To solve this, we’ll need to examine the two options: one square enclosure with two widths and two lengths, and two separate square enclosures, using the given 200 meters of fencing.

Option 1: One Square Enclosure

If you create one square enclosure with a perimeter of 200 meters, the perimeter equation for a square is:

$P = 4 \times \text{side length}$

Setting $P = 200$, we get:

$200 = 4 \times \text{side length}$

Solving for the side length:

$\text{side length} = \frac{200}{4} = 50 \, \text{m}$

The area of a square is:

$A = \text{side length}^2 = 50^2 = 2500 \, \text{m}^2$

Option 2: Two Separate Square Enclosures

If you create two separate square enclosures, the total perimeter used is still 200 meters, but now the perimeter is split between the two squares. Let the side length of the first square be $x$, and the side length of the second square be $y$. The perimeter of each square is $4x$ and $4y$, and the total perimeter equation is:

$4x + 4y = 200$

Simplifying:

$x + y = 50$

Now, the area of the two squares is:

$A_1 = x^2 \quad \text{and} \quad A_2 = y^2$

The total area of both squares combined is:

$A_{\text{total}} = x^2 + y^2$

Since $y = 50 - x$, we substitute this into the area formula:

$A_{\text{total}} = x^2 + (50 - x)^2$

Expanding:

$A_{\text{total}} = x^2 + 2500 - 100x + x^2 = 2x^2 - 100x + 2500$

This is a quadratic function, and to minimize or maximize the area, we can complete the square or use the vertex formula. The vertex formula for a quadratic equation $ax^2 + bx + c$ gives the value of $x$ at the vertex as:

$x = \frac{-b}{2a}$

For our equation, $a = 2$, $b = -100$, and $c = 2500$, so:

$x = \frac{-(-100)}{2(2)} = \frac{100}{4} = 25$

Thus, the side length of the first square that minimizes the area is 25 meters, and the side length of the second square is $y = 50 - 25 = 25$ meters.

So, the minimum area occurs when both squares have side lengths of 25 meters, and the total area is:

$A_{\text{total}} = 25^2 + 25^2 = 625 + 625 = 1250 \, \text{m}^2$

Conclusion:

• The least area occurs when both squares have side lengths of 25 meters, with a total area of 1250 m².
• The most area occurs when there is only one square enclosure with side length 50 meters, giving an area of 2500 m².

Recap:

1. Most Area: 2500 m² (One square enclosure).
2. Least Area: 1250 m² (Two separate square enclosures, each 25 meters in side length).

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Would you like further clarification on the steps, or is there any other detail you'd like to explore? Here are a few more questions to expand on the topic:

1. What would happen if the perimeter was different, say 100 meters, for each option?
2. How would the area change if we used rectangular enclosures instead of squares?
3. What is the geometric principle behind minimizing or maximizing the area with a fixed perimeter?
4. How can calculus help in optimizing such enclosure problems?
5. What is the impact of changing the shape of the enclosure (circle, for example) on the area?

Tip: In optimization problems like these, using the properties of quadratics can help find the maximum or minimum values quickly. The vertex formula is a key tool for these situations!