To solve the given linear programming problems step by step, we will follow the standard procedure of identifying the objective function, constraints, plotting the feasible region, finding the corner points, and evaluating the objective function at these points.
### Problem (i)
**Objective Function:**
\[ Z = 20x + 10y \]
**Constraints:**
1. \( x + 2y \leq 28 \)
2. \( 3x + y \leq 24 \)
3. \( x \geq 2 \)
4. \( x, y \geq 0 \)
#### Step 1: Graph the Constraints
1. **For \( x + 2y = 28 \)**:
- \( x \)-intercept: Set \( y = 0 \) → \( x = 28 \) → Point (28, 0)
- \( y \)-intercept: Set \( x = 0 \) → \( 2y = 28 \) → \( y = 14 \) → Point (0, 14)
2. **For \( 3x + y = 24 \)**:
- \( x \)-intercept: Set \( y = 0 \) → \( 3x = 24 \) → \( x = 8 \) → Point (8, 0)
- \( y \)-intercept: Set \( x = 0 \) → \( y = 24 \) → Point (0, 24)
3. **For \( x = 2 \)**: This is a vertical line at \( x = 2 \).
4. **Non-negativity constraints**: \( x \geq 0 \) and \( y \geq 0 \).
#### Step 2: Identify the Feasible Region
The feasible region is the area where all constraints overlap. It will be bounded by the lines we plotted and the axes.
#### Step 3: Find Corner Points
1. Intersection of \( x + 2y = 28 \) and \( x = 2 \):
- Substitute \( x = 2 \) into \( x + 2y = 28 \):
\[ 2 + 2y = 28 \implies 2y = 26 \implies y = 13 \]
- Point: (2, 13)
2. Intersection of \( 3x + y = 24 \) and \( x = 2 \):
- Substitute \( x = 2 \) into \( 3x + y = 24 \):
\[ 3(2) + y = 24 \implies 6 + y = 24 \implies y = 18 \]
- Point: (2, 18) (not feasible since y cannot exceed 14)
3. Intersection of \( x + 2y = 28 \) and \( 3x + y = 24 \):
- Solve the equations simultaneously:
\[ x + 2y = 28 \]
\[ 3x + y = 24 \]
- From the first equation, \( y = \frac{28 - x}{2} \). Substitute into the second:
\[ 3x + \frac{28 - x}{2} = 24 \]
\[ 6x + 28 - x = 48 \]
\[ 5x = 20 \implies x = 4 \]
- Substitute \( x = 4 \) back into \( y \):
\[ y = \frac{28 - 4}{2} = 12 \]
- Point: (4, 12)
4. Intersection of \( x + 2y = 28 \) and \( y = 0 \):
- Set \( y = 0 \):
\[ x = 28 \]
- Point: (28, 0)
5. Intersection of \( 3x + y = 24 \) and \( y = 0 \):
- Set \( y = 0 \):
\[ 3x = 24 \implies x = 8 \]
- Point: (8, 0)
#### Step 4: Evaluate the Objective Function at Corner Points
1. \( Z(2, 13) = 20(2) + 10(13) = 40 + 130 = 170 \)
2. \( Z(4, 12) = 20(4) + 10(12) = 80 + 120 = 200 \)
3. \( Z(8, 0) = 20(8) + 10(0) = 160 + 0 = 160 \)
#### Step 5: Determine Maximum Value
The maximum value of \( Z \) occurs at the point (4, 12):
\[ Z_{\text{max}} = 200 \]
### Final Answer for Problem (i)
The maximum value of \( Z \) is **200** at the point **(4, 12)**.
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### Problem (ii)
**Objective Function:**
\[ Z = 7x + 10y \]
**Constraints:**
1. \( 4x + 6y \leq 240 \)
2. \( 6x + 3y \leq 240 \)
3. \( x \geq 10 \)
4. \( x, y \geq 0 \)
#### Step 1: Graph the Constraints
1. **For \( 4x + 6y = 240 \)**:
- \( x \)-intercept: \( x = 60 \) → Point (60, 0)
- \( y \)-intercept: \( y = 40 \) → Point (0, 40)
2. **For \( 6x + 3y = 240 \)**:
- \( x \)-intercept: \( x = 40 \) → Point (40, 0)
- \( y \)-intercept: \( y = 80 \) → Point (0, 80)
3. **For \( x = 10 \)**: This is a vertical line at \( x = 10 \).
#### Step 2: Identify the Feasible Region
The feasible region is where all constraints overlap, bounded by the lines and the axes.
