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To solve this linear programming problem, we aim to maximize the objective function [tex]\(Z = 4x_1 + 8x_2\)[/tex] subject to the given constraints:
1. [tex]\(2x_1 + 3x_2 \leq 48\)[/tex]
2. [tex]\(x_1 \leq 15\)[/tex]
3. [tex]\(x_2 \leq 10\)[/tex]
4. [tex]\(x_1 \geq 0\)[/tex]
5. [tex]\(x_2 \geq 0\)[/tex]
Here is the step-by-step solution:
1. Formulate the Problem:
- Objective function: [tex]\(Z = 4x_1 + 8x_2\)[/tex]
- Constraints:
- [tex]\(2x_1 + 3x_2 \leq 48\)[/tex]
- [tex]\(x_1 \leq 15\)[/tex]
- [tex]\(x_2 \leq 10\)[/tex]
- [tex]\(x_1 \geq 0\)[/tex]
- [tex]\(x_2 \geq 0\)[/tex]
2. Identify the Feasible Region:
The feasible region is defined by the intersection of the half-planes created by the constraints. The vertices of the feasible region need to be determined as follows:
- The intersection points of the constraints form the boundary of the feasible region.
3. Evaluate the Objective Function at the Vertices:
We need to evaluate the objective function [tex]\(Z = 4x_1 + 8x_2\)[/tex] at each vertex of the feasible region.
- Vertex at (0,0):
[tex]\(Z = 4(0) + 8(0) = 0\)[/tex]
- Vertex at the intersection of [tex]\(x_1 = 15\)[/tex] and [tex]\(2x_1 + 3x_2 \leq 48\)[/tex]:
Substituting [tex]\(x_1 = 15\)[/tex] into [tex]\(2x_1 + 3x_2 = 48\)[/tex], we get:
[tex]\(2(15) + 3x_2 = 48\)[/tex]
[tex]\(30 + 3x_2 = 48\)[/tex]
[tex]\(3x_2 = 18\)[/tex]
[tex]\(x_2 = 6\)[/tex]
So the vertex is (15, 6).
[tex]\(Z = 4(15) + 8(6) = 60 + 48 = 108\)[/tex]
- Vertex at the intersection of [tex]\(x_2 = 10\)[/tex] and [tex]\(2x_1 + 3x_2 \leq 48\)[/tex]:
Substituting [tex]\(x_2 = 10\)[/tex] into [tex]\(2x_1 + 3x_2 = 48\)[/tex], we get:
[tex]\(2x_1 + 3(10) = 48\)[/tex]
[tex]\(2x_1 + 30 = 48\)[/tex]
[tex]\(2x_1 = 18\)[/tex]
[tex]\(x_1 = 9\)[/tex]
So the vertex is (9, 10).
[tex]\(Z = 4(9) + 8(10) = 36 + 80 = 116\)[/tex]
- Vertex at the intersection of [tex]\(x_1 = 0\)[/tex] and [tex]\(x_2 = 10\)[/tex]:
So the vertex is (0, 10).
[tex]\(Z = 4(0) + 8(10) = 0 + 80 = 80\)[/tex]
- Vertex at the intersection of [tex]\(x_2 = 0\)[/tex] and [tex]\(x_1 = 15\)[/tex]:
So the vertex is (15, 0).
[tex]\(Z = 4(15) + 0 = 60\)[/tex]
4. Determine the Maximum Value:
From the above calculations, the objective function [tex]\(Z\)[/tex] achieves its maximum value at the vertex (9, 10).
Therefore, the optimal solution is:
- [tex]\(x_1 = 9\)[/tex]
- [tex]\(x_2 = 10\)[/tex]
- [tex]\(Z_{\max} = 116\)[/tex]
Thus, the maximum value of the objective function [tex]\(Z = 4x_1 + 8x_2\)[/tex] under the given constraints is [tex]\(116\)[/tex], achieved at [tex]\(x_1 = 9\)[/tex] and [tex]\(x_2 = 10\)[/tex].
