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.
