To solve the linear programming problem of minimizing \( Z = -3x + 4y \) subject to the constraints \( x + 2y \leq 8 \), \( 3x + 2y \leq 12 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:
### Step 1: Identify the objective function and constraints
The objective function is:
\[
Z = -3x + 4y
\]
The constraints are:
1. \( x + 2y \leq 8 \)
2. \( 3x + 2y \leq 12 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)
### Step 2: Convert the inequalities into equations for graphing
To graph the constraints, we convert the inequalities into equations:
1. \( x + 2y = 8 \)
2. \( 3x + 2y = 12 \)
### Step 3: Find the intercepts of the equations
**For the first equation \( x + 2y = 8 \):**
- **x-intercept:** Set \( y = 0 \):
\[
x + 2(0) = 8 \implies x = 8 \quad \text{(Point: (8, 0))}
\]
- **y-intercept:** Set \( x = 0 \):
\[
0 + 2y = 8 \implies y = 4 \quad \text{(Point: (0, 4))}
\]
**For the second equation \( 3x + 2y = 12 \):**
- **x-intercept:** Set \( y = 0 \):
\[
3x + 2(0) = 12 \implies x = 4 \quad \text{(Point: (4, 0))}
\]
- **y-intercept:** Set \( x = 0 \):
\[
3(0) + 2y = 12 \implies y = 6 \quad \text{(Point: (0, 6))}
\]
### Step 4: Graph the constraints
Plot the lines \( x + 2y = 8 \) and \( 3x + 2y = 12 \) on a graph. The feasible region is where all the constraints overlap, which is bounded by the axes and the lines.
### Step 5: Identify the vertices of the feasible region
The vertices of the feasible region can be found at the intersection points and intercepts:
1. (0, 0)
2. (0, 4)
3. (4, 0)
4. Intersection of \( x + 2y = 8 \) and \( 3x + 2y = 12 \)
### Step 6: Solve for the intersection point
To find the intersection of \( x + 2y = 8 \) and \( 3x + 2y = 12 \):
- Subtract the first equation from the second:
\[
(3x + 2y) - (x + 2y) = 12 - 8
\]
This simplifies to:
\[
2x = 4 \implies x = 2
\]
Substituting \( x = 2 \) back into the first equation:
\[
2 + 2y = 8 \implies 2y = 6 \implies y = 3
\]
So the intersection point is (2, 3).
### Step 7: Evaluate the objective function at each vertex
Now we evaluate \( Z = -3x + 4y \) at each vertex:
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 the minimum value
The minimum value of \( Z \) occurs at the vertex (4, 0):
\[
\text{Minimum } Z = -12
\]
### Conclusion
The minimum value of \( Z \) is \(-12\) at the point \((4, 0)\).
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