`Z = 3x + 9y`
`x + 3y le 60, x + y ge 10, x le y , x ge 0, y ge 0`

Answer

To solve the linear programming problem given by the objective function \( Z = 3x + 9y \) with the constraints:

1. \( x + 3y \leq 60 \)
2. \( x + y \geq 10 \)
3. \( x \leq y \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

we will follow these steps:

### Step 1: Identify the Constraints
We have the following constraints:
- \( x + 3y \leq 60 \)
- \( x + y \geq 10 \)
- \( x \leq y \)
- \( x \geq 0 \)
- \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines, we convert the inequalities into equations:
1. \( x + 3y = 60 \)
2. \( x + y = 10 \)
3. \( x = y \)

### Step 3: Find Intercepts
For each equation, we will find the x-intercept and y-intercept.

1. **For \( x + 3y = 60 \)**:
- x-intercept: Set \( y = 0 \) → \( x = 60 \) → Point: \( (60, 0) \)
- y-intercept: Set \( x = 0 \) → \( 3y = 60 \) → \( y = 20 \) → Point: \( (0, 20) \)

2. **For \( x + y = 10 \)**:
- x-intercept: Set \( y = 0 \) → \( x = 10 \) → Point: \( (10, 0) \)
- y-intercept: Set \( x = 0 \) → \( y = 10 \) → Point: \( (0, 10) \)

3. **For \( x = y \)**:
- This line passes through the origin and has points like \( (0, 0) \), \( (1, 1) \), etc.

### Step 4: Graph the Constraints
Now we will graph the lines and shade the feasible region based on the inequalities:
- For \( x + 3y \leq 60 \), shade below the line.
- For \( x + y \geq 10 \), shade above the line.
- For \( x \leq y \), shade below the line \( x = y \).
- The feasible region is where all shaded areas overlap and is bounded by the axes.

### Step 5: Identify Corner Points
The corner points of the feasible region can be found by solving the equations at their intersections:
1. Intersection of \( x + 3y = 60 \) and \( x + y = 10 \):
- From \( x + y = 10 \), we can express \( x = 10 - y \).
- Substitute into \( x + 3y = 60 \):
\[ (10 - y) + 3y = 60 \implies 10 + 2y = 60 \implies 2y = 50 \implies y = 25 \implies x = 10 - 25 = -15 \text{ (not feasible)} \]

2. Intersection of \( x + 3y = 60 \) and \( x = y \):
- Substitute \( y \) for \( x \):
\[ y + 3y = 60 \implies 4y = 60 \implies y = 15 \implies x = 15 \text{ (point: (15, 15))} \]

3. Intersection of \( x + y = 10 \) and \( x = y \):
- Substitute \( y \) for \( x \):
\[ y + y = 10 \implies 2y = 10 \implies y = 5 \implies x = 5 \text{ (point: (5, 5))} \]

4. The feasible points are \( (0, 20) \), \( (0, 10) \), \( (5, 5) \), and \( (15, 15) \).

### Step 6: Evaluate the Objective Function at Corner Points
Now we evaluate \( Z = 3x + 9y \) at each corner point:
1. At \( (0, 20) \): \( Z = 3(0) + 9(20) = 180 \)
2. At \( (0, 10) \): \( Z = 3(0) + 9(10) = 90 \)
3. At \( (5, 5) \): \( Z = 3(5) + 9(5) = 60 \)
4. At \( (15, 15) \): \( Z = 3(15) + 9(15) = 180 \)

### Step 7: Determine Maximum and Minimum Values
- Maximum \( Z = 180 \) at points \( (0, 20) \) and \( (15, 15) \).
- Minimum \( Z = 60 \) at point \( (5, 5) \).

### Conclusion
- Maximum value of \( Z \) is \( 180 \) at points \( (0, 20) \) and \( (15, 15) \).
- Minimum value of \( Z \) is \( 60 \) at point \( (5, 5) \).