The minimum and maximum possible values of `x/y`, where `2 le x le 8` and `16 le y le 32`, respectively are :

Options

A. `1/8, 1/4`
B. `1/16, 1/2`
C. 2, 16
D. not unique

Answer

To find the minimum and maximum possible values of \( \frac{x}{y} \) given the constraints \( 2 \leq x \leq 8 \) and \( 16 \leq y \leq 32 \), we can follow these steps:

### Step 1: Identify the maximum value of \( \frac{x}{y} \)
To maximize \( \frac{x}{y} \), we need to maximize \( x \) and minimize \( y \).
- The maximum value of \( x \) is \( 8 \).
- The minimum value of \( y \) is \( 16 \).

Now, we can calculate the maximum value:
\[ \text{Maximum value of } \frac{x}{y} = \frac{8}{16} = \frac{1}{2} \]

### Step 2: Identify the minimum value of \( \frac{x}{y} \)
To minimize \( \frac{x}{y} \), we need to minimize \( x \) and maximize \( y \).
- The minimum value of \( x \) is \( 2 \).
- The maximum value of \( y \) is \( 32 \).

Now, we can calculate the minimum value:
\[ \text{Minimum value of } \frac{x}{y} = \frac{2}{32} = \frac{1}{16} \]

### Conclusion
The minimum and maximum possible values of \( \frac{x}{y} \) are:
- Minimum: \( \frac{1}{16} \)
- Maximum: \( \frac{1}{2} \)

### Final Answer
\[ \text{Minimum: } \frac{1}{16}, \text{ Maximum: } \frac{1}{2} \]