1. [tex]\(2x_1 + 3x_2 \leq 48\)[/tex]
2. [tex]\(x_1 \leq 15\)[/tex]
3. [tex]\(x_2 \leq 10\)[/tex]
4. [tex]\(x_1 \geq 0\)[/tex]
5. [tex]\(x_2 \geq 0\)[/tex]
Here is the step-by-step solution:
1. Formulate the Problem:
- Objective function: [tex]\(Z = 4x_1 + 8x_2\)[/tex]
- Constraints:
- [tex]\(2x_1 + 3x_2 \leq 48\)[/tex]
- [tex]\(x_1 \leq 15\)[/tex]
- [tex]\(x_2 \leq 10\)[/tex]
- [tex]\(x_1 \geq 0\)[/tex]
- [tex]\(x_2 \geq 0\)[/tex]
2. Identify the Feasible Region:
The feasible region is defined by the intersection of the half-planes created by the constraints. The vertices of the feasible region need to be determined as follows:
- The intersection points of the constraints form the boundary of the feasible region.
3. Evaluate the Objective Function at the Vertices:
We need to evaluate the objective function [tex]\(Z = 4x_1 + 8x_2\)[/tex] at each vertex of the feasible region.
- Vertex at (0,0):
[tex]\(Z = 4(0) + 8(0) = 0\)[/tex]
- Vertex at the intersection of [tex]\(x_1 = 15\)[/tex] and [tex]\(2x_1 + 3x_2 \leq 48\)[/tex]:
Substituting [tex]\(x_1 = 15\)[/tex] into [tex]\(2x_1 + 3x_2 = 48\)[/tex], we get:
[tex]\(2(15) + 3x_2 = 48\)[/tex]
[tex]\(30 + 3x_2 = 48\)[/tex]
[tex]\(3x_2 = 18\)[/tex]
[tex]\(x_2 = 6\)[/tex]
So the vertex is (15, 6).
[tex]\(Z = 4(15) + 8(6) = 60 + 48 = 108\)[/tex]
- Vertex at the intersection of [tex]\(x_2 = 10\)[/tex] and [tex]\(2x_1 + 3x_2 \leq 48\)[/tex]:
Substituting [tex]\(x_2 = 10\)[/tex] into [tex]\(2x_1 + 3x_2 = 48\)[/tex], we get:
[tex]\(2x_1 + 3(10) = 48\)[/tex]
[tex]\(2x_1 + 30 = 48\)[/tex]
[tex]\(2x_1 = 18\)[/tex]
[tex]\(x_1 = 9\)[/tex]
So the vertex is (9, 10).
[tex]\(Z = 4(9) + 8(10) = 36 + 80 = 116\)[/tex]
- Vertex at the intersection of [tex]\(x_1 = 0\)[/tex] and [tex]\(x_2 = 10\)[/tex]:
So the vertex is (0, 10).
[tex]\(Z = 4(0) + 8(10) = 0 + 80 = 80\)[/tex]
- Vertex at the intersection of [tex]\(x_2 = 0\)[/tex] and [tex]\(x_1 = 15\)[/tex]:
So the vertex is (15, 0).
[tex]\(Z = 4(15) + 0 = 60\)[/tex]
4. Determine the Maximum Value:
From the above calculations, the objective function [tex]\(Z\)[/tex] achieves its maximum value at the vertex (9, 10).
Therefore, the optimal solution is:
- [tex]\(x_1 = 9\)[/tex]
- [tex]\(x_2 = 10\)[/tex]
- [tex]\(Z_{\max} = 116\)[/tex]
Thus, the maximum value of the objective function [tex]\(Z = 4x_1 + 8x_2\)[/tex] under the given constraints is [tex]\(116\)[/tex], achieved at [tex]\(x_1 = 9\)[/tex] and [tex]\(x_2 = 10\)[/tex].
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