Answer :
To solve the linear programming problem [tex]\(\operatorname{Min} Z=4 X_1+2 X_2\)[/tex] subject to the given constraints, we can use the graphical method. Let's walk through the steps in detail:
### Step 1: Rewrite the Inequality Constraints as Equalities
First, convert the inequalities to equalities to find the boundary lines.
1. [tex]\(3X_1 + X_2 \geq 27\)[/tex]
[tex]\[ 3X_1 + X_2 = 27 \][/tex]
2. [tex]\(-X_1 - X_2 \leq -27\)[/tex] (which simplifies to [tex]\(X_1 + X_2 \geq 27\)[/tex])
[tex]\[ X_1 + X_2 = 27 \][/tex]
3. [tex]\(X_1 + 2X_2 \leq 30\)[/tex]
[tex]\[ X_1 + 2X_2 = 30 \][/tex]
4. [tex]\(X_1 \geq 0\)[/tex]
5. [tex]\(X_2 \geq 0\)[/tex]
### Step 2: Plot the Lines
Next, we will plot these lines on a graph.
1. For [tex]\(3X_1 + X_2 = 27\)[/tex]:
- When [tex]\(X_1 = 0\)[/tex], [tex]\(X_2 = 27\)[/tex].
- When [tex]\(X_2 = 0\)[/tex], [tex]\(X_1 = 9\)[/tex].
Plot points [tex]\((0, 27)\)[/tex] and [tex]\((9, 0)\)[/tex].
2. For [tex]\(X_1 + X_2 = 27\)[/tex]:
- When [tex]\(X_1 = 0\)[/tex], [tex]\(X_2 = 27\)[/tex].
- When [tex]\(X_2 = 0\)[/tex], [tex]\(X_1 = 27\)[/tex].
Plot points [tex]\((0, 27)\)[/tex] and [tex]\((27, 0)\)[/tex].
3. For [tex]\(X_1 + 2X_2 = 30\)[/tex]:
- When [tex]\(X_1 = 0\)[/tex], [tex]\(X_2 = 15\)[/tex].
- When [tex]\(X_2 = 0\)[/tex], [tex]\(X_1 = 30\)[/tex].
Plot points [tex]\((0, 15)\)[/tex] and [tex]\((30, 0)\)[/tex].
### Step 3: Identify Feasible Region
The feasible region is determined by the intersection of these constraints:
- [tex]\(3X_1 + X_2 \geq 27\)[/tex]
- [tex]\(X_1 + X_2 \geq 27\)[/tex]
- [tex]\(X_1 + 2X_2 \leq 30\)[/tex]
- [tex]\(X_1 \geq 0\)[/tex]
- [tex]\(X_2 \geq 0\)[/tex]
### Step 4: Find Vertices of the Feasible Region
To find the vertices, solve the intersections of the lines:
1. Intersection of [tex]\(3X_1 + X_2 = 27\)[/tex] and [tex]\(X_1 + X_2 = 27\)[/tex]:
[tex]\[ 3X_1 + X_2 = 27 \][/tex]
[tex]\[ X_1 + X_2 = 27 \][/tex]
Subtract the second equation from the first:
[tex]\[ 2X_1 = 0 \implies X_1 = 0 \implies X_2 = 27 \][/tex]
Intersection point: [tex]\((0, 27)\)[/tex]
2. Intersection of [tex]\(3X_1 + X_2 = 27\)[/tex] and [tex]\(X_1 + 2X_2 = 30\)[/tex]:
[tex]\[ 3X_1 + X_2 = 27 \][/tex]
[tex]\[ X_1 + 2X_2 = 30 \][/tex]
Multiply the second equation by 3:
[tex]\[ 3X_1 + 6X_2 = 90 \][/tex]
Subtract the first equation from this:
[tex]\[ 5X_2 = 63 \implies X_2 = 12.6 \implies X_1 = 30 - 2(12.6) = 4.8 \][/tex]
Intersection point: [tex]\((4.8, 12.6)\)[/tex]
3. Intersection of [tex]\(X_1 + X_2 = 27\)[/tex] and [tex]\(X_1 + 2X_2 = 30\)[/tex]:
[tex]\[ X_1 + X_2 = 27 \][/tex]
[tex]\[ X_1 + 2X_2 = 30 \][/tex]
Subtract the first equation from the second:
[tex]\[ X_2 = 3 \implies X_1 = 24 \][/tex]
Intersection point: [tex]\((24, 3)\)[/tex]
### Step 5: Evaluate the Objective Function at Vertices
Calculate [tex]\(Z\)[/tex] at each vertex:
1. At [tex]\((0, 27)\)[/tex]:
[tex]\[ Z = 4 \times 0 + 2 \times 27 = 54 \][/tex]
2. At [tex]\((4.8, 12.6)\)[/tex]:
[tex]\[ Z = 4 \times 4.8 + 2 \times 12.6 = 19.2 + 25.2 = 44.4 \][/tex]
3. At [tex]\((24, 3)\)[/tex]:
[tex]\[ Z = 4 \times 24 + 2 \times 3 = 96 + 6 = 102 \][/tex]
### Step 6: Identify the Minimum Value
The minimum value of [tex]\(Z\)[/tex] is [tex]\(44.4\)[/tex], occurring at the point [tex]\((4.8, 12.6)\)[/tex].
