Let's solve the given linear programming problems step by step.
### (i) Maximize \( Z = 2x + 3y \) subject to:
- \( x + 2y \leq 6 \)
- \( x \geq 4 \)
- \( y \geq 0 \)
**Step 1: Identify the constraints and plot them.**
- The first constraint is \( x + 2y = 6 \).
- When \( x = 0 \), \( y = 3 \) (y-intercept).
- When \( y = 0 \), \( x = 6 \) (x-intercept).
- The second constraint is \( x = 4 \).
- The third constraint is \( y = 0 \).
**Step 2: Plot the lines on the graph.**
- Draw the line for \( x + 2y = 6 \) connecting points (0, 3) and (6, 0).
- Draw the vertical line \( x = 4 \).
- The x-axis represents \( y = 0 \).
**Step 3: Identify the feasible region.**
- The feasible region is bounded by the lines and is where all constraints overlap.
- The vertices of the feasible region are (4, 0), (6, 0), and (4, 1).
**Step 4: Evaluate the objective function at each vertex.**
- At (4, 0): \( Z = 2(4) + 3(0) = 8 \)
- At (6, 0): \( Z = 2(6) + 3(0) = 12 \)
- At (4, 1): \( Z = 2(4) + 3(1) = 11 \)
**Step 5: Determine the maximum value.**
- The maximum value of \( Z \) is 12 at the point (6, 0).
### Conclusion for (i):
- **Maximum \( Z = 12 \) at \( (6, 0) \)**.
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### (ii) Maximize \( Z = 4x + y \) subject to:
- \( x + y \leq 50 \)
- \( y \geq 0 \)
**Step 1: Identify the constraints and plot them.**
- The constraint \( x + y = 50 \) has intercepts:
- When \( x = 0 \), \( y = 50 \).
- When \( y = 0 \), \( x = 50 \).
**Step 2: Plot the line on the graph.**
- Draw the line connecting (0, 50) and (50, 0).
**Step 3: Identify the feasible region.**
- The feasible region is below the line and above the x-axis.
**Step 4: Evaluate the objective function at the vertices.**
- The vertices of the feasible region are (0, 0), (0, 50), and (50, 0).
- At (0, 0): \( Z = 4(0) + 0 = 0 \)
- At (0, 50): \( Z = 4(0) + 50 = 50 \)
- At (50, 0): \( Z = 4(50) + 0 = 200 \)
**Step 5: Determine the maximum value.**
- The maximum value of \( Z \) is 200 at the point (50, 0).
### Conclusion for (ii):
- **Maximum \( Z = 200 \) at \( (50, 0) \)**.
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### (iii) Maximize \( Z = x + 2y \) subject to:
- \( 2x + y \leq 6 \)
- \( x, y \geq 0 \)
**Step 1: Identify the constraints and plot them.**
- The constraint \( 2x + y = 6 \) has intercepts:
- When \( x = 0 \), \( y = 6 \).
- When \( y = 0 \), \( x = 3 \).
**Step 2: Plot the line on the graph.**
- Draw the line connecting (0, 6) and (3, 0).
**Step 3: Identify the feasible region.**
- The feasible region is below the line and above the axes.
**Step 4: Evaluate the objective function at the vertices.**
- The vertices of the feasible region are (0, 0), (3, 0), and (0, 6).
- At (0, 0): \( Z = 0 + 2(0) = 0 \)
- At (3, 0): \( Z = 3 + 2(0) = 3 \)
- At (0, 6): \( Z = 0 + 2(6) = 12 \)
**Step 5: Determine the maximum value.**
- The maximum value of \( Z \) is 12 at the point (0, 6).
### Conclusion for (iii):
- **Maximum \( Z = 12 \) at \( (0, 6) \)**.
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### (iv) Maximize \( Z = 5x + 3y \) subject to:
- \( 3x + 5y \leq 15 \)
- \( x, y \geq 0 \)
**Step 1: Identify the constraints and plot them.**
- The constraint \( 3x + 5y = 15 \) has intercepts:
- When \( x = 0 \), \( y = 3 \).
- When \( y = 0 \), \( x = 5 \).
**Step 2: Plot the line on the graph.**
- Draw the line connecting (0, 3) and (5, 0).
**Step 3: Identify the feasible region.**
- The feasible region is below the line and above the axes.
**Step 4: Evaluate the objective function at the vertices.**
- The vertices of the feasible region are (0, 0), (0, 3), and (5, 0).
- At (0, 0): \( Z = 5(0) + 3(0) = 0 \)
- At (0, 3): \( Z = 5(0) + 3(3) = 9 \)
- At (5, 0): \( Z = 5(5) + 3(0) = 25 \)
**Step 5: Determine the maximum value.**
- The maximum value of \( Z \) is 25 at the point (5, 0).
### Conclusion for (iv):
- **Maximum \( Z = 25 \) at \( (5, 0) \)**.
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