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Let's break down the problem to solve it step by step.

### Given:
- The total fencing available is **414 meters**.
- The field is rectangular, with **two sides of width \(w\)** and **two ends of length \(L\)**.
- Additionally, there are **three cross fences** dividing the field into four plots, each of width \(w\).

### Fencing Equation:
The total amount of fencing used includes:
- The **two lengths** of the field: \(2L\).
- The **two widths** of the field: \(2w\).
- The **three cross fences**, which together add up to \(3w\) since each cross fence is also of width \(w\).

Thus, the total fencing constraint can be expressed as:
\[
2L + 5w = 414
\]

### Area of the Field:
The area \(A\) of the field is given by the formula:
\[
A = L \times w
\]

### Objective:
We need to **maximize the area** while adhering to the fencing constraint. To do this, we first solve for \(L\) in terms of \(w\) from the fencing equation.

\[
2L = 414 - 5w
\]
\[
L = \frac{414 - 5w}{2}
\]

Now, substitute this expression for \(L\) into the area formula:
\[
A = w \times \frac{414 - 5w}{2}
\]
\[
A = \frac{w(414 - 5w)}{2}
\]
\[
A = \frac{414w - 5w^2}{2}
\]
\[
A = 207w - \frac{5}{2}w^2
\]

### Maximizing the Area:
To maximize the area, we take the derivative of \(A\) with respect to \(w\), and set it equal to zero to find the critical points.
\[
\frac{dA}{dw} = 207 - 5w = 0
\]
\[
5w = 207
\]
\[
w = \frac{207}{5} = 41.4
\]

### Finding the Corresponding Length \(L\):
Now, substitute \(w = 41.4\) back into the equation for \(L\):
\[
L = \frac{414 - 5(41.4)}{2}
\]
\[
L = \frac{414 - 207}{2}
\]
\[
L = \frac{207}{2} = 103.5
\]

Thus, the length of the field is **103.5 meters**.

### Final Answer:
The length of the field that maximizes the area, with 414 meters of fencing, is **103.5 meters**.

Would you like further details or have any questions?

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Here are five questions to explore this problem further:
1. How do you determine when a function reaches its maximum or minimum value?
2. Why do we take the derivative to find the maximum area?
3. What would happen if more or fewer cross fences were added?
4. How does the area change if we increase the total amount of fencing?
5. Could we use another optimization method to solve this problem?

**Tip:** In optimization problems, always check if you're maximizing or minimizing the function, and apply constraints accordingly.

A rectangular field is to be fenced along two sides and two ends. It is also required to have three additional cross fences from side to side to form four plots. Find the length of the field to the nearest tenth that will give the entire field a maximum area if 414 meters of fencing is available.

Learn how to maximize the area of a rectangular field, given 414 meters of fencing and specific constraints. This problem involves quadratic equations and optimization techniques, suitable for students in Grades 11-12.

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