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### Let’s solve the problem step by step:

#### Part (a): **Find a function for the total area \( A(x) \) in terms of \( x \):**
1. The wire has a total length of \( 48 \) cm. It is cut into two pieces, one of which forms a square with side length \( x \). The perimeter of this square is \( 4x \), so the length of this wire piece is \( 4x \).
2. The remaining length of the wire is \( 48 - 4x \). This piece forms another square. Let the side length of this square be \( y \). Thus:
   \[
   4y = 48 - 4x \quad \Rightarrow \quad y = 12 - x.
   \]
3. The areas of the squares are:
   - Area of the first square: \( x^2 \),
   - Area of the second square: \( y^2 = (12 - x)^2 \).
4. The total area \( A(x) \) is:
   \[
   A(x) = x^2 + (12 - x)^2.
   \]

#### Simplify \( A(x) \):
\[
A(x) = x^2 + (12 - x)^2 = x^2 + (144 - 24x + x^2) = 2x^2 - 24x + 144.
\]

Thus, the function is:
\[
A(x) = 2x^2 - 24x + 144.
\]

---

#### Part (b): **Find the side length \( x \) that minimizes the total area \( A(x) \):**

To find the value of \( x \) that minimizes \( A(x) \), take the derivative of \( A(x) \) and set it to zero:
\[
A'(x) = 4x - 24.
\]

Set \( A'(x) = 0 \):
\[
4x - 24 = 0 \quad \Rightarrow \quad x = 6.
\]

Thus, the side length \( x \) that minimizes the total area is:
\[
x = 6 \, \text{cm}.
\]

---

#### Part (c): **Find the minimum total area:**

Substitute \( x = 6 \) into the formula for \( A(x) \):
\[
A(6) = 2(6^2) - 24(6) + 144.
\]

Simplify:
\[
A(6) = 2(36) - 144 + 144 = 72.
\]

Thus, the minimum area enclosed by the two squares is:
\[
\text{Minimum Area} = 72 \, \text{square centimeters}.
\]

---

### Final Results:
1. \( A(x) = 2x^2 - 24x + 144 \),
2. Side length \( x = 6 \, \text{cm} \),
3. Minimum area = \( 72 \, \text{square centimeters} \).

---

Would you like more detailed explanations or examples?  
Here are 5 questions to deepen understanding:
1. Why is the derivative used to find the minimum value?
2. What happens if \( x > 12 \) or \( x < 0 \)?
3. Can you solve the problem graphically to confirm the solution?
4. How does the total wire length affect the area function?
5. What are the implications of minimizing area in real-world applications?

**Tip:** Always verify your solutions by substituting values into the original equations!

A wire that is 48 centimeters long is cut into two pieces, and each piece is bent to form two squares. Given that the side length of one square is x cm, find the function for the total area of both squares, determine the side length x that minimizes this area, and calculate the minimum total area.

Solve an optimization problem involving a 48 cm wire cut into two pieces to form two squares. Learn how to find the function for the total area in terms of x, determine the side length that minimizes this area, and calculate the minimum possible area. Ideal for high school students studying quadratic functions and optimization.

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