Maximize:
`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 linear programming problem of maximizing the objective function \( Z = -x + 2y \) subject to the given constraints, we will follow these steps:

### Step 1: Write down the objective function and constraints
The objective function is:
\[ Z = -x + 2y \]
The constraints are:
1. \( x \geq 3 \)
2. \( x + y \geq 5 \)
3. \( x + 2y \geq 6 \)
4. \( y \geq 0 \)

### Step 2: Convert the inequalities into equations
To graph the constraints, we convert each inequality into an equation:
1. \( x = 3 \)
2. \( x + y = 5 \)
3. \( x + 2y = 6 \)
4. \( y = 0 \)

### Step 3: Find the intercepts for each equation
- For \( x = 3 \): This is a vertical line at \( x = 3 \).
- For \( x + y = 5 \):
- \( x \)-intercept: Set \( y = 0 \) → \( x = 5 \) (point (5, 0))
- \( y \)-intercept: Set \( x = 0 \) → \( y = 5 \) (point (0, 5))
- For \( x + 2y = 6 \):
- \( x \)-intercept: Set \( y = 0 \) → \( x = 6 \) (point (6, 0))
- \( y \)-intercept: Set \( x = 0 \) → \( 2y = 6 \) → \( y = 3 \) (point (0, 3))

### Step 4: Graph the constraints
Plot the lines on a graph:
- Draw the vertical line for \( x = 3 \).
- Draw the line for \( x + y = 5 \) connecting points (5, 0) and (0, 5).
- Draw the line for \( x + 2y = 6 \) connecting points (6, 0) and (0, 3).
- Draw the horizontal line for \( y = 0 \) (the x-axis).

### Step 5: Identify the feasible region
The feasible region is where all the constraints overlap. Since we have inequalities, we will shade the area that satisfies all the constraints:
- The area must be to the right of \( x = 3 \).
- Above the line \( x + y = 5 \).
- Above the line \( x + 2y = 6 \).
- Above the x-axis (where \( y \geq 0 \)).

### Step 6: Determine the corner points of the feasible region
The corner points of the feasible region can be found by solving the equations of the lines where they intersect:
1. Intersection of \( x + y = 5 \) and \( x + 2y = 6 \):
- Solve the system:
\[ x + y = 5 \quad (1) \]
\[ x + 2y = 6 \quad (2) \]
- From (1): \( y = 5 - x \)
- Substitute into (2):
\[ x + 2(5 - x) = 6 \\ x + 10 - 2x = 6 \\ -x + 10 = 6 \\ x = 4 \quad \Rightarrow \quad y = 5 - 4 = 1 \]
- Corner point: \( (4, 1) \)

2. Intersection of \( x + y = 5 \) and \( x = 3 \):
- Substitute \( x = 3 \) into \( x + y = 5 \):
\[ 3 + y = 5 \\ y = 2 \]
- Corner point: \( (3, 2) \)

3. Intersection of \( x + 2y = 6 \) and \( x = 3 \):
- Substitute \( x = 3 \) into \( x + 2y = 6 \):
\[ 3 + 2y = 6 \\ 2y = 3 \\ y = 1.5 \]
- Corner point: \( (3, 1.5) \)

### Step 7: Evaluate the objective function at each corner point
Now we evaluate \( Z = -x + 2y \) at each corner point:
1. At \( (4, 1) \):
\[ Z = -4 + 2(1) = -4 + 2 = -2 \]
2. At \( (3, 2) \):
\[ Z = -3 + 2(2) = -3 + 4 = 1 \]
3. At \( (3, 1.5) \):
\[ Z = -3 + 2(1.5) = -3 + 3 = 0 \]

### Step 8: Determine the maximum value
The maximum value of \( Z \) occurs at the point \( (3, 2) \) where \( Z = 1 \).

### Final Result
The maximum value of the objective function \( Z \) is:
\[ \text{Maximum } Z = 1 \text{ at the point } (3, 2). \]