Minimize `Z = 3x + 2y` subject to the constraints : `x + y ge 8, 3x + 5y le 15, x ge 0, y ge 0`.

Answer

To solve the problem of minimizing \( Z = 3x + 2y \) subject to the constraints \( x + y \geq 8 \), \( 3x + 5y \leq 15 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:

### Step 1: Identify the Objective Function and Constraints
- The objective function is \( Z = 3x + 2y \).
- The constraints are:
1. \( x + y \geq 8 \)
2. \( 3x + 5y \leq 15 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( x + y = 8 \)
2. \( 3x + 5y = 15 \)

### Step 3: Find Intercepts for Each Constraint
- For \( x + y = 8 \):
- \( x \)-intercept: Set \( y = 0 \) → \( x = 8 \) (Point: (8, 0))
- \( y \)-intercept: Set \( x = 0 \) → \( y = 8 \) (Point: (0, 8))

- For \( 3x + 5y = 15 \):
- \( x \)-intercept: Set \( y = 0 \) → \( 3x = 15 \) → \( x = 5 \) (Point: (5, 0))
- \( y \)-intercept: Set \( x = 0 \) → \( 5y = 15 \) → \( y = 3 \) (Point: (0, 3))

### Step 4: Graph the Constraints
- Plot the lines on a graph:
- Line for \( x + y = 8 \) passes through (8, 0) and (0, 8).
- Line for \( 3x + 5y = 15 \) passes through (5, 0) and (0, 3).

### Step 5: Determine the Feasible Region
- The line \( x + y = 8 \) indicates that the feasible region is above this line (since \( x + y \geq 8 \)).
- The line \( 3x + 5y = 15 \) indicates that the feasible region is below this line (since \( 3x + 5y \leq 15 \)).
- Both \( x \) and \( y \) must be non-negative, meaning we are limited to the first quadrant.

### Step 6: Identify the Vertices of the Feasible Region
- The vertices of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( x + y = 8 \) and \( 3x + 5y = 15 \):
- From \( x + y = 8 \), we have \( y = 8 - x \).
- Substitute into \( 3x + 5(8 - x) = 15 \):
\[ 3x + 40 - 5x = 15 \implies -2x + 40 = 15 \implies -2x = -25 \implies x = 12.5 \text{ (not feasible)} \]
- Hence, this intersection does not yield a feasible solution.

2. Check the intercepts:
- (8, 0) from \( x + y = 8 \)
- (0, 3) from \( 3x + 5y = 15 \)
- (5, 0) from \( 3x + 5y = 15 \)

### Step 7: Evaluate the Objective Function at Each Vertex
- Evaluate \( Z \) at feasible points:
1. At (8, 0): \( Z = 3(8) + 2(0) = 24 \)
2. At (5, 0): \( Z = 3(5) + 2(0) = 15 \)
3. At (0, 3): \( Z = 3(0) + 2(3) = 6 \)

### Step 8: Determine the Minimum Value
- The minimum value of \( Z \) occurs at the point (0, 3) where \( Z = 6 \).

### Conclusion
The minimum value of \( Z = 3x + 2y \) subject to the given constraints is \( Z = 6 \) at the point \( (0, 3) \).