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Let's break down how we derive \( L = 0.25L \) in the context of the problem.

### Step 1: Understanding the Perimeter Formula
The perimeter \( P \) of a rectangle is given by:
\[
P = 2L + 2W
\]
where \( L \) is the length and \( W \) is the width of the rectangle.

### Step 2: Given Conditions
We are given that the perimeter is 120 meters, so:
\[
2L + 2W = 120
\]
Dividing the entire equation by 2:
\[
L + W = 60
\]

### Step 3: Maximizing the Area
To maximize the area of the rectangle, we can apply the **optimization technique** or observe the relationship geometrically.

- The area \( A \) of the rectangle is given by:
\[
A = L \times W
\]
From the equation \( L + W = 60 \), we solve for \( W \):
\[
W = 60 - L
\]
Thus, the area in terms of \( L \) is:
\[
A = L \times (60 - L) = 60L - L^2
\]
This is a quadratic function, and the maximum area occurs at the vertex of this parabola. The vertex for a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \).

Here, \( a = -1 \) and \( b = 60 \), so the value of \( L \) that maximizes the area is:
\[
L = \frac{-60}{2(-1)} = 30
\]
Thus, the length \( L \) and width \( W \) are both 30 meters when the area is maximized.

### Step 4: Generalizing for Any Perimeter \( L \)
If the perimeter is represented as \( P = L \) units, then the equation becomes:
\[
L + W = \frac{L}{2}
\]
For the rectangle to have the maximum area, we need \( L = W \), which gives us:
\[
2L = \frac{L}{2}
\]
So, \( L = W = 0.25L \).

### Conclusion
Thus, the value \( L = 0.25L \) arises when the perimeter is divided into four equal parts, implying that the rectangle is a square, which yields the maximum area for a given perimeter.

Would you like more details on any step? Here are five follow-up questions:

1. Can you explain why a square gives the maximum area for a given perimeter?
2. How do we derive the vertex of a quadratic function to find the maximum area?
3. What is the geometric significance of having \( L = W \)?
4. Can this method be applied to other shapes for maximizing area?
5. How would the problem change if the perimeter was different?

**Tip:** When optimizing geometric problems, quadratic functions often appear, and their maximum or minimum values are found at the vertex!

A rectangle has a perimeter of 120 meters. What length and width yield the maximum area? What is the result when the perimeter is L units?

Learn how to maximize the area of a rectangle with a perimeter of 120 meters using quadratic equations. Understand why the maximum area occurs when the rectangle is a square. Explore optimization techniques, relevant formulas, and concepts suitable for high school students.

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