`Z = -3x + 4y`
`x + 2y le 8, 3x + 2y le 12, x ge 0 , y ge 0`.

Answer

To solve the linear programming problem given by the objective function \( Z = -3x + 4y \) with the constraints:

1. \( x + 2y \leq 8 \)
2. \( 3x + 2y \leq 12 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

we will follow these steps:

### Step 1: Identify the Constraints
We have the following constraints:
- \( x + 2y \leq 8 \)
- \( 3x + 2y \leq 12 \)
- \( x \geq 0 \)
- \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines for the constraints, we convert the inequalities into equations:
1. \( x + 2y = 8 \)
2. \( 3x + 2y = 12 \)

### Step 3: Find Intercepts for Each Line
**For the line \( x + 2y = 8 \):**
- When \( x = 0 \): \( 2y = 8 \) → \( y = 4 \) (Point: \( (0, 4) \))
- When \( y = 0 \): \( x = 8 \) (Point: \( (8, 0) \))

**For the line \( 3x + 2y = 12 \):**
- When \( x = 0 \): \( 2y = 12 \) → \( y = 6 \) (Point: \( (0, 6) \))
- When \( y = 0 \): \( 3x = 12 \) → \( x = 4 \) (Point: \( (4, 0) \))

### Step 4: Plot the Lines
Plot the points \( (0, 4) \), \( (8, 0) \), \( (0, 6) \), and \( (4, 0) \) on a graph. Draw the lines for the equations and shade the feasible region that satisfies all constraints.

### Step 5: Find the Intersection Points
To find the intersection of the two lines:
Set the equations equal to each other:
1. From \( x + 2y = 8 \) → \( 2y = 8 - x \) → \( y = 4 - \frac{x}{2} \)
2. Substitute into \( 3x + 2y = 12 \):
\[ 3x + 2(4 - \frac{x}{2}) = 12 \]
\[ 3x + 8 - x = 12 \]
\[ 2x = 4 \Rightarrow x = 2 \]
Substitute \( x = 2 \) back into \( y = 4 - \frac{x}{2} \):
\[ y = 4 - 1 = 3 \]
So, the intersection point is \( (2, 3) \).

### Step 6: Identify Corner Points
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 4) \)
3. \( (4, 0) \)
4. \( (2, 3) \)

### Step 7: Evaluate the Objective Function at Each Corner Point
Now, we evaluate \( Z = -3x + 4y \) at each corner point:
1. At \( (0, 0) \): \( Z = -3(0) + 4(0) = 0 \)
2. At \( (0, 4) \): \( Z = -3(0) + 4(4) = 16 \)
3. At \( (4, 0) \): \( Z = -3(4) + 4(0) = -12 \)
4. At \( (2, 3) \): \( Z = -3(2) + 4(3) = -6 + 12 = 6 \)

### Step 8: Determine Maximum and Minimum Values
From the evaluations:
- Maximum value of \( Z \) is \( 16 \) at \( (0, 4) \).
- Minimum value of \( Z \) is \( -12 \) at \( (4, 0) \).

### Conclusion
The solution to the linear programming problem is:
- Maximum value of \( Z \) is \( 16 \) at the point \( (0, 4) \).
- Minimum value of \( Z \) is \( -12 \) at the point \( (4, 0) \).

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