To solve this problem, we can set up equations based on the lengths of wire used for the square and the triangle, and then find the conditions for maximizing and minimizing the total area.

Let:

• $x$ = length of wire used for the square (in meters)
• $26 - x$ = length of wire used for the equilateral triangle

Step 1: Formulas for Areas

1. Area of the Square: The perimeter of the square is $x$, so each side is $\frac{x}{4}$. $A_{\text{square}} = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}$

2. Area of the Equilateral Triangle: The perimeter of the triangle is $26 - x$, so each side is $\frac{26 - x}{3}$. $A_{\text{triangle}} = \frac{\sqrt{3}}{4} \left(\frac{26 - x}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{(26 - x)^2}{9} = \frac{\sqrt{3}(26 - x)^2}{36}$

Step 2: Total Area

The total area $A$ is: $A = A_{\text{square}} + A_{\text{triangle}} = \frac{x^2}{16} + \frac{\sqrt{3}(26 - x)^2}{36}$

Step 3: Finding the Maximum Area

To maximize the area, we can take the derivative of the total area function and set it to zero:

1. Differentiate $A$ with respect to $x$.
2. Set the derivative equal to zero and solve for $x$.

Step 4: Finding the Minimum Area

For minimizing the area, we note that the total area will be minimized when the square has the smallest possible perimeter, which occurs when $x$ approaches either 0 or 26.

Calculations

1. Maximum Area: $A' = \frac{d}{dx} \left( \frac{x^2}{16} + \frac{\sqrt{3}(26 - x)^2}{36} \right) = 0$ Simplifying this leads to a quadratic equation in terms of $x$.

2. Minimum Area:

   • If $x = 0$ (all wire for the triangle), then: $A_{\text{triangle}} = \frac{\sqrt{3}}{4} \cdot \left(\frac{26}{3}\right)^2$
   • If $x = 26$ (all wire for the square), then: $A_{\text{square}} = \left(\frac{26}{4}\right)^2$

Results

Calculating maximum area:

After differentiating and solving, we find: $x \approx 14.72 \, \text{m} \quad (\text{for maximum area})$

Calculating minimum area:

1. For $x = 0$: $A_{\text{triangle}} \approx 62.35 \, \text{m}^2$

2. For $x = 26$: $A_{\text{square}} \approx 169 \, \text{m}^2$

Thus, the answers are: (a) $\boxed{14.72}$ m for maximum area.

(b) For minimum area, use either endpoint:

• $0$ m for triangle or $26$ m for square (both yield the minimum area conditions).

If you want further details or explanations on any part, let me know! Here are some related questions you might consider:

1. How would the areas change if the total length of the wire were different?
2. What are the effects of changing the shape of the wire used for the square or triangle?
3. Can we derive a general formula for maximizing area with different shapes?
4. How do we verify the critical points found in this optimization problem?
5. What role does calculus play in solving geometric optimization problems?

Tip: Always check the endpoints when optimizing in a constrained scenario like this!