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To solve the given linear programming problem, we aim to minimize the objective function [tex]\( 3x_2 + 2x_1 \)[/tex] subject to the constraints:
1. [tex]\( 8 + x_1 - 3x_2 \geq 3 \)[/tex]
2. [tex]\( 2x_2 \leq 7 \)[/tex]
3. [tex]\( x_1 + 2x_2 \leq 5 \)[/tex]
4. [tex]\( x_1 \geq 0 \)[/tex]
5. [tex]\( x_2 \geq 0 \)[/tex]
Let's break it down step-by-step:
### Step 1: Convert the Constraints
Firstly, we need to convert the inequalities into a standard form, which would look like [tex]\( \mathbf{A}x \leq \mathbf{b} \)[/tex].
1. [tex]\( 8 + x_1 - 3x_2 \geq 3 \)[/tex] can be rewritten by subtracting 8 from both sides:
[tex]\[ x_1 - 3x_2 \geq -5 \quad \text{or by multiplying by -1 we get} \quad -x_1 + 3x_2 \leq 5 \][/tex]
2. [tex]\( 2x_2 \leq 7 \)[/tex]
3. [tex]\( x_1 + 2x_2 \leq 5 \)[/tex]
### Step 2: Standard Form and Coefficients
We have converted constraints in standard form:
[tex]\[ \begin{cases} -x_1 + 3x_2 \leq 5 \\ 2x_2 \leq 7 \\ x_1 + 2x_2 \leq 5 \end{cases} \][/tex]
### Step 3: Set Up the Objective Function
The objective function is already in the correct form, and here we aim to minimize:
[tex]\[ 3x_2 + 2x_1 \][/tex]
### Step 4: Define the Bounds for Variables
The non-negativity constraints for the variables are [tex]\( x_1 \geq 0 \)[/tex] and [tex]\( x_2 \geq 0 \)[/tex], or in standard form:
[tex]\[ \begin{cases} x_1 \geq 0 \\ x_2 \geq 0 \end{cases} \][/tex]
### Step 5: Solve using Linear Programming
Given our constraints and bounds:
Inequality Constraints Coefficients Matrix A:
[tex]\[ \mathbf{A} = \begin{bmatrix} -1 & 3 \\ 0 & 2 \\ 1 & 2 \\ \end{bmatrix} \][/tex]
Right-hand Side Vector b:
[tex]\[ \mathbf{b} = \begin{bmatrix} 5 \\ 7 \\ 5 \\ \end{bmatrix} \][/tex]
Objective Function Coefficients Vector c:
[tex]\[ \mathbf{c} = \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix} \][/tex]
Bounds for Variables:
[tex]\[ \begin{cases} 0 \leq x_1 < \infty \\ 0 \leq x_2 < \infty \\ \end{cases} \][/tex]
### Solution Interpretation
By solving this set of linear inequalities and the objective function with the specified constraints and bounds, the optimal values obtained are:
[tex]\[ x_1 = 0.0, \quad x_2 = 0.0 \][/tex]
And the minimized value of the objective function:
[tex]\[ 3x_2 + 2x_1 = 3 \cdot 0.0 + 2 \cdot 0.0 = 0.0 \][/tex]
Hence, the optimal solution values are [tex]\( x_1 = 0 \)[/tex], [tex]\( x_2 = 0 \)[/tex], and the minimum value of the objective function is 0.
1. [tex]\( 8 + x_1 - 3x_2 \geq 3 \)[/tex]
2. [tex]\( 2x_2 \leq 7 \)[/tex]
3. [tex]\( x_1 + 2x_2 \leq 5 \)[/tex]
4. [tex]\( x_1 \geq 0 \)[/tex]
5. [tex]\( x_2 \geq 0 \)[/tex]
Let's break it down step-by-step:
### Step 1: Convert the Constraints
Firstly, we need to convert the inequalities into a standard form, which would look like [tex]\( \mathbf{A}x \leq \mathbf{b} \)[/tex].
1. [tex]\( 8 + x_1 - 3x_2 \geq 3 \)[/tex] can be rewritten by subtracting 8 from both sides:
[tex]\[ x_1 - 3x_2 \geq -5 \quad \text{or by multiplying by -1 we get} \quad -x_1 + 3x_2 \leq 5 \][/tex]
2. [tex]\( 2x_2 \leq 7 \)[/tex]
3. [tex]\( x_1 + 2x_2 \leq 5 \)[/tex]
### Step 2: Standard Form and Coefficients
We have converted constraints in standard form:
[tex]\[ \begin{cases} -x_1 + 3x_2 \leq 5 \\ 2x_2 \leq 7 \\ x_1 + 2x_2 \leq 5 \end{cases} \][/tex]
### Step 3: Set Up the Objective Function
The objective function is already in the correct form, and here we aim to minimize:
[tex]\[ 3x_2 + 2x_1 \][/tex]
### Step 4: Define the Bounds for Variables
The non-negativity constraints for the variables are [tex]\( x_1 \geq 0 \)[/tex] and [tex]\( x_2 \geq 0 \)[/tex], or in standard form:
[tex]\[ \begin{cases} x_1 \geq 0 \\ x_2 \geq 0 \end{cases} \][/tex]
### Step 5: Solve using Linear Programming
Given our constraints and bounds:
Inequality Constraints Coefficients Matrix A:
[tex]\[ \mathbf{A} = \begin{bmatrix} -1 & 3 \\ 0 & 2 \\ 1 & 2 \\ \end{bmatrix} \][/tex]
Right-hand Side Vector b:
[tex]\[ \mathbf{b} = \begin{bmatrix} 5 \\ 7 \\ 5 \\ \end{bmatrix} \][/tex]
Objective Function Coefficients Vector c:
[tex]\[ \mathbf{c} = \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix} \][/tex]
Bounds for Variables:
[tex]\[ \begin{cases} 0 \leq x_1 < \infty \\ 0 \leq x_2 < \infty \\ \end{cases} \][/tex]
### Solution Interpretation
By solving this set of linear inequalities and the objective function with the specified constraints and bounds, the optimal values obtained are:
[tex]\[ x_1 = 0.0, \quad x_2 = 0.0 \][/tex]
And the minimized value of the objective function:
[tex]\[ 3x_2 + 2x_1 = 3 \cdot 0.0 + 2 \cdot 0.0 = 0.0 \][/tex]
Hence, the optimal solution values are [tex]\( x_1 = 0 \)[/tex], [tex]\( x_2 = 0 \)[/tex], and the minimum value of the objective function is 0.
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