Answer :
To solve the given linear programming problem using the simplex method, follow these detailed steps:
### Problem Formulation
The problem is to maximize the objective function:
[tex]\[ z = 3x_1 + 2x_2 + x_3 \][/tex]
subject to the constraints:
[tex]\[ 2x_1 + 2x_2 + x_3 \leq 10 \][/tex]
[tex]\[ x_1 + 2x_2 + 2x_3 \leq 11 \][/tex]
with non-negativity constraints:
[tex]\[ x_1 \geq 0, x_2 \geq 0, x_3 \geq 0 \][/tex]
### Step 1: Convert Inequalities to Equations with Slack Variables
To use the simplex method, we first convert the inequality constraints into equations by introducing slack variables [tex]\( s_1 \)[/tex] and [tex]\( s_2 \)[/tex]:
[tex]\[ 2x_1 + 2x_2 + x_3 + s_1 = 10 \][/tex]
[tex]\[ x_1 + 2x_2 + 2x_3 + s_2 = 11 \][/tex]
### Step 2: Initial Simplex Tableau
We set up the initial simplex tableau. The objective function can be rewritten as:
[tex]\[ z - 3x_1 - 2x_2 - x_3 = 0 \][/tex]
Our tableau looks like this:
[tex]\[ \begin{array}{c|cccccc|c} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \\ \hline s_1 & 2 & 2 & 1 & 1 & 0 & 10 \\ s_2 & 1 & 2 & 2 & 0 & 1 & 11 \\ \hline z & -3 & -2 & -1 & 0 & 0 & 0 \\ \end{array} \][/tex]
### Step 3: Perform Simplex Iterations
Let's perform the iterations:
1. Identify the entering variable: The most negative coefficient in the objective function row is [tex]\(-3\)[/tex] for [tex]\(x_1\)[/tex].
2. Identify the departing variable: Calculate the ratios of the RHS to the coefficients of [tex]\(x_1\)[/tex]:
- For [tex]\(s_1\)[/tex]: [tex]\( \frac{10}{2} = 5 \)[/tex]
- For [tex]\(s_2\)[/tex]: [tex]\( \frac{11}{1} = 11 \)[/tex]
The smallest non-negative ratio is 5, so [tex]\(s_1\)[/tex] will leave the basis, and [tex]\(x_1\)[/tex] will enter.
3. Pivot on the element [tex]\( a_{11} = 2 \)[/tex] and update the tableau.
- New row for [tex]\(x_1\)[/tex]: Divide the entire row [tex]\(s_1\)[/tex] by 2.
- Update the rest of the tableau accordingly.
After performing the required pivot operations, we eventually reach the final optimal tableau:
[tex]\[ \begin{array}{c|cccccc|c} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \\ \hline x_1 & 1 & 1 & 0.5 & 0.5 & 0 & 5 \\ s_2 & 0 & 1 & 1 & -0.5 & 1 & 6 \\ \hline z & 0 & -1 & 0.5 & 1.5 & 0 & 15 \\ \end{array} \][/tex]
### Step 4: Read off the Solution
The optimal solution from the final tableau:
- [tex]\( x_1 = 5 \)[/tex]
- [tex]\( x_2 = 0 \)[/tex]
- [tex]\( x_3 = 0 \)[/tex]
The maximum value of the objective function is:
[tex]\[ z = 15 \][/tex]
The values of the slack variables:
- [tex]\( s_1 = 0 \)[/tex]
- [tex]\( s_2 = 6 \)[/tex]
### Final Answer
A. The maximum is [tex]\( 15.0 \)[/tex] when [tex]\( x_1 = 5.0 \)[/tex], [tex]\( x_2 = 0.0 \)[/tex], [tex]\( x_3 = 0.0 \)[/tex], [tex]\( s_1 = 0.0 \)[/tex], and [tex]\( s_2 = 6.0 \)[/tex].
### Problem Formulation
The problem is to maximize the objective function:
[tex]\[ z = 3x_1 + 2x_2 + x_3 \][/tex]
subject to the constraints:
[tex]\[ 2x_1 + 2x_2 + x_3 \leq 10 \][/tex]
[tex]\[ x_1 + 2x_2 + 2x_3 \leq 11 \][/tex]
with non-negativity constraints:
[tex]\[ x_1 \geq 0, x_2 \geq 0, x_3 \geq 0 \][/tex]
### Step 1: Convert Inequalities to Equations with Slack Variables
To use the simplex method, we first convert the inequality constraints into equations by introducing slack variables [tex]\( s_1 \)[/tex] and [tex]\( s_2 \)[/tex]:
[tex]\[ 2x_1 + 2x_2 + x_3 + s_1 = 10 \][/tex]
[tex]\[ x_1 + 2x_2 + 2x_3 + s_2 = 11 \][/tex]
### Step 2: Initial Simplex Tableau
We set up the initial simplex tableau. The objective function can be rewritten as:
[tex]\[ z - 3x_1 - 2x_2 - x_3 = 0 \][/tex]
Our tableau looks like this:
[tex]\[ \begin{array}{c|cccccc|c} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \\ \hline s_1 & 2 & 2 & 1 & 1 & 0 & 10 \\ s_2 & 1 & 2 & 2 & 0 & 1 & 11 \\ \hline z & -3 & -2 & -1 & 0 & 0 & 0 \\ \end{array} \][/tex]
### Step 3: Perform Simplex Iterations
Let's perform the iterations:
1. Identify the entering variable: The most negative coefficient in the objective function row is [tex]\(-3\)[/tex] for [tex]\(x_1\)[/tex].
2. Identify the departing variable: Calculate the ratios of the RHS to the coefficients of [tex]\(x_1\)[/tex]:
- For [tex]\(s_1\)[/tex]: [tex]\( \frac{10}{2} = 5 \)[/tex]
- For [tex]\(s_2\)[/tex]: [tex]\( \frac{11}{1} = 11 \)[/tex]
The smallest non-negative ratio is 5, so [tex]\(s_1\)[/tex] will leave the basis, and [tex]\(x_1\)[/tex] will enter.
3. Pivot on the element [tex]\( a_{11} = 2 \)[/tex] and update the tableau.
- New row for [tex]\(x_1\)[/tex]: Divide the entire row [tex]\(s_1\)[/tex] by 2.
- Update the rest of the tableau accordingly.
After performing the required pivot operations, we eventually reach the final optimal tableau:
[tex]\[ \begin{array}{c|cccccc|c} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \\ \hline x_1 & 1 & 1 & 0.5 & 0.5 & 0 & 5 \\ s_2 & 0 & 1 & 1 & -0.5 & 1 & 6 \\ \hline z & 0 & -1 & 0.5 & 1.5 & 0 & 15 \\ \end{array} \][/tex]
### Step 4: Read off the Solution
The optimal solution from the final tableau:
- [tex]\( x_1 = 5 \)[/tex]
- [tex]\( x_2 = 0 \)[/tex]
- [tex]\( x_3 = 0 \)[/tex]
The maximum value of the objective function is:
[tex]\[ z = 15 \][/tex]
The values of the slack variables:
- [tex]\( s_1 = 0 \)[/tex]
- [tex]\( s_2 = 6 \)[/tex]
### Final Answer
A. The maximum is [tex]\( 15.0 \)[/tex] when [tex]\( x_1 = 5.0 \)[/tex], [tex]\( x_2 = 0.0 \)[/tex], [tex]\( x_3 = 0.0 \)[/tex], [tex]\( s_1 = 0.0 \)[/tex], and [tex]\( s_2 = 6.0 \)[/tex].
