Question 1185698
Here's how to solve this optimization problem:

**Understanding the Problem**

We have a fixed perimeter (200m of fencing) and want to maximize the area of a rectangle.

**a. Possible Dimensions:**

Let *l* be the length and *w* be the width of the rectangle.  The perimeter is given by:

2*l + 2*w = 200

We can simplify this to:

l + w = 100

Since length and width must be positive, the possible dimensions are any values of *l* and *w* that satisfy this equation, where 0 < l < 100 and 0 < w < 100.  It's impossible to list *all* possible dimensions because they are continuous, not discrete, but we can list some examples:

*   l = 10, w = 90
*   l = 20, w = 80
*   l = 30, w = 70
*   l = 40, w = 60
*   l = 50, w = 50
*   l = 60, w = 40
*   l = 70, w = 30
*   l = 80, w = 20
*   l = 90, w = 10

**b. Table of Values:**

It's helpful to organize the dimensions and areas in a table. Here are some examples:

| Length (l) | Width (w) | Area (A) |
|---|---|---|
| 10 | 90 | 900 |
| 20 | 80 | 1600 |
| 30 | 70 | 2100 |
| 40 | 60 | 2400 |
| 50 | 50 | 2500 |
| 60 | 40 | 2400 |
| 70 | 30 | 2100 |
| 80 | 20 | 1600 |
| 90 | 10 | 900 |

**c. Computing the Area:**

The area of a rectangle is given by:

A = l * w

We've already calculated the areas in the table above.

**d. Maximum Area:**

From the table, we can see that the maximum area is 2500 m².

**e. Dimensions of Maximum Area:**

The dimensions that give the maximum area are l = 50m and w = 50m.  This means the rectangle with the maximum area is a square.

**General Solution (using calculus):**

While the table gives the answer, calculus can be used for a more general approach.

1.  Express *w* in terms of *l*:  w = 100 - l

2.  Area formula: A = l(100 - l) = 100l - l²

3.  To find the maximum area, take the derivative of A with respect to *l* and set it equal to zero:

    dA/dl = 100 - 2l = 0
    2l = 100
    l = 50

4.  Since w = 100 - l, then w = 100 - 50 = 50.

This confirms that the maximum area occurs when l = 50 and w = 50.