To solve the linear programming problem given by the objective function \( Z = 30x + 20y \) subject to the constraints:
1. \( x + y \leq 8 \)
2. \( x + 2y \geq 4 \)
3. \( 6x + 4y \geq 12 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
we will follow these steps:
### Step 1: Graph the Constraints
We need to graph the constraints to find the feasible region.
1. **For \( x + y \leq 8 \)**:
- When \( x = 0 \), \( y = 8 \) (Point A: (0, 8))
- When \( y = 0 \), \( x = 8 \) (Point B: (8, 0))
- The line will connect points A and B.
2. **For \( x + 2y \geq 4 \)**:
- When \( x = 0 \), \( y = 2 \) (Point C: (0, 2))
- When \( y = 0 \), \( x = 4 \) (Point D: (4, 0))
- The line will connect points C and D.
3. **For \( 6x + 4y \geq 12 \)**:
- When \( x = 0 \), \( y = 3 \) (Point E: (0, 3))
- When \( y = 0 \), \( x = 2 \) (Point F: (2, 0))
- The line will connect points E and F.
### Step 2: Determine the Feasible Region
Next, we identify the feasible region that satisfies all the constraints. This is the area where all the shaded regions from the inequalities overlap.
### Step 3: Identify Corner Points
The feasible region will typically be a polygon. We need to find the corner points (vertices) of this polygon, which are the intersection points of the lines.
1. **Intersection of \( x + y = 8 \) and \( x + 2y = 4 \)**:
- Solve the equations:
\[
x + y = 8 \quad (1)
\]
\[
x + 2y = 4 \quad (2)
\]
- From (1), \( y = 8 - x \).
- Substitute into (2):
\[
x + 2(8 - x) = 4 \implies x + 16 - 2x = 4 \implies -x + 16 = 4 \implies x = 12 \quad \text{(not feasible)}
\]
2. **Intersection of \( x + y = 8 \) and \( 6x + 4y = 12 \)**:
- Solve the equations:
\[
x + y = 8 \quad (1)
\]
\[
6x + 4y = 12 \quad (3)
\]
- From (1), \( y = 8 - x \).
- Substitute into (3):
\[
6x + 4(8 - x) = 12 \implies 6x + 32 - 4x = 12 \implies 2x + 32 = 12 \implies 2x = -20 \quad \text{(not feasible)}
\]
3. **Intersection of \( x + 2y = 4 \) and \( 6x + 4y = 12 \)**:
- Solve the equations:
\[
x + 2y = 4 \quad (2)
\]
\[
6x + 4y = 12 \quad (3)
\]
- From (2), \( y = \frac{4 - x}{2} \).
- Substitute into (3):
\[
6x + 4\left(\frac{4 - x}{2}\right) = 12 \implies 6x + 8 - 2x = 12 \implies 4x + 8 = 12 \implies 4x = 4 \implies x = 1
\]
- Substitute \( x = 1 \) back into (2):
\[
1 + 2y = 4 \implies 2y = 3 \implies y = 1.5
\]
- Intersection point is \( (1, 1.5) \).
### Step 4: Evaluate the Objective Function at the Corner Points
Evaluate \( Z = 30x + 20y \) at the feasible corner points:
1. At \( (0, 8) \):
\[
Z = 30(0) + 20(8) = 160
\]
2. At \( (4, 0) \):
\[
Z = 30(4) + 20(0) = 120
\]
3. At \( (1, 1.5) \):
\[
Z = 30(1) + 20(1.5) = 30 + 30 = 60
\]
### Step 5: Determine Maximum and Minimum Values
From the evaluations, we find:
- Maximum \( Z = 160 \) at \( (0, 8) \)
- Minimum \( Z = 60 \) at \( (1, 1.5) \)
### Conclusion
The minimum value of \( Z \) is \( 60 \) at the point \( (1, 1.5) \).
