Minimize `Z= 3x + 2y` subject to `x + y le 8, x, y ge 0`.

Answer

To solve the problem of minimizing \( Z = 3x + 2y \) subject to the constraints \( x + y \leq 8 \) and \( x, y \geq 0 \), we will follow these steps:

### Step 1: Identify the constraints and plot them
The constraints are:
1. \( x + y \leq 8 \)
2. \( x \geq 0 \)
3. \( y \geq 0 \)

To plot the line \( x + y = 8 \):
- When \( x = 0 \), \( y = 8 \) (Point B: \( (0, 8) \))
- When \( y = 0 \), \( x = 8 \) (Point A: \( (8, 0) \))

The feasible region is bounded by these points and the axes.

### Step 2: Identify the feasible region
The feasible region is the area that satisfies all constraints. It is bounded by the points:
- \( (0, 0) \) (origin)
- \( (0, 8) \) (Point B)
- \( (8, 0) \) (Point A)

### Step 3: Determine the corner points of the feasible region
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 8) \)
3. \( (8, 0) \)

### Step 4: Evaluate the objective function at each corner point
Now we will calculate \( Z = 3x + 2y \) at each corner point:
1. At \( (0, 0) \):
\[ Z = 3(0) + 2(0) = 0 \]
2. At \( (0, 8) \):
\[ Z = 3(0) + 2(8) = 16 \]
3. At \( (8, 0) \):
\[ Z = 3(8) + 2(0) = 24 \]

### Step 5: Identify the minimum value of \( Z \)
From the calculations:
- \( Z(0, 0) = 0 \)
- \( Z(0, 8) = 16 \)
- \( Z(8, 0) = 24 \)

The minimum value of \( Z \) occurs at the point \( (0, 0) \) and is \( Z = 0 \).

### Conclusion
The minimum value of \( Z = 3x + 2y \) subject to the given constraints is \( \boxed{0} \).

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