### Conclusion
The solution to the given linear programming problem is:
[tex]\[ X_1 = 4.8, \quad X_2 = 12.6, \quad \text{with} \quad Z_{\text{min}} = 44.4. \][/tex]
### Step 1: Rewrite the Inequality Constraints as Equalities
First, convert the inequalities to equalities to find the boundary lines.
1. [tex]\(3X_1 + X_2 \geq 27\)[/tex]
[tex]\[ 3X_1 + X_2 = 27 \][/tex]
2. [tex]\(-X_1 - X_2 \leq -27\)[/tex] (which simplifies to [tex]\(X_1 + X_2 \geq 27\)[/tex])
[tex]\[ X_1 + X_2 = 27 \][/tex]
3. [tex]\(X_1 + 2X_2 \leq 30\)[/tex]
[tex]\[ X_1 + 2X_2 = 30 \][/tex]
4. [tex]\(X_1 \geq 0\)[/tex]
5. [tex]\(X_2 \geq 0\)[/tex]
### Step 2: Plot the Lines
Next, we will plot these lines on a graph.
1. For [tex]\(3X_1 + X_2 = 27\)[/tex]:
- When [tex]\(X_1 = 0\)[/tex], [tex]\(X_2 = 27\)[/tex].
- When [tex]\(X_2 = 0\)[/tex], [tex]\(X_1 = 9\)[/tex].
Plot points [tex]\((0, 27)\)[/tex] and [tex]\((9, 0)\)[/tex].
2. For [tex]\(X_1 + X_2 = 27\)[/tex]:
- When [tex]\(X_1 = 0\)[/tex], [tex]\(X_2 = 27\)[/tex].
- When [tex]\(X_2 = 0\)[/tex], [tex]\(X_1 = 27\)[/tex].
Plot points [tex]\((0, 27)\)[/tex] and [tex]\((27, 0)\)[/tex].
3. For [tex]\(X_1 + 2X_2 = 30\)[/tex]:
- When [tex]\(X_1 = 0\)[/tex], [tex]\(X_2 = 15\)[/tex].
- When [tex]\(X_2 = 0\)[/tex], [tex]\(X_1 = 30\)[/tex].
Plot points [tex]\((0, 15)\)[/tex] and [tex]\((30, 0)\)[/tex].
### Step 3: Identify Feasible Region
The feasible region is determined by the intersection of these constraints:
- [tex]\(3X_1 + X_2 \geq 27\)[/tex]
- [tex]\(X_1 + X_2 \geq 27\)[/tex]
- [tex]\(X_1 + 2X_2 \leq 30\)[/tex]
- [tex]\(X_1 \geq 0\)[/tex]
- [tex]\(X_2 \geq 0\)[/tex]
### Step 4: Find Vertices of the Feasible Region
To find the vertices, solve the intersections of the lines:
1. Intersection of [tex]\(3X_1 + X_2 = 27\)[/tex] and [tex]\(X_1 + X_2 = 27\)[/tex]:
[tex]\[ 3X_1 + X_2 = 27 \][/tex]
[tex]\[ X_1 + X_2 = 27 \][/tex]
Subtract the second equation from the first:
[tex]\[ 2X_1 = 0 \implies X_1 = 0 \implies X_2 = 27 \][/tex]
Intersection point: [tex]\((0, 27)\)[/tex]
2. Intersection of [tex]\(3X_1 + X_2 = 27\)[/tex] and [tex]\(X_1 + 2X_2 = 30\)[/tex]:
[tex]\[ 3X_1 + X_2 = 27 \][/tex]
[tex]\[ X_1 + 2X_2 = 30 \][/tex]
Multiply the second equation by 3:
[tex]\[ 3X_1 + 6X_2 = 90 \][/tex]
Subtract the first equation from this:
[tex]\[ 5X_2 = 63 \implies X_2 = 12.6 \implies X_1 = 30 - 2(12.6) = 4.8 \][/tex]
Intersection point: [tex]\((4.8, 12.6)\)[/tex]
3. Intersection of [tex]\(X_1 + X_2 = 27\)[/tex] and [tex]\(X_1 + 2X_2 = 30\)[/tex]:
[tex]\[ X_1 + X_2 = 27 \][/tex]
[tex]\[ X_1 + 2X_2 = 30 \][/tex]
Subtract the first equation from the second:
[tex]\[ X_2 = 3 \implies X_1 = 24 \][/tex]
Intersection point: [tex]\((24, 3)\)[/tex]
### Step 5: Evaluate the Objective Function at Vertices
Calculate [tex]\(Z\)[/tex] at each vertex:
1. At [tex]\((0, 27)\)[/tex]:
[tex]\[ Z = 4 \times 0 + 2 \times 27 = 54 \][/tex]
2. At [tex]\((4.8, 12.6)\)[/tex]:
[tex]\[ Z = 4 \times 4.8 + 2 \times 12.6 = 19.2 + 25.2 = 44.4 \][/tex]
3. At [tex]\((24, 3)\)[/tex]:
[tex]\[ Z = 4 \times 24 + 2 \times 3 = 96 + 6 = 102 \][/tex]
### Step 6: Identify the Minimum Value
The minimum value of [tex]\(Z\)[/tex] is [tex]\(44.4\)[/tex], occurring at the point [tex]\((4.8, 12.6)\)[/tex].
### Conclusion
The solution to the given linear programming problem is:
[tex]\[ X_1 = 4.8, \quad X_2 = 12.6, \quad \text{with} \quad Z_{\text{min}} = 44.4. \][/tex]
