Maximise `Z= -x + 2y`, subject to the constraints: `x ge 3, x + y ge 5, x + 2y ge 6, y ge 0`.

Answer

To solve the problem of maximizing \( Z = -x + 2y \) subject to the constraints \( x \geq 3 \), \( x + y \geq 5 \), \( x + 2y \geq 6 \), and \( y \geq 0 \), we will follow these steps:

### Step 1: Identify the Constraints
We have the following constraints:
1. \( x \geq 3 \)
2. \( x + y \geq 5 \)
3. \( x + 2y \geq 6 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines of the inequalities, we convert them into equations:
1. \( x = 3 \)
2. \( x + y = 5 \) (or \( y = 5 - x \))
3. \( x + 2y = 6 \) (or \( y = \frac{6 - x}{2} \))
4. \( y = 0 \)

### Step 3: Find Intersection Points
We will find the intersection points of the lines formed by the equations.

**Intersection of \( x + y = 5 \) and \( x + 2y = 6 \):**
- From \( x + y = 5 \), we can express \( y = 5 - x \).
- Substitute into \( x + 2(5 - x) = 6 \):
\[ x + 10 - 2x = 6 \implies -x + 10 = 6 \implies x = 4 \]
- Substitute \( x = 4 \) back into \( y = 5 - x \):
\[ y = 5 - 4 = 1 \]
- So, the intersection point is \( (4, 1) \).

**Intersection of \( x + y = 5 \) and \( y = 0 \):**
- Substitute \( y = 0 \) into \( x + y = 5 \):
\[ x + 0 = 5 \implies x = 5 \]
- So, the intersection point is \( (5, 0) \).

**Intersection of \( x + 2y = 6 \) and \( y = 0 \):**
- Substitute \( y = 0 \) into \( x + 2y = 6 \):
\[ x + 0 = 6 \implies x = 6 \]
- So, the intersection point is \( (6, 0) \).

### Step 4: Identify Feasible Region
Now we need to determine the feasible region defined by the constraints. We will test the origin (0,0) in each inequality:
- For \( x \geq 3 \): Not satisfied.
- For \( x + y \geq 5 \): Not satisfied.
- For \( x + 2y \geq 6 \): Not satisfied.
- For \( y \geq 0 \): Satisfied.

Since the origin does not satisfy the inequalities, we check the region above the lines defined by the constraints.

### Step 5: Evaluate the Objective Function at Corner Points
The corner points of the feasible region are \( (4, 1) \), \( (5, 0) \), and \( (6, 0) \). We will evaluate \( Z = -x + 2y \) at these points:

1. At \( (4, 1) \):
\[ Z = -4 + 2(1) = -4 + 2 = -2 \]

2. At \( (5, 0) \):
\[ Z = -5 + 2(0) = -5 + 0 = -5 \]

3. At \( (6, 0) \):
\[ Z = -6 + 2(0) = -6 + 0 = -6 \]

### Step 6: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( (4, 1) \):
\[ \text{Maximum } Z = -2 \]

### Conclusion
Thus, the maximum value of \( Z \) is \( -2 \) at the point \( (4, 1) \).

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