`Z = 5x + 10y`
`x + y ge 60, x + 2y le 120, x -2y ge 0, x, y ge 0`

Answer

To solve the linear programming problem given by the objective function \( Z = 5x + 10y \) with the constraints:

1. \( x + y \geq 60 \)
2. \( x + 2y \leq 120 \)
3. \( x - 2y \geq 0 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

we will follow these steps:

### Step 1: Identify the Constraints
We have the following inequalities:
- \( x + y \geq 60 \)
- \( x + 2y \leq 120 \)
- \( x - 2y \geq 0 \)
- \( x \geq 0 \)
- \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines, we convert the inequalities into equations:
1. \( x + y = 60 \)
2. \( x + 2y = 120 \)
3. \( x - 2y = 0 \) (or \( x = 2y \))

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

1. For \( x + y = 60 \):
- x-intercept: \( (60, 0) \)
- y-intercept: \( (0, 60) \)

2. For \( x + 2y = 120 \):
- x-intercept: \( (120, 0) \)
- y-intercept: \( (0, 60) \)

3. For \( x = 2y \):
- When \( y = 0 \), \( x = 0 \) (point \( (0, 0) \))
- When \( x = 60 \), \( y = 30 \) (point \( (60, 30) \))

### Step 4: Graph the Constraints
Plot the lines on a graph. The feasible region is determined by the inequalities:
- The line \( x + y = 60 \) is shaded above.
- The line \( x + 2y = 120 \) is shaded below.
- The line \( x = 2y \) is shaded above.
- The first quadrant is considered due to \( x \geq 0 \) and \( y \geq 0 \).

### Step 5: Identify the Feasible Region
The feasible region is the area where all shaded regions overlap. This region is bounded by the points where the lines intersect.

### Step 6: Find the Corner Points
To find the corner points of the feasible region, we solve the equations:
1. Intersection of \( x + y = 60 \) and \( x + 2y = 120 \):
- Substitute \( x = 60 - y \) into \( x + 2y = 120 \):
\[ 60 - y + 2y = 120 \implies y = 60 \implies x = 0 \quad \text{(Point A: (0, 60))} \]

2. Intersection of \( x + 2y = 120 \) and \( x = 2y \):
- Substitute \( x = 2y \) into \( x + 2y = 120 \):
\[ 2y + 2y = 120 \implies 4y = 120 \implies y = 30 \implies x = 60 \quad \text{(Point B: (60, 30))} \]

3. Intersection of \( x + y = 60 \) and \( x = 2y \):
- Substitute \( x = 2y \) into \( x + y = 60 \):
\[ 2y + y = 60 \implies 3y = 60 \implies y = 20 \implies x = 40 \quad \text{(Point C: (40, 20))} \]

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = 5x + 10y \) at each corner point:
1. At \( (0, 60) \):
\[ Z = 5(0) + 10(60) = 600 \]
2. At \( (60, 30) \):
\[ Z = 5(60) + 10(30) = 300 + 300 = 600 \]
3. At \( (40, 20) \):
\[ Z = 5(40) + 10(20) = 200 + 200 = 400 \]

### Step 8: Determine Maximum and Minimum Values
From the evaluations:
- Maximum \( Z = 600 \) occurs at points \( (0, 60) \) and \( (60, 30) \).
- Minimum \( Z = 400 \) occurs at point \( (40, 20) \).

### Final Answer
The maximum value of \( Z \) is \( 600 \) at points \( (0, 60) \) and \( (60, 30) \), and the minimum value of \( Z \) is \( 400 \) at point \( (40, 20) \).