To maximize the area of a rectangular space with a fixed perimeter, we can apply the concept that a square (where all sides are equal) gives the maximum area for a given perimeter.

Here's the step-by-step solution:

1. Define the Variables: Let the length of the rectangle be $L$ and the width be $W$.

2. Set Up the Perimeter Equation: Since the perimeter is fixed at 100 meters, we have: $2L + 2W = 100$ Simplifying, we find: $L + W = 50$ So, $W = 50 - L$.

3. Write the Area Formula: The area $A$ of the rectangle is given by: $A = L \times W$ Substituting $W = 50 - L$: $A = L \times (50 - L) = 50L - L^2$

4. Maximize the Area: To find the value of $L$ that maximizes the area, take the derivative of $A$ with respect to $L$ and set it to zero: $\frac{dA}{dL} = 50 - 2L$ Setting $\frac{dA}{dL} = 0$: $50 - 2L = 0$ Solving for $L$: $L = 25$ Since $L + W = 50$, it follows that $W = 25$.

5. Conclusion: The dimensions that give the largest area are $L = 25$ meters and $W = 25$ meters, meaning the rectangle is actually a square.

6. Calculate the Maximum Area: $A = 25 \times 25 = 625 \text{ square meters}$

Thus, the largest possible area is 625 square meters, achieved when both the length and width are 25 meters.

Would you like more details on the calculus used, or do you have any other questions?

Related Questions

1. How would the dimensions change if the perimeter were different, say 200 meters?
2. What if the area needed to be minimized instead of maximized?
3. How does this approach generalize to other shapes with a fixed perimeter?
4. What is the maximum area for a rectangular field if only three sides are fenced, with one side along a river?
5. How does changing one dimension affect the area while keeping the perimeter fixed?

Tip

Remember, for any fixed perimeter, a square always provides the maximum area among all rectangles!