Answer :
Sure, let's solve this problem using graphical methods step-by-step. We want to minimize the objective function [tex]\( Z = 4X_1 + 2X_2 \)[/tex] subject to the given constraints:
[tex]\[ \begin{array}{r} 3X_1 + X_2 \geq 27 \\ -X_1 - X_2 \leq -27 \\ X_1 + 2X_2 \leq 30 \\ X_1, X_2 \geq 0 \end{array} \][/tex]
### Step 1: Convert Inequalities to Equations
We will start by converting the inequalities to equations to find the boundary lines.
1. [tex]\( 3X_1 + X_2 = 27 \)[/tex]
2. [tex]\( -X_1 - X_2 = -27 \rightarrow X_1 + X_2 = 27 \)[/tex]
3. [tex]\( X_1 + 2X_2 = 30 \)[/tex]
### Step 2: Graph the Feasible Region
Plot these lines on graph paper or using graphing software.
1. For [tex]\( 3X_1 + X_2 = 27 \)[/tex]:
- Let [tex]\( X_1 = 0 \)[/tex]: [tex]\( X_2 = 27 \)[/tex]
- Let [tex]\( X_2 = 0 \)[/tex]: [tex]\( X_1 = 9 \)[/tex]
2. For [tex]\( X_1 + X_2 = 27 \)[/tex]:
- Let [tex]\( X_1 = 0 \)[/tex]: [tex]\( X_2 = 27 \)[/tex]
- Let [tex]\( X_2 = 0 \)[/tex]: [tex]\( X_1 = 27 \)[/tex]
3. For [tex]\( X_1 + 2X_2 = 30 \)[/tex]:
- Let [tex]\( X_1 = 0 \)[/tex]: [tex]\( X_2 = 15 \)[/tex]
- Let [tex]\( X_2 = 0 \)[/tex]: [tex]\( X_1 = 30 \)[/tex]
Also, include the non-negativity constraints [tex]\( X_1 \geq 0 \)[/tex] and [tex]\( X_2 \geq 0 \)[/tex] which confines the solution to the first quadrant.
### Step 3: Find the Intersection Points
Find the intersection points of the lines to determine the vertices of the feasible region.
1. Intersection of [tex]\( 3X_1 + X_2 = 27 \)[/tex] and [tex]\( X_1 + X_2 = 27 \)[/tex]:
- Set the equations equal: [tex]\( 3X_1 + X_2 = X_1 + X_2 \)[/tex]
- Simplifying: [tex]\( 2X_1 = 0 \rightarrow X_1 = 0 \)[/tex]
- Plug [tex]\( X_1 = 0 \)[/tex] into [tex]\( X_1 + X_2 = 27 \)[/tex]: [tex]\( X_2 = 27 \)[/tex]
- Point [tex]\( (0, 27) \)[/tex]
2. Intersection of [tex]\( 3X_1 + X_2 = 27 \)[/tex] and [tex]\( X_1 + 2X_2 = 30 \)[/tex]:
- Set the equations: [tex]\( 3X_1 + X_2 = 27 \)[/tex] and [tex]\( X_1 + 2X_2 = 30 \)[/tex]
- Solve these simultaneously:
[tex]\[ 3X_1 + X_2 = 27 \quad \text{(i)} \][/tex]
[tex]\[ X_1 + 2X_2 = 30 \quad \text{(ii)} \][/tex]
Multiply equation (ii) by 3: [tex]\( 3X_1 + 6X_2 = 90 \)[/tex]
- Subtract equation (i) from the new equation:
[tex]\[ 5X_2 = 63 \rightarrow X_2 = \frac{63}{5} = 12.6 \][/tex]
- Solve for [tex]\( X_1 \)[/tex] in equation (ii):
[tex]\[ X_1 + 2(12.6) = 30 \rightarrow X_1 + 25.2 = 30 \rightarrow X_1 = 4.8 \][/tex]
- 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]:
- Set equations: [tex]\( X_1 + X_2 = 27 \)[/tex] and [tex]\( X_1 + 2X_2 = 30 \)[/tex]
- Solve these simultaneously:
[tex]\[ X_1 + X_2 = 27 \quad \text{(i)} \][/tex]
[tex]\[ X_1 + 2X_2 = 30 \quad \text{(ii)} \][/tex]
- Subtract equation (i) from (ii):
[tex]\[ X_2 = 3 \][/tex]
- Solve for [tex]\( X_1 \)[/tex]:
[tex]\[ X_1 + 3 = 27 \rightarrow X_1 = 24 \][/tex]
- Point [tex]\( (24, 3) \)[/tex]
### Step 4: Evaluate the Objective Function at Each Vertex
Evaluate [tex]\( Z = 4X_1 + 2X_2 \)[/tex] at each of these points to find the minimum value:
1. At [tex]\( (0, 27) \)[/tex]: [tex]\( Z = 4(0) + 2(27) = 54 \)[/tex]
2. At [tex]\( (4.8, 12.6) \)[/tex]: [tex]\( Z = 4(4.8) + 2(12.6) = 19.2 + 25.2 = 44.4 \)[/tex]
3. At [tex]\( (24, 3) \)[/tex]: [tex]\( Z = 4(24) + 2(3) = 96 + 6 = 102 \)[/tex]
### Conclusion
The minimum value of [tex]\( Z \)[/tex] occurs at the point [tex]\( (4.8, 12.6) \)[/tex] with [tex]\( Z = 44.4 \)[/tex].
Therefore, the solved optimization problem is:
Minimum [tex]\( Z = 44.4 \)[/tex] at [tex]\( X_1 = 4.8 \)[/tex] and [tex]\( X_2 = 12.6 \)[/tex].