#### Step 3: Find Corner Points
1. Intersection of \( 4x + 6y = 240 \) and \( x = 10 \):
- Substitute \( x = 10 \):
\[ 4(10) + 6y = 240 \implies 40 + 6y = 240 \implies 6y = 200 \implies y = \frac{200}{6} \approx 33.33 \]
- Point: (10, 33.33)
2. Intersection of \( 6x + 3y = 240 \) and \( x = 10 \):
- Substitute \( x = 10 \):
\[ 6(10) + 3y = 240 \implies 60 + 3y = 240 \implies 3y = 180 \implies y = 60 \]
- Point: (10, 60)
3. Intersection of \( 4x + 6y = 240 \) and \( 6x + 3y = 240 \):
- Solve simultaneously:
\[ 4x + 6y = 240 \]
\[ 6x + 3y = 240 \]
- From the second equation, \( y = 80 - 2x \). Substitute into the first:
\[ 4x + 6(80 - 2x) = 240 \]
\[ 4x + 480 - 12x = 240 \]
\[ -8x = -240 \implies x = 30 \]
- Substitute \( x = 30 \) back into \( y \):
\[ y = 80 - 2(30) = 20 \]
- Point: (30, 20)
4. Intersection of \( 4x + 6y = 240 \) and \( y = 0 \):
- Set \( y = 0 \):
\[ 4x = 240 \implies x = 60 \]
- Point: (60, 0)
5. Intersection of \( 6x + 3y = 240 \) and \( y = 0 \):
- Set \( y = 0 \):
\[ 6x = 240 \implies x = 40 \]
- Point: (40, 0)
#### Step 4: Evaluate the Objective Function at Corner Points
1. \( Z(10, 33.33) = 7(10) + 10(33.33) = 70 + 333.33 = 403.33 \)
2. \( Z(10, 60) = 7(10) + 10(60) = 70 + 600 = 670 \)
3. \( Z(30, 20) = 7(30) + 10(20) = 210 + 200 = 410 \)
4. \( Z(40, 0) = 7(40) + 10(0) = 280 + 0 = 280 \)
5. \( Z(60, 0) = 7(60) + 10(0) = 420 + 0 = 420 \)
#### Step 5: Determine Maximum Value
The maximum value of \( Z \) occurs at the point (10, 60):
\[ Z_{\text{max}} = 670 \]
### Final Answer for Problem (ii)
The maximum value of \( Z \) is **670** at the point **(10, 60)**.
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### Problem (iii)
**Objective Function:**
\[ Z = 22x + 44y \]
**Constraints:**
1. \( x + y \geq 3 \)
2. \( 3x + 8y \leq 24 \)
3. \( x - y \geq 0 \)
4. \( x, y \geq 0 \)
#### Step 1: Graph the Constraints
1. **For \( x + y = 3 \)**:
- \( x \)-intercept: \( x = 3 \) → Point (3, 0)
- \( y \)-intercept: \( y = 3 \) → Point (0, 3)
2. **For \( 3x + 8y = 24 \)**:
- \( x \)-intercept: \( x = 8 \) → Point (8, 0)
- \( y \)-intercept: \( y = 3 \) → Point (0, 3)
3. **For \( x - y = 0 \)**: This is the line \( y = x \).
#### Step 2: Identify the Feasible Region
The feasible region is where all constraints overlap, bounded by the lines and the axes.
#### Step 3: Find Corner Points
1. Intersection of \( x + y = 3 \) and \( x - y = 0 \):
- Set \( y = x \):
\[ x + x = 3 \implies 2x = 3 \implies x = \frac{3}{2} \]
- Point: \(\left(\frac{3}{2}, \frac{3}{2}\right)\)
2. Intersection of \( 3x + 8y = 24 \) and \( x - y = 0 \):
- Set \( y = x \):
\[ 3x + 8x = 24 \implies 11x = 24 \implies x = \frac{24}{11} \]
- Point: \(\left(\frac{24}{11}, \frac{24}{11}\right)\)
3. Intersection of \( 3x + 8y = 24 \) and \( x + y = 3 \):
- Solve simultaneously:
\[ 3x + 8(3 - x) = 24 \]
\[ 3x + 24 - 8x = 24 \]
\[ -5x = 0 \implies x = 0 \]
- Substitute \( x = 0 \) into \( x + y = 3 \):
\[ y = 3 \]
- Point: (0, 3)
4. Intersection of \( x + y = 3 \) and \( y = 0 \):
- Set \( y = 0 \):
\[ x = 3 \]
- Point: (3, 0)
5. Intersection of \( 3x + 8y = 24 \) and \( y = 0 \):
- Set \( y = 0 \):
\[ 3x = 24 \implies x = 8 \]
- Point: (8, 0)
#### Step 4: Evaluate the Objective Function at Corner Points
1. \( Z\left(\frac{3}{2}, \frac{3}{2}\right) = 22\left(\frac{3}{2}\right) + 44\left(\frac{3}{2}\right) = 33 + 66 = 99 \)
2. \( Z\left(\frac{24}{11}, \frac{24}{11}\right) = 22\left(\frac{24}{11}\right) + 44\left(\frac{24}{11}\right) = \frac{528}{11} + \frac{1056}{11} = \frac{1584}{11} \approx 144 \)
3. \( Z(0, 3) = 22(0) + 44(3) = 0 + 132 = 132 \)
4. \( Z(3, 0) = 22(3) + 44(0) = 66 + 0 = 66 \)
#### Step 5: Determine Maximum Value
The maximum value of \( Z \) occurs at the point \(\left(\frac{24}{11}, \frac{24}{11}\right)\):
\[ Z_{\text{max}} \approx 144 \]
### Final Answer for Problem (iii)
The maximum value of \( Z \) is approximately **144** at the point **\(\left(\frac{24}{11}, \frac{24}{11}\right)\)**.
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