Answer :
Certainly! Let's follow the step-by-step simplex method to solve the given linear programming problem.
The initial tableau is:
[tex]\[ \begin{array}{rrrrrr|r} x_1 & x_2 & s_1 & s_2 & s_3 & z & \\ 1 & 5 & 1 & 0 & 0 & 0 & 12 \\ 4 & 1 & 0 & 1 & 0 & 0 & 28 \\ 1 & 1 & 0 & 0 & 1 & 0 & 2 \\ \hline -3 & -1 & 0 & 0 & 0 & 1 & 0 \end{array} \][/tex]
### Step 1: Identify the Pivot Column
Look at the bottom row (objective function coefficients). Find the most negative entry, since we want to maximize the objective function [tex]\(z\)[/tex]. Here, the most negative entry is -3 in column [tex]\(x_1\)[/tex]. Thus, [tex]\(x_1\)[/tex] is the entering variable.
### Step 2: Identify the Pivot Row
Next, determine the pivot row by dividing the right-hand side of the constraint equations by the positive coefficients in the pivot column:
[tex]\[ \begin{align*} \text{Row 1:} & \quad \frac{12}{1} = 12 \\ \text{Row 2:} & \quad \frac{28}{4} = 7 \\ \text{Row 3:} & \quad \frac{2}{1} = 2 \end{align*} \][/tex]
The smallest ratio is 2 in row 3, so row 3 is the pivot row. The pivot element is 1 (intersection of column [tex]\(x_1\)[/tex] and row 3).
### Step 3: Pivot Operation
Perform row operations to make the pivot element 1 and all other entries in the pivot column 0.
- Pivot Row (Row 3):
Keep row 3 as it is since the pivot element is already 1.
- Adjust Row 1:
[tex]\[ \text{Row 1} - 1 \times \text{Row 3} \][/tex]
[tex]\[ \begin{array}{rrrrrr|r} 1 - 1 & 5 - 1 & 1 - 0 & 0 - 0 & 0 - 1 & 0 - 0 & 12 - 2 \\ 0 & 4 & 1 & 0 & -1 & 0 & 10 \end{array} \][/tex]
- Adjust Row 2:
[tex]\[ \text{Row 2} - 4 \times \text{Row 3} \][/tex]
[tex]\[ \begin{array}{rrrrrr|r} 4 - 4 & 1 - 4 & 0 - 0 & 1 - 0 & 0 - 4 & 0 - 0 & 28 - 8 \\ 0 & -3 & 0 & 1 & -4 & 0 & 20 \end{array} \][/tex]
- Adjust Objective Function Row (Row 4):
[tex]\[ -3 \times \text{Row 3} + \text{Objective Row} \][/tex]
[tex]\[ \begin{array}{rrrrrr|r} -3 + 3 & -1 + 3 & 0 + 0 & 0 + 0 & 0 + 3 & 1 + 0 & 0 + 6 \\ 0 & 2 & 0 & 0 & 3 & 1 & 6 \end{array} \][/tex]
The new tableau looks like this:
[tex]\[ \begin{array}{rrrrrr|r} x_1 & x_2 & s_1 & s_2 & s_3 & z & \\ 0 & 4 & 1 & 0 & -1 & 0 & 10 \\ 0 & -3 & 0 & 1 & -4 & 0 & 20 \\ 1 & 1 & 0 & 0 & 1 & 0 & 2 \\ \hline 0 & 2 & 0 & 0 & 3 & 1 & 6 \end{array} \][/tex]
### Step 4: Repeat Until Optimal
Identify the next pivot column, which is [tex]\(x_2\)[/tex]. The ratios are not relevant anymore, because there are no positive numbers in the objective function row - meaning we reached the optimal solution.
### Optimal Solution:
The maximum value of [tex]\( z \)[/tex] is 6 when [tex]\( x_1 = 2 \)[/tex], [tex]\( x_2 = 0 \)[/tex]. The values of slack variables [tex]\( s_1 \)[/tex], [tex]\( s_2 \)[/tex], and [tex]\( s_3 \)[/tex] are, [tex]\( s_1 = 10 \)[/tex], [tex]\( s_2 = 20 \)[/tex], and [tex]\( s_3 = 0 \)[/tex].
Final answer:
- The maximum is [tex]\( 6 \)[/tex]
- when [tex]\( x_1 = 2 \)[/tex], [tex]\( x_2 = 0 \)[/tex]. [tex]\( s_1 = 10 \)[/tex], [tex]\( s_2 = 20 \)[/tex], and [tex]\( s_3 = 0 \)[/tex].
