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To solve this problem, let's denote:

- The total length of the wire as \( L = 7 \) meters.
- Let \( x \) be the length of the wire used for the square.
- The remaining length, \( 7 - x \), will be used for the circle.

### Step-by-Step Solution

1. **Square Formation**:
   - If \( x \) meters of wire is used to form a square, then each side of the square is \( \frac{x}{4} \).
   - The area of the square, \( A_{\text{square}} \), is:
     \[
     A_{\text{square}} = \left( \frac{x}{4} \right)^2 = \frac{x^2}{16}
     \]

2. **Circle Formation**:
   - If \( 7 - x \) meters of wire is used to form a circle, this length represents the circumference of the circle:
     \[
     2\pi r = 7 - x
     \]
     Solving for \( r \), the radius of the circle:
     \[
     r = \frac{7 - x}{2\pi}
     \]
   - The area of the circle, \( A_{\text{circle}} \), is:
     \[
     A_{\text{circle}} = \pi r^2 = \pi \left( \frac{7 - x}{2\pi} \right)^2 = \frac{(7 - x)^2}{4\pi}
     \]

3. **Total Area**:
   - The total area \( A_{\text{total}} \) of both figures is the sum of the square's area and the circle's area:
     \[
     A_{\text{total}} = \frac{x^2}{16} + \frac{(7 - x)^2}{4\pi}
     \]

### Part (a): Maximizing the Total Area
To maximize the total area, we need to find the value of \( x \) that maximizes \( A_{\text{total}} \). 

### Part (b): Minimizing the Total Area
To minimize the total area, we need to find the value of \( x \) that minimizes \( A_{\text{total}} \).

### Solution Steps

1. Take the derivative of \( A_{\text{total}} \) with respect to \( x \).
2. Set the derivative equal to zero and solve for \( x \).
3. Check the second derivative or use endpoints (i.e., \( x = 0 \) or \( x = 7 \)) to confirm whether it’s a maximum or minimum.

I'll proceed with the calculations and provide the specific values of \( x \) for both maximum and minimum cases.The results for the total area \( A_{\text{total}} \) at various points are:

1. **For \( x = \frac{28}{\pi + 4} \approx 2.23 \) meters**:
   - The total area \( A_{\text{total}} \approx 1.72 \) square meters, which is the minimum.

2. **For \( x = 0 \) meters** (all wire used for the circle):
   - The total area \( A_{\text{total}} \approx 3.90 \) square meters, which is the maximum.

3. **For \( x = 7 \) meters** (all wire used for the square):
   - The total area \( A_{\text{total}} \approx 3.06 \) square meters.

### Final Answers:
- **(a) To maximize the sum of the areas**, the wire should be cut such that the square uses \( x = 0 \) meters (i.e., all wire is used for the circle), resulting in \( A_{\text{total}} \approx 3.90 \) square meters.
- **(b) To minimize the sum of the areas**, the wire should be cut such that the square uses \( x \approx 2.23 \) meters, resulting in \( A_{\text{total}} \approx 1.72 \) square meters.

Would you like a detailed breakdown of the derivative process? 

Here are some related questions:

1. How would the result change if the total wire length were different?
2. What if both shapes had to use equal wire lengths?
3. How does using other shapes (e.g., triangle instead of square) impact area maximization?
4. What happens if a restriction is added to use at least half the wire for each shape?
5. How would this approach change with a different wire thickness?

**Tip:** When dealing with optimization problems, always check both critical points and endpoints to ensure finding the true maximum or minimum values.

A 7-meter long iron wire is cut in two. One piece is to form a square and the other a circle.

Where should you cut the wire to maximize the sum of the areas from the two figures that have been produced?
Where should you cut the wire to minimize the sum of the areas from the two figures that have been produced?

Learn how to solve an optimization problem involving a 7-meter iron wire cut to form a square and a circle. Discover where to cut the wire to maximize or minimize the total area of the two shapes. This problem covers key calculus and geometry concepts and includes formulas for area calculation and optimization.

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