Maximize `Z = x + 2y` subject to `x + y ge 5, x ge 0, y ge 0`.

Answer

To solve the problem of maximizing \( Z = x + 2y \) subject to the constraints \( x + y \geq 5 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:

### Step 1: Identify the Constraints
The constraints given are:
1. \( x + y \geq 5 \)
2. \( x \geq 0 \)
3. \( y \geq 0 \)

### Step 2: Graph the Constraints
1. **Graph \( x + y = 5 \)**:
- When \( x = 0 \), \( y = 5 \) (point (0, 5)).
- When \( y = 0 \), \( x = 5 \) (point (5, 0)).
- Plot these points on the coordinate axis and draw the line.

2. **Identify the Feasible Region**:
- Since we need \( x + y \geq 5 \), the feasible region will be above the line \( x + y = 5 \).
- The other constraints \( x \geq 0 \) and \( y \geq 0 \) restrict the feasible region to the first quadrant.

### Step 3: Determine the Feasible Region
- The feasible region is the area above the line \( x + y = 5 \) in the first quadrant. This region is unbounded as it extends infinitely upwards and to the right.

### Step 4: Find the Objective Function
- The objective function is \( Z = x + 2y \). To maximize \( Z \), we need to assess how \( Z \) behaves in the feasible region.

### Step 5: Analyze the Objective Function
- As we move along the line \( x + y = 5 \), we can express \( y \) in terms of \( x \):
\[ y = 5 - x \]
- Substitute into the objective function:
\[ Z = x + 2(5 - x) = x + 10 - 2x = 10 - x \]
- This shows that as \( x \) decreases, \( Z \) increases.

### Step 6: Determine the Maximum Value
- Since \( x \) can be decreased indefinitely (approaching 0), \( Z \) can increase indefinitely as well. Therefore, the maximum value of \( Z \) is not defined (it approaches infinity).

### Conclusion
Thus, the maximum value of \( Z = x + 2y \) subject to the given constraints is **not defined** as it approaches infinity.

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