Maximum of `Z = 3x + 2y` subject to:
`x + y ge 88, x, y ge 0` is .........at .....................

Answer

To solve the problem of maximizing \( Z = 3x + 2y \) subject to the constraints \( x + y \geq 88 \) and \( x, y \geq 0 \), we can follow these steps:

### Step 1: Identify the Constraints
The constraints given are:
1. \( x + y \geq 88 \)
2. \( x \geq 0 \)
3. \( y \geq 0 \)

### Step 2: Graph the Constraints
To graph the constraint \( x + y = 88 \):
- When \( x = 0 \), \( y = 88 \) (point A: (0, 88))
- When \( y = 0 \), \( x = 88 \) (point B: (88, 0))

These points can be plotted on a graph. The line \( x + y = 88 \) divides the plane into two regions. The area above this line represents the feasible region where \( x + y \geq 88 \).

### Step 3: Determine the Feasible Region
Since \( x \) and \( y \) must both be greater than or equal to 0, we only consider the first quadrant of the graph. The feasible region is therefore the area above the line \( x + y = 88 \) that is also within the first quadrant.

### Step 4: Analyze the Objective Function
The objective function \( Z = 3x + 2y \) is a linear function. To maximize \( Z \), we need to consider the direction of the gradient of \( Z \). The lines of constant \( Z \) will be parallel to the line \( 3x + 2y = k \) for various values of \( k \).

### Step 5: Check for Unboundedness
As we move in the direction of increasing \( Z \), we find that we can continue to increase \( x \) and \( y \) indefinitely while still satisfying the constraint \( x + y \geq 88 \). This indicates that the feasible region is unbounded.

### Conclusion
Since the feasible region is unbounded, the maximum value of \( Z \) can increase indefinitely. Thus, we conclude that:

- The maximum of \( Z = 3x + 2y \) is **unbounded**.