A piece of wire 26 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Round your answers to two decimal places.)
(a)
How much wire (in meters) should be used for the square in order to maximize the total area?
 
 m
(b)
How much wire (in meters) should be used for the square in order to minimize the total area?

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!

A piece of wire 26 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (a) How much wire (in meters) should be used for the square in order to maximize the total area? (b) How much wire (in meters) should be used for the square in order to minimize the total area?

Explore how to optimize the area of a square and an equilateral triangle formed from a 26 m wire. Learn the calculations for maximizing and minimizing the area, including results of approximately 14.72 m for maximum area and conditions for minimum area.

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