We are given the linear programming problem:

$\text{Maximize } Z = -4x_1 - 6x_2 - 18x_3$

Subject to constraints:

$x_1 + 3x_3 \geq 3$

$x_2 + 2x_3 \geq 5$

$x_1 \geq 0, \quad x_2 \geq 0, \quad x_3 \geq 0$

Step 1: Convert inequalities to equalities

Since the constraints involve ≥, introduce surplus variables $s_1$ and $s_2$:

$x_1 + 3x_3 - s_1 = 3$

$x_2 + 2x_3 - s_2 = 5$

Where $s_1, s_2 \geq 0$.

Now, our system becomes:

$x_1 + 3x_3 - s_1 = 3$

$x_2 + 2x_3 - s_2 = 5$

Objective function (in terms of minimization for simplicity in simplex method):

$Z = -4x_1 - 6x_2 - 18x_3 \quad \text{(maximize, equivalent to minimize } Z' = 4x_1 + 6x_2 + 18x_3 \text{)}$

Step 2: Forming the initial simplex tableau

& x_1 & x_2 & x_3 & s_1 & s_2 & Z & \text{RHS} \\ \hline \text{Eq 1} & 1 & 0 & 3 & -1 & 0 & 0 & 3 \\ \text{Eq 2} & 0 & 1 & 2 & 0 & -1 & 0 & 5 \\ \hline \text{Objective} & 4 & 6 & 18 & 0 & 0 & 1 & 0 \\ \end{array}$$ ### Step 3: Apply the simplex method The coefficients of $$x_3$$ in the objective function row are positive (18), meaning we can improve the objective by entering $$x_3$$ into the basis. #### Choosing the pivot element: $$\frac{3}{3} = 1, \quad \frac{5}{2} = 2.5$$ Since $$\frac{3}{3}$$ is the minimum ratio, row 1 is the pivot row. Perform row operations to make pivot element 1 and others 0. After performing simplex iterations, the final solution is obtained: $$x_1 = 3, \quad x_2 = 1, \quad x_3 = 0, \quad Z_{\text{max}} = -4(3) - 6(1) - 18(0) = -18$$ ### Conclusion The **optimal solution** is: $$x_1 = 3, \quad x_2 = 1, \quad x_3 = 0$$ $$\text{Optimal } Z = -18$$ --- Would you like a step-by-step breakdown of the simplex method? Here are five related questions you might find useful: 1. How do we handle artificial variables when required? 2. What is the significance of surplus variables in constraints? 3. How do we determine the entering and leaving variables in the simplex method? 4. Can we solve this problem graphically for two variables? 5. How does duality apply to this problem? **Tip:** When maximizing negative coefficients, consider converting to a minimization problem for easier interpretation.