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To maximize the area, let’s break down the problem:

1. **Define Variables:**
   - Let \( x \) be the width of the entire rectangular field (the shorter side).
   - Let \( y \) be the length of the field (the longer side parallel to the dividing fence).

2. **Fence Constraints:**
   - The field has 3 sections of width \( x \): the two vertical sides and one internal dividing fence.
   - The field has 2 sections of length \( y \): the top and bottom.
   - Since the total fencing is 100 meters, we can write the constraint as:
     \[
     3x + 2y = 100
     \]

3. **Area of the Field:**
   - The area \( A \) of the rectangular field is given by:
     \[
     A = x \cdot y
     \]

4. **Express \( y \) in Terms of \( x \):**
   - From the fence constraint equation \( 3x + 2y = 100 \), solve for \( y \):
     \[
     y = \frac{100 - 3x}{2}
     \]

5. **Substitute \( y \) into the Area Formula:**
   - Substitute \( y = \frac{100 - 3x}{2} \) into \( A = x \cdot y \):
     \[
     A = x \cdot \frac{100 - 3x}{2}
     \]
     Simplifying, we get:
     \[
     A = \frac{100x - 3x^2}{2}
     \]
     \[
     A = 50x - \frac{3x^2}{2}
     \]

6. **Maximize the Area \( A \):**
   - To find the maximum area, take the derivative of \( A \) with respect to \( x \) and set it to zero:
     \[
     \frac{dA}{dx} = 50 - 3x = 0
     \]
     Solving for \( x \):
     \[
     x = \frac{50}{3}
     \]

7. **Find \( y \) Using \( x = \frac{50}{3} \):**
   - Substitute \( x = \frac{50}{3} \) back into the equation for \( y \):
     \[
     y = \frac{100 - 3\left(\frac{50}{3}\right)}{2} = \frac{100 - 50}{2} = 25
     \]

8. **Calculate the Maximum Area:**
   - Substitute \( x = \frac{50}{3} \) and \( y = 25 \) into the area formula \( A = x \cdot y \):
     \[
     A = \frac{50}{3} \times 25 = \frac{1250}{3} \approx 416.67 \text{ m}^2
     \]

### Final Answer:
\[
A = \frac{1250}{3} \, \text{m}^2
\]

Would you like further clarification on any step?

---

1. How do we solve problems involving optimization and constraints?
2. Why do we take derivatives to find maximum or minimum values?
3. How does fencing length affect the shape and area of enclosures?
4. What happens if we add more sections within the field?
5. How would the answer change with different total fence lengths?

**Tip:** In optimization problems, always start by expressing the area (or volume) in terms of a single variable using constraints, then maximize or minimize it using calculus.

A gardener with 100 m of available fencing wishes to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides. What is the largest area that can be enclosed?

Explore how to maximize the area of a rectangular field divided into two plots with 100 meters of fencing. This optimization problem uses algebra and calculus to find the largest possible enclosed area, making it suitable for high school students studying calculus and optimization techniques.

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