Problem Breakdown:

• The problem gives that the perimeter of the rectangle is 20 meters.
• We need to express the area $A(L)$ of the rectangle as a function of the length $L$ of one of its sides.

Step 1: Expressing the Width in Terms of Length $L$

For a rectangle, the perimeter $P$ is given by:

$P = 2L + 2W$

where $L$ is the length and $W$ is the width. Given $P = 20$ meters:

$2L + 2W = 20$

Dividing the entire equation by 2:

$L + W = 10$

This implies:

$W = 10 - L$

Step 2: Expressing the Area $A(L)$

The area $A$ of the rectangle is given by:

$A = L \times W$

Substituting the expression for $W$:

$A(L) = L \times (10 - L)$

So, the area function becomes:

$A(L) = 10L - L^2$

Thus:

$A(L) = 10L - L^2 \text{ m}^2$

Step 3: Stating the Domain of $A(L)$

For the area $A(L)$ to be positive, both $L$ and $W = 10 - L$ must be positive. This implies:

$0 < L < 10$

So the domain of $A(L)$ is:

$\text{Domain: } (0, 10) \text{ meters}$

Summary:

1. Area function: $A(L) = 10L - L^2$ m$^2$
2. Domain: $(0, 10)$ meters

Would you like further clarification on how the domain or the area function was determined?

Here are five related questions:

1. How does the perimeter constraint affect the shape of the rectangle?
2. What is the maximum possible area for this rectangle, and how is it calculated?
3. What happens to the area function if the perimeter changes?
4. Can the function $A(L) = 10L - L^2$ be used to find the area for any value of $L$?
5. How do you determine the dimensions of the rectangle that give the maximum area?

Tip: When working with quadratic functions like $A(L) = 10L - L^2$, remember that the vertex form can help you easily find the maximum or minimum values.