Maximise `Z = x+y`, subject to `x - y le -1, - x + y le 0, x, y ge 0`.

Answer

To solve the linear programming problem of maximizing \( Z = x + y \) subject to the constraints:

1. \( x - y \leq -1 \)
2. \( -x + y \leq 0 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

we will follow these steps:

### Step 1: Rewrite the Constraints
We can rewrite the constraints in a more standard form for analysis:
- From \( x - y \leq -1 \), we can rewrite it as \( x - y + 1 \leq 0 \).
- From \( -x + y \leq 0 \), we can rewrite it as \( y - x \leq 0 \) or \( y \leq x \).

### Step 2: Find the Intersection Points
Next, we will find the points of intersection of the lines formed by the constraints.

1. **For the first constraint \( x - y = -1 \)**:
- When \( x = 0 \): \( 0 - y = -1 \) → \( y = 1 \) → Point (0, 1)
- When \( y = 0 \): \( x - 0 = -1 \) → \( x = -1 \) (not valid since \( x \geq 0 \))

2. **For the second constraint \( -x + y = 0 \)**:
- This simplifies to \( y = x \).
- When \( x = 0 \): \( y = 0 \) → Point (0, 0)
- When \( y = 0 \): \( 0 = x \) → Point (0, 0)

### Step 3: Plot the Lines
We can plot the lines based on the points we found:
- The line \( x - y = -1 \) passes through (0, 1) and has a slope of 1.
- The line \( y = x \) passes through the origin (0, 0).

### Step 4: Identify the Feasible Region
Next, we need to identify the feasible region defined by the constraints:
- For \( x - y \leq -1 \), the region is below the line.
- For \( y \leq x \), the region is below the line \( y = x \).
- Additionally, we must consider the non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \).

### Step 5: Check the Feasibility
We check the origin (0, 0) to see if it satisfies all constraints:
- For \( x - y \leq -1 \): \( 0 - 0 \leq -1 \) (False)
- For \( -x + y \leq 0 \): \( -0 + 0 \leq 0 \) (True)

Since the origin does not satisfy the first constraint, we need to look for other points.

### Step 6: Determine the Unbounded Region
The feasible region is unbounded. The line \( y = x \) continues indefinitely in the first quadrant, and the region defined by the constraints does not close off.

### Step 7: Conclusion
Since the feasible region is unbounded, the value of \( Z = x + y \) can increase indefinitely as we move along the line \( y = x \) towards infinity. Therefore, we conclude that:

**The maximum value of \( Z \) is unbounded.**