Find the area enclosed by the curves
`max(|x+y|,|x-y|)=1`

Answer

To find the area enclosed by the curves given by the equation \( \max(|x+y|, |x-y|) = 1 \), we will break down the problem step by step.

### Step 1: Understand the Equation
The equation \( \max(|x+y|, |x-y|) = 1 \) means that either \( |x+y| = 1 \) or \( |x-y| = 1 \). We will analyze both cases separately.

### Step 2: Solve for \( |x+y| = 1 \)
The absolute value equation \( |x+y| = 1 \) can be split into two linear equations:
1. \( x + y = 1 \)
2. \( x + y = -1 \)

### Step 3: Solve for \( |x-y| = 1 \)
Similarly, the absolute value equation \( |x-y| = 1 \) can also be split into two linear equations:
1. \( x - y = 1 \)
2. \( x - y = -1 \)

### Step 4: Graph the Lines
Now we will graph the four lines obtained from the equations:
1. \( x + y = 1 \) (Line 1)
2. \( x + y = -1 \) (Line 2)
3. \( x - y = 1 \) (Line 3)
4. \( x - y = -1 \) (Line 4)

### Step 5: Find Intersections
Next, we will find the points of intersection of these lines:
- Intersection of Line 1 and Line 3:
- \( x + y = 1 \)
- \( x - y = 1 \)

Solving these, we add the equations:
\[ 2x = 2 \implies x = 1 \implies y = 0 \quad \text{(Point A: (1, 0))} \]

- Intersection of Line 1 and Line 4:
- \( x + y = 1 \)
- \( x - y = -1 \)

Solving these:
\[ 2x = 0 \implies x = 0 \implies y = 1 \quad \text{(Point B: (0, 1))} \]

- Intersection of Line 2 and Line 3:
- \( x + y = -1 \)
- \( x - y = 1 \)

Solving these:
\[ 2x = 0 \implies x = 0 \implies y = -1 \quad \text{(Point C: (0, -1))} \]

- Intersection of Line 2 and Line 4:
- \( x + y = -1 \)
- \( x - y = -1 \)

Solving these:
\[ 2x = -2 \implies x = -1 \implies y = 0 \quad \text{(Point D: (-1, 0))} \]

### Step 6: Identify the Shape
The points of intersection are:
- Point A: (1, 0)
- Point B: (0, 1)
- Point C: (0, -1)
- Point D: (-1, 0)

These points form a square centered at the origin with vertices at these coordinates.

### Step 7: Calculate the Area
The distance between any two adjacent vertices (for example, from (1, 0) to (0, 1)) can be calculated using the distance formula:
\[ \text{Distance} = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]
Since the shape is a square, the area \( A \) is given by:
\[ A = \text{side}^2 = (\sqrt{2})^2 = 2 \text{ square units} \]

### Conclusion
The area enclosed by the curves \( \max(|x+y|, |x-y|) = 1 \) is \( 2 \) square units.

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