Answer :
To solve this linear programming problem using graphical methods, we will follow these steps:
1. Identify and plot the constraint lines.
2. Identify the feasible region.
3. Determine the vertices (corner points) of the feasible region.
4. Evaluate the objective function at each vertex to find the minimum value.
Let's begin with the constraints and plot them on a graph.
### 1. Constraints:
We have four constraints:
1. [tex]\( 3x_1 + x_2 \geq 27 \)[/tex]
2. [tex]\( -x_1 - x_2 \leq -27 \)[/tex]
3. [tex]\( x_1 + 2x_2 \leq 30 \)[/tex]
4. [tex]\( x_1 \geq 0 \)[/tex]
5. [tex]\( x_2 \geq 0 \)[/tex]
To make it easier, we can convert the inequalities to equalities to find the boundary lines:
First Constraint:
[tex]\[ 3x_1 + x_2 = 27 \][/tex]
To find intercepts:
- Set [tex]\( x_1 = 0 \)[/tex], then [tex]\( x_2 = 27 \)[/tex].
- Set [tex]\( x_2 = 0 \)[/tex], then [tex]\( x_1 = 9 \)[/tex].
So, the line passes through points [tex]\((0, 27)\)[/tex] and [tex]\((9, 0)\)[/tex].
Second Constraint:
[tex]\[ -x_1 - x_2 = -27 \][/tex]
[tex]\[ x_1 + x_2 = 27 \][/tex] (after multiplying both sides by -1)
To find intercepts:
- Set [tex]\( x_1 = 0 \)[/tex], then [tex]\( x_2 = 27 \)[/tex].
- Set [tex]\( x_2 = 0 \)[/tex], then [tex]\( x_1 = 27 \)[/tex].
So, the line passes through points [tex]\((0, 27)\)[/tex] and [tex]\((27, 0)\)[/tex].
Third Constraint:
[tex]\[ x_1 + 2x_2 = 30 \][/tex]
To find intercepts:
- Set [tex]\( x_1 = 0 \)[/tex], then [tex]\( x_2 = 15 \)[/tex].
- Set [tex]\( x_2 = 0 \)[/tex], then [tex]\( x_1 = 30 \)[/tex].
So, the line passes through points [tex]\((0, 15)\)[/tex] and [tex]\((30, 0)\)[/tex].
Non-negativity constraints:
[tex]\[ x_1 \geq 0 \][/tex]
[tex]\[ x_2 \geq 0 \][/tex]
### 2. Plotting the Constraints:
We create a graph and plot these lines:
- [tex]\( 3x_1 + x_2 = 27 \)[/tex] as a line.
- [tex]\( x_1 + x_2 = 27 \)[/tex] as a line.
- [tex]\( x_1 + 2x_2 = 30 \)[/tex] as a line.
- The [tex]\( x_1 \geq 0 \)[/tex] region is to the right of the y-axis.
- The [tex]\( x_2 \geq 0 \)[/tex] region is above the x-axis.
These lines will form a polygon (feasible region).
### 3. Finding the Feasible Region:
We need to shade the region that satisfies all inequalities. This intersection will give us the feasible region. The coordinates where the lines intersect each other or the axes give the vertices (points).
### 4. Points of Intersection:
Identifying intersections of lines:
1. Intersection of [tex]\( 3x_1 + x_2 = 27 \)[/tex] and [tex]\( x_1 + x_2 = 27 \)[/tex]:
Solve simultaneously:
[tex]\[ 3x_1 + x_2 = 27 \][/tex]
[tex]\[ x_1 + x_2 = 27 \][/tex]
Subtract the second equation from the first:
[tex]\[ 2x_1 = 0 \][/tex]
[tex]\[ x_1 = 0 \][/tex]
Substitute [tex]\( x_1 = 0 \)[/tex] in [tex]\( x_1 + x_2 = 27 \)[/tex]:
[tex]\[ x_2 = 27 \][/tex]
So, one point is [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 first equation by 2:
[tex]\[ 6x_1 + 2x_2 = 54 \][/tex]
Subtract the second equation from this:
[tex]\[ 5x_1 = 24 \][/tex]
[tex]\[ x_1 = \frac{24}{5} = 4.8 \][/tex]
Substitute [tex]\( x_1 = 4.8 \)[/tex] in [tex]\( x_1 + 2x_2 = 30 \)[/tex]:
[tex]\[ 4.8 + 2x_2 = 30 \][/tex]
[tex]\[ 2x_2 = 25.2 \][/tex]
[tex]\[ x_2 = 12.6 \][/tex]
So, the second point is [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]\[ x_2 = 3 \][/tex]
Substitute [tex]\( x_2 = 3 \)[/tex] in [tex]\( x_1 + x_2 = 27 \)[/tex]:
[tex]\[ x_1 = 24 \][/tex]
So, the third point is [tex]\((24, 3)\)[/tex].
Combine with intercepts unless any point is negative in value as we want [tex]\( x_1, x_2 \geq 0 \)[/tex].
### 5. Evaluating the Objective Function:
Evaluate [tex]\( Z = 4x_1 + 2x_2 \)[/tex] at each corner point:
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]
Thus, minimization identifies the minimum.
### Conclusion:
The minimum value of [tex]\( Z \)[/tex] occurs at [tex]\((4.8, 12.6)\)[/tex] with [tex]\(\min Z = 44.4\)[/tex].
