We will solve the following Linear Programming (LP) problem using the Simplex Method:

Problem:

Maximize $Z = 3x_1 + 2x_2 + 2x_3$

Subject to: 5x_1 + 7x_2 + 4x_3 \leq 7 \tag{1} 4x_1 - 7x_2 - 5x_3 \leq 2 \tag{2} 3x_1 + 4x_2 - 6x_3 \geq 3 \tag{3} $x_1, x_2, x_3 \geq 0$

Step 1: Convert the inequalities into equalities by introducing slack, surplus, and artificial variables.

1. For the first constraint $5x_1 + 7x_2 + 4x_3 \leq 7$, we introduce a slack variable $s_1 \geq 0$: $5x_1 + 7x_2 + 4x_3 + s_1 = 7$

2. For the second constraint $4x_1 - 7x_2 - 5x_3 \leq 2$, we introduce another slack variable $s_2 \geq 0$: $4x_1 - 7x_2 - 5x_3 + s_2 = 2$

3. For the third constraint $3x_1 + 4x_2 - 6x_3 \geq 3$, we need to convert it into an equality by subtracting a surplus variable $s_3 \geq 0$ and introducing an artificial variable $a_1 \geq 0$ to ensure the constraint is met: $3x_1 + 4x_2 - 6x_3 - s_3 + a_1 = 3$

Now the LP problem becomes: Maximize $Z = 3x_1 + 2x_2 + 2x_3$

Subject to: 5x_1 + 7x_2 + 4x_3 + s_1 = 7 \tag{4} 4x_1 - 7x_2 - 5x_3 + s_2 = 2 \tag{5} 3x_1 + 4x_2 - 6x_3 - s_3 + a_1 = 3 \tag{6} $x_1, x_2, x_3, s_1, s_2, s_3, a_1 \geq 0$

Step 2: Set up the initial Simplex tableau.

We introduce the following variables in the objective function: $Z = 3x_1 + 2x_2 + 2x_3$. The artificial variable $a_1$ will be handled in the Big M method.

Objective Function (Z):

$Z - 3x_1 - 2x_2 - 2x_3 = 0$

The initial tableau with all the variables is:

\text{Basic Var} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & a_1 & \text{RHS} \\ \hline s_1 & 5 & 7 & 4 & 1 & 0 & 0 & 0 & 7 \\ s_2 & 4 & -7 & -5 & 0 & 1 & 0 & 0 & 2 \\ a_1 & 3 & 4 & -6 & 0 & 0 & -1 & 1 & 3 \\ \hline Z & -3 & -2 & -2 & 0 & 0 & 0 & M & 0 \\ \end{array}$$ Where $$M$$ is a large positive constant (for the artificial variable). ### Step 3: Apply the Simplex Method. 1. **Select pivot column**: Identify the most negative value in the bottom row (except the RHS). The pivot column corresponds to $$x_1$$ because it has the most negative value (-3). 2. **Select pivot row**: To find the pivot row, divide the RHS values by the corresponding values in the pivot column. The smallest positive ratio gives the pivot row. - For row 1: $$\frac{7}{5} = 1.4$$ - For row 2: $$\frac{2}{4} = 0.5$$ - For row 3: $$\frac{3}{3} = 1$$ - The pivot row is row 2 (because 0.5 is the smallest). 3. **Perform row operations**: Perform Gaussian elimination to make all other entries in the pivot column zero, except for the pivot. #### After the first pivot step: $$\begin{array}{c|ccccccc|c} \text{Basic Var} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & a_1 & \text{RHS} \\ \hline s_1 & 0 & 11 & 7 & 1 & 0 & 0 & 0 & 5 \\ x_1 & 1 & -\frac{7}{4} & -\frac{5}{4} & 0 & \frac{1}{4} & 0 & 0 & \frac{1}{2} \\ a_1 & 0 & \frac{5}{4} & -\frac{13}{4} & 0 & \frac{7}{4} & -1 & 1 & \frac{5}{2} \\ \hline Z & 0 & \frac{5}{2} & \frac{5}{2} & 0 & \frac{3}{4} & 0 & M & \frac{3}{2} \\ \end{array}$$ At this point, we would repeat the process of selecting pivot columns, identifying pivot rows, and performing Gaussian elimination until all the values in the objective function row (Z-row) are non-negative. Would you like me to continue with further steps or would you prefer an explanation of any specific part?