Find the least number $A$ such that for any two squares of combined area $1$, a rectangle of area $A$ exists such that the two squares can be packed in the rectangle (without the interiors of the squares overlapping) . You may assume the sides of the squares will be parallel to the sides of the rectangle.
1. **Initial Consideration**:
   - We need to find the least number \( A \) such that any two squares with a combined area of 1 can be packed into a rectangle of area \( A \).
   - Each side of the rectangle must have a length of at least 1 because we can take a \( 1 \times 1 \) square and an empty square.

2. **Case Analysis**:
   - Consider the case when we have two squares of area \( \frac{1}{2} \). The side length of each square is \( \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \).
   - Since the squares are parallel to the sides of the rectangle, the larger side of the rectangle must have a length of at least \( 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \).

3. **Area Calculation**:
   - The area of the rectangle must be at least \( 1 \times \sqrt{2} = \sqrt{2} \). Thus, \( A \geq \sqrt{2} \).

4. **Verification**:
   - We now claim that a \( 1 \times \sqrt{2} \) rectangle works.
   - Assume we have an \( a \times a \) square and a \( b \times b \) square with \( a^2 + b^2 = 1 \).
   - We need to show that these squares can fit into a \( 1 \times \sqrt{2} \) rectangle.

5. **Inequality Proof**:
   - We need to prove that \( a + b \leq \sqrt{2} \) given that \( a^2 + b^2 = 1 \).
   - By the Cauchy-Schwarz inequality, we have:
     \[
     (a + b)^2 \leq (1^2 + 1^2)(a^2 + b^2) = 2 \cdot 1 = 2
     \]
   - Taking the square root of both sides, we get:
     \[
     a + b \leq \sqrt{2}
     \]
   - Therefore, the two squares can indeed fit into a \( 1 \times \sqrt{2} \) rectangle.

Conclusion:
- The least number \( A \) such that any two squares with a combined area of 1 can be packed into a rectangle of area \( A \) is \( \sqrt{2} \).

The final answer is \( \boxed{\sqrt{2}} \).