1. Identify and plot the constraint lines.
2. Identify the feasible region.
3. Determine the vertices (corner points) of the feasible region.
4. Evaluate the objective function at each vertex to find the minimum value.
Let's begin with the constraints and plot them on a graph.
### 1. Constraints:
We have four constraints:
1. [tex]\( 3x_1 + x_2 \geq 27 \)[/tex]
2. [tex]\( -x_1 - x_2 \leq -27 \)[/tex]
3. [tex]\( x_1 + 2x_2 \leq 30 \)[/tex]
4. [tex]\( x_1 \geq 0 \)[/tex]
5. [tex]\( x_2 \geq 0 \)[/tex]
To make it easier, we can convert the inequalities to equalities to find the boundary lines:
First Constraint:
[tex]\[ 3x_1 + x_2 = 27 \][/tex]
To find intercepts:
- Set [tex]\( x_1 = 0 \)[/tex], then [tex]\( x_2 = 27 \)[/tex].
- Set [tex]\( x_2 = 0 \)[/tex], then [tex]\( x_1 = 9 \)[/tex].
So, the line passes through points [tex]\((0, 27)\)[/tex] and [tex]\((9, 0)\)[/tex].
Second Constraint:
[tex]\[ -x_1 - x_2 = -27 \][/tex]
[tex]\[ x_1 + x_2 = 27 \][/tex] (after multiplying both sides by -1)
To find intercepts:
- Set [tex]\( x_1 = 0 \)[/tex], then [tex]\( x_2 = 27 \)[/tex].
- Set [tex]\( x_2 = 0 \)[/tex], then [tex]\( x_1 = 27 \)[/tex].
So, the line passes through points [tex]\((0, 27)\)[/tex] and [tex]\((27, 0)\)[/tex].
Third Constraint:
[tex]\[ x_1 + 2x_2 = 30 \][/tex]
To find intercepts:
- Set [tex]\( x_1 = 0 \)[/tex], then [tex]\( x_2 = 15 \)[/tex].
- Set [tex]\( x_2 = 0 \)[/tex], then [tex]\( x_1 = 30 \)[/tex].
So, the line passes through points [tex]\((0, 15)\)[/tex] and [tex]\((30, 0)\)[/tex].
Non-negativity constraints:
[tex]\[ x_1 \geq 0 \][/tex]
[tex]\[ x_2 \geq 0 \][/tex]
### 2. Plotting the Constraints:
We create a graph and plot these lines:
- [tex]\( 3x_1 + x_2 = 27 \)[/tex] as a line.
- [tex]\( x_1 + x_2 = 27 \)[/tex] as a line.
- [tex]\( x_1 + 2x_2 = 30 \)[/tex] as a line.
- The [tex]\( x_1 \geq 0 \)[/tex] region is to the right of the y-axis.
- The [tex]\( x_2 \geq 0 \)[/tex] region is above the x-axis.
These lines will form a polygon (feasible region).
### 3. Finding the Feasible Region:
We need to shade the region that satisfies all inequalities. This intersection will give us the feasible region. The coordinates where the lines intersect each other or the axes give the vertices (points).
### 4. Points of Intersection:
Identifying intersections of lines:
1. Intersection of [tex]\( 3x_1 + x_2 = 27 \)[/tex] and [tex]\( x_1 + x_2 = 27 \)[/tex]:
Solve simultaneously:
[tex]\[ 3x_1 + x_2 = 27 \][/tex]
[tex]\[ x_1 + x_2 = 27 \][/tex]
Subtract the second equation from the first:
[tex]\[ 2x_1 = 0 \][/tex]
[tex]\[ x_1 = 0 \][/tex]
Substitute [tex]\( x_1 = 0 \)[/tex] in [tex]\( x_1 + x_2 = 27 \)[/tex]:
[tex]\[ x_2 = 27 \][/tex]
So, one point is [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 first equation by 2:
[tex]\[ 6x_1 + 2x_2 = 54 \][/tex]
Subtract the second equation from this:
[tex]\[ 5x_1 = 24 \][/tex]
[tex]\[ x_1 = \frac{24}{5} = 4.8 \][/tex]
Substitute [tex]\( x_1 = 4.8 \)[/tex] in [tex]\( x_1 + 2x_2 = 30 \)[/tex]:
[tex]\[ 4.8 + 2x_2 = 30 \][/tex]
[tex]\[ 2x_2 = 25.2 \][/tex]
[tex]\[ x_2 = 12.6 \][/tex]
So, the second point is [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]\[ x_2 = 3 \][/tex]
Substitute [tex]\( x_2 = 3 \)[/tex] in [tex]\( x_1 + x_2 = 27 \)[/tex]:
[tex]\[ x_1 = 24 \][/tex]
So, the third point is [tex]\((24, 3)\)[/tex].
Combine with intercepts unless any point is negative in value as we want [tex]\( x_1, x_2 \geq 0 \)[/tex].
### 5. Evaluating the Objective Function:
Evaluate [tex]\( Z = 4x_1 + 2x_2 \)[/tex] at each corner point:
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]
Thus, minimization identifies the minimum.
### Conclusion:
The minimum value of [tex]\( Z \)[/tex] occurs at [tex]\((4.8, 12.6)\)[/tex] with [tex]\(\min Z = 44.4\)[/tex].
