Minimise and Maximise `Z = 5x + 10y` subject to `x + 2y le 120, x + y ge 60, x - 2y ge 0, x, y ge 0`.

Answer

To solve the problem of minimizing and maximizing the objective function \( Z = 5x + 10y \) subject to the given constraints, we will follow these steps:

### Step 1: Identify the Constraints
The constraints given are:
1. \( x + 2y \leq 120 \)
2. \( x + y \geq 60 \)
3. \( x - 2y \geq 0 \) (which can be rewritten as \( x \geq 2y \))
4. \( x \geq 0 \)
5. \( y \geq 0 \)

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

### Step 3: Find Intercepts
For each equation, we find the intercepts:
1. For \( x + 2y = 120 \):
- \( x \)-intercept: Set \( y = 0 \) → \( x = 120 \)
- \( y \)-intercept: Set \( x = 0 \) → \( 2y = 120 \) → \( y = 60 \)

2. For \( x + y = 60 \):
- \( x \)-intercept: Set \( y = 0 \) → \( x = 60 \)
- \( y \)-intercept: Set \( x = 0 \) → \( y = 60 \)

3. For \( x = 2y \):
- \( x \)-intercept: Set \( y = 0 \) → \( x = 0 \)
- \( y \)-intercept: Set \( x = 0 \) → \( y = 0 \)

### Step 4: Graph the Constraints
We plot the lines on a graph:
- The line \( x + 2y = 120 \) intersects the axes at (120, 0) and (0, 60).
- The line \( x + y = 60 \) intersects the axes at (60, 0) and (0, 60).
- The line \( x = 2y \) passes through the origin and has a slope of 2.

### Step 5: Identify the Feasible Region
The feasible region is where all the constraints overlap. This region is bounded by the lines and must satisfy all inequalities.

### Step 6: Find Corner Points
To find the corner points of the feasible region, we need to solve the equations pairwise:
1. Intersection of \( x + 2y = 120 \) and \( x + y = 60 \):
- Substitute \( x = 60 - y \) into \( x + 2y = 120 \):
- \( 60 - y + 2y = 120 \) → \( y = 60 \) → \( x = 0 \) → Point (0, 60)

2. Intersection of \( x + 2y = 120 \) and \( x = 2y \):
- Substitute \( x = 2y \) into \( x + 2y = 120 \):
- \( 2y + 2y = 120 \) → \( 4y = 120 \) → \( y = 30 \) → \( x = 60 \) → Point (60, 30)

3. Intersection of \( x + y = 60 \) and \( x = 2y \):
- Substitute \( x = 2y \) into \( x + y = 60 \):
- \( 2y + y = 60 \) → \( 3y = 60 \) → \( y = 20 \) → \( x = 40 \) → Point (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
- The maximum value of \( Z \) is 600 at points (0, 60) and (60, 30).
- The minimum value of \( Z \) is 400 at point (40, 20).

### Final Solution
- **Maximum \( Z = 600 \)** at points (0, 60) and (60, 30).
- **Minimum \( Z = 400 \)** at point (40, 20).