To maximize the area of a rectangle with a given perimeter, the rectangle must be a square. Here's how we can solve for the length and width:

Step 1: Perimeter formula

The formula for the perimeter $P$ of a rectangle is: $P = 2L + 2W$ where $L$ is the length and $W$ is the width. Given the perimeter is 56 meters: $2L + 2W = 56$ Simplifying: $L + W = 28$ (Equation 1)

Step 2: Area formula

The area $A$ of a rectangle is given by: $A = L \times W$ We want to maximize this area while satisfying $L + W = 28$.

Step 3: Condition for maximum area

For a given perimeter, the area of a rectangle is maximized when the rectangle is a square. This means $L = W$.

Step 4: Solving for $L$ and $W$

Using $L = W$ in Equation (1): $L + L = 28$ $2L = 28$ $L = 14$ Since $L = W$, the width is also 14 meters.

Final answer:

The length and width are both 14 meters. Therefore, the dimensions are: $\boxed{14, 14}$

Would you like more details or have any further questions?

Related Questions:

1. How does the shape of the rectangle affect its area when the perimeter is fixed?
2. What is the formula to calculate the area of a rectangle?
3. How can we derive the formula for the perimeter of any polygon?
4. What if the rectangle’s length is twice its width? What would the area and dimensions be then?
5. What happens to the area when the perimeter of a shape increases?

Tip:

Maximizing the area for a given perimeter often leads to shapes that are symmetrical, such as squares in the case of rectangles!