Minimize `Z= 3x + 9y` subject to ` x + 3y le 60, x le y `and `x, y ge 0`.

Answer

To solve the linear programming problem of minimizing \( Z = 3x + 9y \) subject to the constraints \( x + 3y \leq 60 \), \( x \leq y \), and \( x, y \geq 0 \), we will follow these steps:

### Step 1: Identify the Constraints
The constraints given are:
1. \( x + 3y \leq 60 \)
2. \( x \leq y \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines of the feasible region, we convert the inequalities into equations:
1. \( x + 3y = 60 \)
2. \( x - y = 0 \) or \( x = y \)

### Step 3: Find Points of Intersection
Next, we find the points where these lines intersect with each other and the axes.

1. **Finding the intersection of \( x + 3y = 60 \) and \( x = y \)**:
Substitute \( x = y \) into \( x + 3y = 60 \):
\[ y + 3y = 60 \implies 4y = 60 \implies y = 15 \implies x = 15 \]
So, one intersection point is \( (15, 15) \).

2. **Finding the x-intercept of \( x + 3y = 60 \)**:
Set \( y = 0 \):
\[ x + 3(0) = 60 \implies x = 60 \]
So, the x-intercept is \( (60, 0) \).

3. **Finding the y-intercept of \( x + 3y = 60 \)**:
Set \( x = 0 \):
\[ 0 + 3y = 60 \implies y = 20 \]
So, the y-intercept is \( (0, 20) \).

### Step 4: Identify the Feasible Region
The feasible region is bounded by:
- The line \( x + 3y = 60 \)
- The line \( x = y \)
- The axes \( x = 0 \) and \( y = 0 \)

The vertices of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 20) \)
3. \( (15, 15) \)

### Step 5: Evaluate the Objective Function at Each Vertex
Now, we calculate \( Z = 3x + 9y \) at each vertex:

1. At \( (0, 0) \):
\[ Z = 3(0) + 9(0) = 0 \]

2. At \( (0, 20) \):
\[ Z = 3(0) + 9(20) = 180 \]

3. At \( (15, 15) \):
\[ Z = 3(15) + 9(15) = 45 + 135 = 180 \]

### Step 6: Determine the Minimum Value
From the values calculated:
- At \( (0, 0) \), \( Z = 0 \)
- At \( (0, 20) \), \( Z = 180 \)
- At \( (15, 15) \), \( Z = 180 \)

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

### Conclusion
Thus, the minimum value of \( Z \) is:
\[ \boxed{0} \]