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The problem asks to find the length and width of a rectangle that has a perimeter of 68 meters and a maximum area.

### Key details:
- **Perimeter** \( P = 68 \) meters
- **Maximum Area** condition

### Formula for the perimeter of a rectangle:
\[
P = 2L + 2W
\]
where \( L \) is the length and \( W \) is the width. Given that \( P = 68 \), we have:
\[
68 = 2L + 2W \quad \Rightarrow \quad L + W = 34
\]

### Maximizing the area:
The area \( A \) of a rectangle is given by:
\[
A = L \times W
\]
To maximize the area under the constraint \( L + W = 34 \), the rectangle should be a **square**. This is a property of geometry: a square encloses the maximum area for a given perimeter.

Thus, the length \( L \) and width \( W \) must both be equal:
\[
L = W
\]
Substituting into \( L + W = 34 \), we get:
\[
2L = 34 \quad \Rightarrow \quad L = 17
\]
Thus, the length and width are both 17 meters.

### Final answer:
The length and width are \( \mathbf{17, 17} \) meters.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. How would the area change if the rectangle were not a square?
2. Can you derive the formula for the perimeter of a rectangle?
3. What is the maximum possible area for the rectangle in this case?
4. How does the shape of a rectangle affect the area for a fixed perimeter?
5. What if the perimeter was 80 meters instead of 68?

**Tip:** For any rectangle with a fixed perimeter, making it a square gives the maximum possible area.

Find the length and width (in meters) of a rectangle that has a given perimeter and a maximum area. Perimeter: 68 meters.

Solve for the dimensions of a rectangle with a perimeter of 68 meters that maximizes its area. This problem uses key geometric formulas and involves finding when a rectangle becomes a square to achieve maximum area.

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