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!