Area bounded by the min. `{|x|,|y|}=1` and the max. `{|x|,|y|}=2` is

Options

A. 4
B. 8
C. 16
D. 9

Explanation

To find the area bounded by the minimum of \(|x|\) and \(|y| = 1\) and the maximum of \(|x|\) and \(|y| = 2\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Conditions**:
- The condition \(\min\{|x|, |y|\} = 1\) means that either \(|x| = 1\) or \(|y| = 1\).
- The condition \(\max\{|x|, |y|\} = 2\) means that either \(|x| = 2\) or \(|y| = 2\).

2. **Identifying the Lines**:
- From \(|x| = 1\), we have the vertical lines \(x = 1\) and \(x = -1\).
- From \(|y| = 1\), we have the horizontal lines \(y = 1\) and \(y = -1\).
- From \(|x| = 2\), we have the vertical lines \(x = 2\) and \(x = -2\).
- From \(|y| = 2\), we have the horizontal lines \(y = 2\) and \(y = -2\).

3. **Graphing the Lines**:
- We can plot these lines on a Cartesian plane. The lines \(x = 1\), \(x = -1\), \(y = 1\), and \(y = -1\) will form a square with vertices at \((1, 1)\), \((1, -1)\), \((-1, 1)\), and \((-1, -1)\).
- The lines \(x = 2\), \(x = -2\), \(y = 2\), and \(y = -2\) will form a larger square with vertices at \((2, 2)\), \((2, -2)\), \((-2, 2)\), and \((-2, -2)\).

4. **Calculating the Areas**:
- The area of the smaller square (bounded by \(|x|, |y| = 1\)) is:
\[ \text{Area}_{\text{small}} = \text{side}^2 = 2 \times 2 = 4 \text{ square units} \]
- The area of the larger square (bounded by \(|x|, |y| = 2\)) is:
\[ \text{Area}_{\text{large}} = \text{side}^2 = 4 \times 4 = 16 \text{ square units} \]

5. **Finding the Area Between the Squares**:
- The area bounded between the two conditions is:
\[ \text{Area}_{\text{between}} = \text{Area}_{\text{large}} - \text{Area}_{\text{small}} = 16 - 4 = 12 \text{ square units} \]

### Final Answer:
The area bounded by the minimum \(\{|x|, |y|\} = 1\) and the maximum \(\{|x|, |y|\} = 2\) is \(12\) square units.