The problem asks to find the dimensions (length and width) of a rectangle that has a perimeter of 64 meters and a maximum area.

Steps:

1. Perimeter Formula: The perimeter $P$ of a rectangle is given by: $P = 2L + 2W$ where $L$ is the length and $W$ is the width.

   Given that the perimeter is 64 meters: $64 = 2L + 2W \quad \text{or} \quad L + W = 32$

2. Maximizing Area: The area $A$ of a rectangle is given by: $A = L \times W$ To maximize the area, we use the relationship $L + W = 32$. Solve for one variable in terms of the other. Let’s solve for $L$: $L = 32 - W$

   Substitute this into the area equation: $A = (32 - W) \times W = 32W - W^2$ This is a quadratic equation in terms of $W$, and it represents a parabola that opens downward. The maximum area occurs at the vertex of the parabola, which is found by the formula: $W = \frac{-b}{2a}$ where $a = -1$ and $b = 32$. So: $W = \frac{-32}{2(-1)} = 16$

3. Finding $L$: Since $L + W = 32$ and $W = 16$: $L = 32 - 16 = 16$

Conclusion:

The length and width that give a maximum area for the rectangle are both 16 meters, meaning the rectangle is actually a square.

Would you like further details or have any other questions?

Related Questions:

1. How do you find the perimeter of a rectangle given the area?
2. What are the steps for solving optimization problems in calculus?
3. Why is the area maximized when the rectangle is a square?
4. How can you verify the maximum area using derivatives?
5. What is the difference between maximizing and minimizing functions in geometry?

Tip:

For rectangles, a square always gives the maximum area when the perimeter is fixed!