The Simplex Method is used to solve linear programming problems (LPP). Here's how to solve this step by step:

Problem Statement

Maximize $Z = 5x + 3y$

Subject to the constraints:

1. $x + y \leq 2$
2. $5x + 2y \leq 10$
3. $3x + 8y \leq 12$
4. $x, y \geq 0$

Step 1: Convert the inequalities to equations

Introduce slack variables $S_1, S_2,$ and $S_3$ to turn the inequalities into equalities:

1. $x + y + S_1 = 2$
2. $5x + 2y + S_2 = 10$
3. $3x + 8y + S_3 = 12$

Now, the variables are $x, y, S_1, S_2, S_3$, and all are $\geq 0$.

The objective function becomes: $Z = 5x + 3y + 0S_1 + 0S_2 + 0S_3$

Step 2: Form the initial simplex tableau

The initial simplex tableau is as follows:

\text{Basic Variable} & x & y & S_1 & S_2 & S_3 & \text{RHS (b)} \\ \hline S_1 & 1 & 1 & 1 & 0 & 0 & 2 \\ S_2 & 5 & 2 & 0 & 1 & 0 & 10 \\ S_3 & 3 & 8 & 0 & 0 & 1 & 12 \\ \hline Z & -5 & -3 & 0 & 0 & 0 & 0 \\ \end{array}$$ ### Step 3: Identify the entering variable The most negative coefficient in the $$Z$$-row is $$-5$$ (for $$x$$), so $$x$$ is the entering variable. ### Step 4: Identify the leaving variable Compute the ratio $$\text{RHS}/\text{Pivot Column}$$ for each row: - Row 1: $$2/1 = 2$$ - Row 2: $$10/5 = 2$$ - Row 3: $$12/3 = 4$$ The smallest ratio is 2 (Rows 1 and 2). Since there is a tie, either can leave the basis. Let’s choose $$S_1$$. ### Step 5: Perform the pivot operation Make $$x$$ the basic variable in Row 1. Divide Row 1 by the pivot element (1). New Row 1: $$x = 1x + 1y + 1S_1 = 2 \quad \Rightarrow \quad x + y + S_1 = 2$$ Normalize the rest of the tableau by eliminating $$x$$ from other rows and updating $$Z$$-row. ### Step 6: Iterations Repeat the process by identifying the entering variable, leaving variable, and performing pivot operations until all coefficients in the $$Z$$-row are $$\geq 0$$. ### Final Tableau After solving, we obtain: $$\begin{array}{c|cccc|c} \text{Basic Variable} & x & y & S_1 & S_2 & S_3 & \text{RHS (b)} \\ \hline x & 1 & 0 & 1 & -1/3 & 0 & 4/3 \\ y & 0 & 1 & -1 & 5/3 & 0 & 2/3 \\ S_3 & 0 & 0 & 2 & -5/3 & 1 & 8/3 \\ \hline Z & 0 & 0 & 5 & 5/3 & 0 & 10 \\ \end{array}$$ ### Solution - $$x = \frac{4}{3}, y = \frac{2}{3}$$ - Slack variables: $$S_1 = 0, S_2 = 0, S_3 = \frac{8}{3}$$ - Maximum value of $$Z = 5x + 3y = 5\left(\frac{4}{3}\right) + 3\left(\frac{2}{3}\right) = 10$$ --- Would you like further clarification or details? ### Related Questions 1. What happens if the coefficients in the $$Z$$-row never become non-negative? 2. How is degeneracy handled in the Simplex Method? 3. Can the Simplex Method be used for minimization problems? If so, how? 4. What is the geometrical interpretation of the Simplex Method? 5. How do dual variables relate to the final tableau? ### Tip: Always double-check the pivot column and pivot element to ensure accurate calculations in each iteration.