[tex]\[ \begin{array}{r} 3X_1 + X_2 \geq 27 \\ -X_1 - X_2 \leq -27 \\ X_1 + 2X_2 \leq 30 \\ X_1, X_2 \geq 0 \end{array} \][/tex]
### Step 1: Convert Inequalities to Equations
We will start by converting the inequalities to equations to find the boundary lines.
1. [tex]\( 3X_1 + X_2 = 27 \)[/tex]
2. [tex]\( -X_1 - X_2 = -27 \rightarrow X_1 + X_2 = 27 \)[/tex]
3. [tex]\( X_1 + 2X_2 = 30 \)[/tex]
### Step 2: Graph the Feasible Region
Plot these lines on graph paper or using graphing software.
1. For [tex]\( 3X_1 + X_2 = 27 \)[/tex]:
- Let [tex]\( X_1 = 0 \)[/tex]: [tex]\( X_2 = 27 \)[/tex]
- Let [tex]\( X_2 = 0 \)[/tex]: [tex]\( X_1 = 9 \)[/tex]
2. For [tex]\( X_1 + X_2 = 27 \)[/tex]:
- Let [tex]\( X_1 = 0 \)[/tex]: [tex]\( X_2 = 27 \)[/tex]
- Let [tex]\( X_2 = 0 \)[/tex]: [tex]\( X_1 = 27 \)[/tex]
3. For [tex]\( X_1 + 2X_2 = 30 \)[/tex]:
- Let [tex]\( X_1 = 0 \)[/tex]: [tex]\( X_2 = 15 \)[/tex]
- Let [tex]\( X_2 = 0 \)[/tex]: [tex]\( X_1 = 30 \)[/tex]
Also, include the non-negativity constraints [tex]\( X_1 \geq 0 \)[/tex] and [tex]\( X_2 \geq 0 \)[/tex] which confines the solution to the first quadrant.
### Step 3: Find the Intersection Points
Find the intersection points of the lines to determine the vertices of the feasible region.
1. Intersection of [tex]\( 3X_1 + X_2 = 27 \)[/tex] and [tex]\( X_1 + X_2 = 27 \)[/tex]:
- Set the equations equal: [tex]\( 3X_1 + X_2 = X_1 + X_2 \)[/tex]
- Simplifying: [tex]\( 2X_1 = 0 \rightarrow X_1 = 0 \)[/tex]
- Plug [tex]\( X_1 = 0 \)[/tex] into [tex]\( X_1 + X_2 = 27 \)[/tex]: [tex]\( X_2 = 27 \)[/tex]
- Point [tex]\( (0, 27) \)[/tex]
2. Intersection of [tex]\( 3X_1 + X_2 = 27 \)[/tex] and [tex]\( X_1 + 2X_2 = 30 \)[/tex]:
- Set the equations: [tex]\( 3X_1 + X_2 = 27 \)[/tex] and [tex]\( X_1 + 2X_2 = 30 \)[/tex]
- Solve these simultaneously:
[tex]\[ 3X_1 + X_2 = 27 \quad \text{(i)} \][/tex]
[tex]\[ X_1 + 2X_2 = 30 \quad \text{(ii)} \][/tex]
Multiply equation (ii) by 3: [tex]\( 3X_1 + 6X_2 = 90 \)[/tex]
- Subtract equation (i) from the new equation:
[tex]\[ 5X_2 = 63 \rightarrow X_2 = \frac{63}{5} = 12.6 \][/tex]
- Solve for [tex]\( X_1 \)[/tex] in equation (ii):
[tex]\[ X_1 + 2(12.6) = 30 \rightarrow X_1 + 25.2 = 30 \rightarrow X_1 = 4.8 \][/tex]
- 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]:
- Set equations: [tex]\( X_1 + X_2 = 27 \)[/tex] and [tex]\( X_1 + 2X_2 = 30 \)[/tex]
- Solve these simultaneously:
[tex]\[ X_1 + X_2 = 27 \quad \text{(i)} \][/tex]
[tex]\[ X_1 + 2X_2 = 30 \quad \text{(ii)} \][/tex]
- Subtract equation (i) from (ii):
[tex]\[ X_2 = 3 \][/tex]
- Solve for [tex]\( X_1 \)[/tex]:
[tex]\[ X_1 + 3 = 27 \rightarrow X_1 = 24 \][/tex]
- Point [tex]\( (24, 3) \)[/tex]
### Step 4: Evaluate the Objective Function at Each Vertex
Evaluate [tex]\( Z = 4X_1 + 2X_2 \)[/tex] at each of these points to find the minimum value:
1. At [tex]\( (0, 27) \)[/tex]: [tex]\( Z = 4(0) + 2(27) = 54 \)[/tex]
2. At [tex]\( (4.8, 12.6) \)[/tex]: [tex]\( Z = 4(4.8) + 2(12.6) = 19.2 + 25.2 = 44.4 \)[/tex]
3. At [tex]\( (24, 3) \)[/tex]: [tex]\( Z = 4(24) + 2(3) = 96 + 6 = 102 \)[/tex]
### Conclusion
The minimum value of [tex]\( Z \)[/tex] occurs at the point [tex]\( (4.8, 12.6) \)[/tex] with [tex]\( Z = 44.4 \)[/tex].
Therefore, the solved optimization problem is:
Minimum [tex]\( Z = 44.4 \)[/tex] at [tex]\( X_1 = 4.8 \)[/tex] and [tex]\( X_2 = 12.6 \)[/tex].