The initial tableau is:
[tex]\[ \begin{array}{rrrrrr|r} x_1 & x_2 & s_1 & s_2 & s_3 & z & \\ 1 & 5 & 1 & 0 & 0 & 0 & 12 \\ 4 & 1 & 0 & 1 & 0 & 0 & 28 \\ 1 & 1 & 0 & 0 & 1 & 0 & 2 \\ \hline -3 & -1 & 0 & 0 & 0 & 1 & 0 \end{array} \][/tex]
### Step 1: Identify the Pivot Column
Look at the bottom row (objective function coefficients). Find the most negative entry, since we want to maximize the objective function [tex]\(z\)[/tex]. Here, the most negative entry is -3 in column [tex]\(x_1\)[/tex]. Thus, [tex]\(x_1\)[/tex] is the entering variable.
### Step 2: Identify the Pivot Row
Next, determine the pivot row by dividing the right-hand side of the constraint equations by the positive coefficients in the pivot column:
[tex]\[ \begin{align*} \text{Row 1:} & \quad \frac{12}{1} = 12 \\ \text{Row 2:} & \quad \frac{28}{4} = 7 \\ \text{Row 3:} & \quad \frac{2}{1} = 2 \end{align*} \][/tex]
The smallest ratio is 2 in row 3, so row 3 is the pivot row. The pivot element is 1 (intersection of column [tex]\(x_1\)[/tex] and row 3).
### Step 3: Pivot Operation
Perform row operations to make the pivot element 1 and all other entries in the pivot column 0.
- Pivot Row (Row 3):
Keep row 3 as it is since the pivot element is already 1.
- Adjust Row 1:
[tex]\[ \text{Row 1} - 1 \times \text{Row 3} \][/tex]
[tex]\[ \begin{array}{rrrrrr|r} 1 - 1 & 5 - 1 & 1 - 0 & 0 - 0 & 0 - 1 & 0 - 0 & 12 - 2 \\ 0 & 4 & 1 & 0 & -1 & 0 & 10 \end{array} \][/tex]
- Adjust Row 2:
[tex]\[ \text{Row 2} - 4 \times \text{Row 3} \][/tex]
[tex]\[ \begin{array}{rrrrrr|r} 4 - 4 & 1 - 4 & 0 - 0 & 1 - 0 & 0 - 4 & 0 - 0 & 28 - 8 \\ 0 & -3 & 0 & 1 & -4 & 0 & 20 \end{array} \][/tex]
- Adjust Objective Function Row (Row 4):
[tex]\[ -3 \times \text{Row 3} + \text{Objective Row} \][/tex]
[tex]\[ \begin{array}{rrrrrr|r} -3 + 3 & -1 + 3 & 0 + 0 & 0 + 0 & 0 + 3 & 1 + 0 & 0 + 6 \\ 0 & 2 & 0 & 0 & 3 & 1 & 6 \end{array} \][/tex]
The new tableau looks like this:
[tex]\[ \begin{array}{rrrrrr|r} x_1 & x_2 & s_1 & s_2 & s_3 & z & \\ 0 & 4 & 1 & 0 & -1 & 0 & 10 \\ 0 & -3 & 0 & 1 & -4 & 0 & 20 \\ 1 & 1 & 0 & 0 & 1 & 0 & 2 \\ \hline 0 & 2 & 0 & 0 & 3 & 1 & 6 \end{array} \][/tex]
### Step 4: Repeat Until Optimal
Identify the next pivot column, which is [tex]\(x_2\)[/tex]. The ratios are not relevant anymore, because there are no positive numbers in the objective function row - meaning we reached the optimal solution.
### Optimal Solution:
The maximum value of [tex]\( z \)[/tex] is 6 when [tex]\( x_1 = 2 \)[/tex], [tex]\( x_2 = 0 \)[/tex]. The values of slack variables [tex]\( s_1 \)[/tex], [tex]\( s_2 \)[/tex], and [tex]\( s_3 \)[/tex] are, [tex]\( s_1 = 10 \)[/tex], [tex]\( s_2 = 20 \)[/tex], and [tex]\( s_3 = 0 \)[/tex].
Final answer:
- The maximum is [tex]\( 6 \)[/tex]
- when [tex]\( x_1 = 2 \)[/tex], [tex]\( x_2 = 0 \)[/tex]. [tex]\( s_1 = 10 \)[/tex], [tex]\( s_2 = 20 \)[/tex], and [tex]\( s_3 = 0 \)[/tex].
