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.