\]
\[
w = \frac{1000}{6} = 166.67 \text{ feet}
\]

### Step 5: Find the corresponding \( l \)
Substitute \( w = 166.67 \) into the expression for \( l \):
\[
l = \frac{1000 - 3(166.67)}{2} = \frac{1000 - 500}{2} = \frac{500}{2} = 250 \text{ feet}
\]

### Step 6: Calculate the area of each enclosure
The area of each enclosure is:
\[
A = 166.67 \times 250 = 41667.5 \text{ square feet}
\]

### Final Answer:
- Width (shortest side) \( w = 166.67 \) feet
- Length (longest side) \( l = 250 \) feet
- Area of each enclosure: \( 41667.5 \) square feet

Would you like more details on any part of this solution?

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Here are 5 related questions:
1. How would the solution change if there was no fence between the enclosures?
2. What if the farmer had only 800 feet of fencing?
3. Can you solve this problem using Lagrange multipliers?
4. How does the shape of the enclosures (square vs. rectangle) affect the total area?
5. What would happen if the farmer wanted three enclosures instead of two?

**Tip:** When dealing with optimization problems, always check the boundary conditions to ensure you've found the maximum or minimum.

A farmer has 1,000 feet of fence and wishes to build two identical rectangular enclosures. What should be the dimensions of each enclosure if the total area is to be a maximum? What is the area of each enclosure?

This problem explores how to maximize the area of two identical rectangular enclosures using 1,000 feet of fencing. It involves algebra, optimization, and calculus concepts to determine the optimal dimensions and area of each enclosure. Suitable for students in Grades 10-12.

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