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

Problem:

Maximize $Z = 5x_1 + 3x_2 + 2x_3$

Subject to the constraints:

1. $x_1 + 4x_2 + 5x_3 \leq 59$
2. $x_1 + 4x_2 + 5x_3 \leq 110$
3. $x_1 \geq 0, x_2 \geq 0, x_3 \geq 0$

Step 1: Convert Inequalities to Equalities (Add Slack Variables)

We introduce slack variables $s_1$ and $s_2$ to convert the inequalities to equalities:

$x_1 + 4x_2 + 5x_3 + s_1 = 59$ $x_1 + 4x_2 + 5x_3 + s_2 = 110$

Now, the objective function becomes:

$Z = 5x_1 + 3x_2 + 2x_3 + 0s_1 + 0s_2$

Step 2: Set Up the Initial Simplex Tableau

\text{Basic} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{Z} & \text{RHS} \\ \hline s_1 & 1 & 4 & 5 & 1 & 0 & 0 & 59 \\ s_2 & 1 & 4 & 5 & 0 & 1 & 0 & 110 \\ \hline \text{Z} & -5 & -3 & -2 & 0 & 0 & 1 & 0 \\ \end{array}$$ Where: - The first two rows correspond to the constraint equations. - The last row represents the objective function, where the coefficients are negated because we're maximizing $$Z$$. ### Step 3: Identify the Entering Variable The entering variable is the one with the most negative coefficient in the objective function row. Here, $$x_1$$ has the most negative coefficient (-5), so it will enter the basis. ### Step 4: Identify the Leaving Variable To identify the leaving variable, we calculate the ratio of the right-hand side (RHS) to the corresponding coefficient of $$x_1$$ for each row: $$\frac{59}{1} = 59 \quad \text{and} \quad \frac{110}{1} = 110$$ Since 59 is the smaller ratio, $$s_1$$ will leave the basis. ### Step 5: Pivot to Form a New Tableau We now pivot on the element in the first row and first column (the pivot element is 1). This will make $$x_1$$ enter the basis and $$s_1$$ leave the basis. We perform row operations to get 1 in the pivot position and 0s in the other positions of the column. #### New Tableau: $$\begin{array}{c|cccccc|c} \text{Basic} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{Z} & \text{RHS} \\ \hline x_1 & 1 & 4 & 5 & 1 & 0 & 0 & 59 \\ s_2 & 0 & 0 & 0 & -1 & 1 & 0 & 51 \\ \hline \text{Z} & 0 & 2 & 3 & 5 & 0 & 1 & 295 \\ \end{array}$$ ### Step 6: Check for Optimality In the objective row (the bottom row), all the coefficients of the decision variables $$x_1$$, $$x_2$$, and $$x_3$$ are non-negative, so the current solution is optimal. ### Step 7: Solution The optimal solution is: $$x_1 = 59, \quad x_2 = 0, \quad x_3 = 0$$ $$Z = 295$$ Thus, the maximum value of $$Z$$ is 295 when $$x_1 = 59$$, $$x_2 = 0$$, and $$x_3 = 0$$. ### Summary of the Steps: 1. Convert inequalities to equalities by introducing slack variables. 2. Set up the initial simplex tableau. 3. Identify the entering variable (most negative coefficient in the objective row). 4. Identify the leaving variable (smallest positive ratio in the RHS). 5. Pivot and repeat until the objective row has no negative coefficients. --- Do you want further details on any step, or have any questions? Here are 5 questions to expand on this topic: 1. What happens if there is a tie in identifying the leaving variable? 2. How do we handle unbounded solutions in the Simplex method? 3. What is the role of artificial variables in the Simplex method? 4. How do we know if the solution is degenerate? 5. How can we implement the Simplex method in matrix form using software? **Tip:** Always double-check the pivot operations to ensure accurate results during the Simplex method.