ve the linear programing problem using Simplex Method. $$ \begin{array}{l}\max Z=6 X_{1}+5 X_{2}+8 x_{3}\end{array} $$ $$ \begin{array}{c}x_{1}+x_{2}+x_{3} \leq 12 \ 2 x_{1}+x_{2}+2 x_{3} \leq 2 \ x_{1}, x_{2}, x_{3} \quad \geq 0\end{array} $$

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To solve the given linear programming problem using the Simplex Method, we will follow these steps:

1. **Identify the objective function and constraints.**
   - Objective function: 
     $$
     \max Z = 6X_1 + 5X_2 + 8X_3
     $$
   - Constraints:
     $$
     \begin{align*}
     1. & \quad X_1 + X_2 + X_3 \leq 12 \
     2. & \quad 2X_1 + X_2 + 2X_3 \leq 2 \
     3. & \quad X_1, X_2, X_3 \geq 0
     \end{align*}
     $$

2. **Convert inequalities to equalities by adding slack variables.**
   - Let $$ S_1 $$ and $$ S_2 $$ be the slack variables for the first and second constraints, respectively.
   - The constraints become:
     $$
     \begin{align*}
     1. & \quad X_1 + X_2 + X_3 + S_1 = 12 \
     2. & \quad 2X_1 + X_2 + 2X_3 + S_2 = 2 \
     \end{align*}
     $$

3. **Set up the initial Simplex tableau.**
   The tableau will include the coefficients of the objective function and the constraints.

$$
\begin{array}{c|cccccc|c}
\text{Basis} & X_1 & X_2 & X_3 & S_1 & S_2 & Z & \text{RHS} \
\hline
S_1 & 1 & 1 & 1 & 1 & 0 & 0 & 12 \
S_2 & 2 & 1 & 2 & 0 & 1 & 0 & 2 \
\hline
-Z & -6 & -5 & -8 & 0 & 0 & 1 & 0 \
\end{array}
$$

4. **Perform the Simplex iterations.**
   - Identify the pivot column (most negative in the Z-row).
   - Identify the pivot row (minimum ratio of RHS to pivot column).
   - Perform row operations to make the pivot column a unit column.

Let's perform the calculations to find the optimal solution.

### Step 1: Identify the pivot column and row
- The most negative value in the Z-row is -8 (for $$ X_3 $$).
- The ratios for the pivot row are:
  - For $$ S_1 $$: $$ \frac{12}{1} = 12 $$
  - For $$ S_2 $$: $$ \frac{2}{2} = 1 $$ (minimum)

Thus, $$ S_2 $$ will be the pivot row, and $$ X_3 $$ will be the pivot column.

### Step 2: Perform row operations
We will pivot on the element in the second row and third column (2).

1. **Row 2 (new)**: Divide by 2:
   $$
   R_2 = \frac{1}{2} R_2 \Rightarrow (1, 0.5, 1, 0, 0.5, 0, 1)
   $$

2. **Row 1 (new)**: Subtract $$ R_2 $$ from $$ R_1 $$:
   $$
   R_1 = R_1 - R_2 \Rightarrow (0, 0.5, 0, 1, -0.5, 0, 11)
   $$

3. **Row 3 (new)**: Add 8 times $$ R_2 $$ to $$ R_3 $$:
   $$
   R_3 = R_3 + 8R_2 \Rightarrow (0, 0, 0, 0, 4, 1, 8)
   $$

The new tableau is:

$$
\begin{array}{c|cccccc|c}
\text{Basis} & X_1 & X_2 & X_3 & S_1 & S_2 & Z & \text{RHS} \
\hline
S_1 & 0 & 0.5 & 0 & 1 & -0.5 & 0 & 11 \
X_3 & 1 & 0.5 & 1 & 0 & 0.5 & 0 & 1 \
\hline
-Z & 0 & -4 & 0 & 0 & 4 & 1 & 8 \
\end{array}
$$

### Step 3: Repeat the process
- The most negative value in the Z-row is -4 (for $$ X_2 $$).
- The ratios for the pivot row are:
  - For $$ S_1 $$: $$ \frac{11}{0.5} = 22 $$
  - For $$ X_3 $$: $$ \frac{1}{0.5} = 2 $$ (minimum)

Thus, $$ X_3 $$ will be the pivot row, and $$ X_2 $$ will be the pivot column.

### Step 4: Perform row operations again
1. **Row 1 (new)**: Divide by 0.5:
   $$
   R_1 = \frac{1}{0.5} R_1 \Rightarrow (0, 1, 0, 2, -1, 0, 22)
   $$

2. **Row 2 (new)**: Subtract $$ 0.5R_1 $$ from $$ R_2 $$:
   $$
   R_2 = R_2 - 0.5R_1 \Rightarrow (1, 0, 1, -1, 0, 0, 0)
   $$

3. **Row 3 (new)**: Add 4 times $$ R_1 $$ to $$ R_3 $$:
   $$
   R_3 = R_3 + 4R_1 \Rightarrow (0, 0, 0, 0, 0, 1, 0)
   $$

The new tableau is:

\[
\begin{array}{c|cccccc|c}
\text{Basis} & X_1 & X_2 & X_3 & S_1 & S_2 & Z & \text{RHS} \
\hline
X_2 & 0 & 1 & 0 & 2 & -1 & 0 & 22 \
X_3 & 1 & 0 & 1 & -1 & 0 & 0 & 0 \
\hline
-Z & 0 & 0 & 0 & 8 & 0 & 1 & 88
The optimal solution is $$ X_1 = 0 $$, $$ X_2 = 22 $$, and $$ X_3 = 0 $$ with a maximum value of $$ Z = 88 $$.

ve the linear programing problem using Simplex Method. \[ \begin{array}{l}\max Z=6 X_{1}+5 X_{2}+8 x_{3}\end{array} \] \[ \begin{array}{c}x_{1}+x_{2}+x_{3} \leq 12 \\ 2 x_{1}+x_{2}+2 x_{3} \leq 2 \\ x_{1}, x_{2}, x_{3} \quad \geq 0\end{array} \]

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