`Z =13x + 3y`
`x + y le 6, 3x + 2y le 15, x ge 0, y ge 0`.

Answer

To solve the linear programming problem given by the objective function \( Z = 13x + 3y \) with the constraints:

1. \( x + y \leq 6 \)
2. \( 3x + 2y \leq 15 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

we will follow these steps:

### Step 1: Identify the Objective Function and Constraints
The objective function is:
\[ Z = 13x + 3y \]

The constraints are:
1. \( x + y \leq 6 \)
2. \( 3x + 2y \leq 15 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Constraints to Equations
To find the boundary lines of the constraints, we convert the inequalities into equations:

1. \( x + y = 6 \)
2. \( 3x + 2y = 15 \)

### Step 3: Find Intercepts of the Constraints
For the first equation \( x + y = 6 \):
- When \( x = 0 \), \( y = 6 \) (y-intercept).
- When \( y = 0 \), \( x = 6 \) (x-intercept).

For the second equation \( 3x + 2y = 15 \):
- When \( x = 0 \), \( y = 7.5 \) (y-intercept).
- When \( y = 0 \), \( x = 5 \) (x-intercept).

### Step 4: Plot the Constraints
Plot the lines on a graph:
- The line \( x + y = 6 \) connects points (6, 0) and (0, 6).
- The line \( 3x + 2y = 15 \) connects points (5, 0) and (0, 7.5).

### Step 5: Determine the Feasible Region
The feasible region is where all constraints overlap. Since both constraints are less than or equal to, we shade the area below both lines in the first quadrant (where \( x \geq 0 \) and \( y \geq 0 \)).

### Step 6: Find Corner Points of the Feasible Region
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 6) \)
3. \( (5, 0) \)
4. The intersection of the two lines.

To find the intersection of \( x + y = 6 \) and \( 3x + 2y = 15 \):
- From \( x + y = 6 \), we can express \( y = 6 - x \).
- Substitute \( y \) in the second equation:
\[ 3x + 2(6 - x) = 15 \]
\[ 3x + 12 - 2x = 15 \]
\[ x + 12 = 15 \implies x = 3 \]
Substituting \( x = 3 \) back into \( y = 6 - x \):
\[ y = 6 - 3 = 3 \]
Thus, the intersection point is \( (3, 3) \).

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = 13x + 3y \) at each corner point:
1. At \( (0, 0) \):
\[ Z = 13(0) + 3(0) = 0 \]
2. At \( (0, 6) \):
\[ Z = 13(0) + 3(6) = 18 \]
3. At \( (5, 0) \):
\[ Z = 13(5) + 3(0) = 65 \]
4. At \( (3, 3) \):
\[ Z = 13(3) + 3(3) = 39 + 9 = 48 \]

### Step 8: Determine the Maximum Value
The maximum value of \( Z \) occurs at \( (5, 0) \):
\[ Z_{\text{max}} = 65 \]

### Conclusion
The optimal solution is:
\[ \text{Maximum } Z = 65 \text{ at } (5, 0) \]