The area of the region on place bounded by max `(|x|,|y|) le 1/2` is

Options

A. `1//2 + ln2`
B. `3 + ln 2`
C. `31//4`
D. `1+2 ln 2`

Answer

To find the area of the region bounded by the condition \( \max(|x|, |y|) \leq \frac{1}{2} \), we can follow these steps:

### Step 1: Understand the Condition
The expression \( \max(|x|, |y|) \leq \frac{1}{2} \) means that both \( |x| \) and \( |y| \) must be less than or equal to \( \frac{1}{2} \). This implies:
- \( |x| \leq \frac{1}{2} \)
- \( |y| \leq \frac{1}{2} \)

### Step 2: Identify the Boundaries
From the inequalities:
- \( |x| \leq \frac{1}{2} \) gives us the lines \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).
- \( |y| \leq \frac{1}{2} \) gives us the lines \( y = \frac{1}{2} \) and \( y = -\frac{1}{2} \).

### Step 3: Draw the Region
The lines \( x = \frac{1}{2} \), \( x = -\frac{1}{2} \), \( y = \frac{1}{2} \), and \( y = -\frac{1}{2} \) form a square in the coordinate plane. The vertices of this square are:
- \( \left(\frac{1}{2}, \frac{1}{2}\right) \)
- \( \left(-\frac{1}{2}, \frac{1}{2}\right) \)
- \( \left(-\frac{1}{2}, -\frac{1}{2}\right) \)
- \( \left(\frac{1}{2}, -\frac{1}{2}\right) \)

### Step 4: Calculate the Area
The side length of the square is the distance from \( -\frac{1}{2} \) to \( \frac{1}{2} \), which is:
\[ \text{Side length} = \frac{1}{2} - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1 \]

The area \( A \) of a square is given by the formula:
\[ A = \text{side}^2 \]
Substituting the side length:
\[ A = 1^2 = 1 \]

### Conclusion
The area of the region bounded by \( \max(|x|, |y|) \leq \frac{1}{2} \) is \( 1 \) square unit.

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