To solve the given linear programming problem (LPP) using the simplex method, we will follow these steps:

1. Convert the inequalities into equations by adding slack variables.
2. Set up the initial simplex tableau.
3. Perform the simplex algorithm to find the optimal solution.

Step 1: Convert inequalities to equations

Given the constraints:

1. $5x_1 + 2x_2 \leq 20$
2. $x_1 \geq 3$
3. $x_2 \leq 5$
4. $x_1 \geq 0$
5. $x_2 \geq 0$

Let's introduce slack variables $s_1$ for the first constraint to convert it into an equation.

$5x_1 + 2x_2 + s_1 = 20 \quad (s_1 \geq 0)$

We do not need to add slack variables for the remaining constraints since they are bounds on $x_1$ and $x_2$.

Step 2: Set up the initial simplex tableau

The objective function to maximize is: $z = 9x_1 + 3x_2$

The initial simplex tableau is:

& x_1 & x_2 & s_1 & \text{RHS} \\ \hline s_1 & 5 & 2 & 1 & 20 \\ \hline z & -9 & -3 & 0 & 0 \\ \end{array}$$ ### Step 3: Perform the simplex algorithm 1. **Identify the entering variable:** The most negative value in the objective row is -9 (corresponding to $$x_1$$). So, $$x_1$$ is the entering variable. 2. **Identify the leaving variable:** Compute the ratios of the right-hand side to the coefficients of $$x_1$$ in the constraints to find the pivot row. - Ratio for $$s_1$$: $$\frac{20}{5} = 4$$ Since $$s_1$$ has the smallest ratio, it is the leaving variable. #### Pivot to make $$x_1$$ the basic variable in the first row: $$\begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & \text{RHS} \\ \hline x_1 & 1 & \frac{2}{5} & \frac{1}{5} & 4 \\ \hline z & 0 & -\frac{3}{5} & \frac{9}{5} & 36 \\ \end{array}$$ 3. **Identify the new entering variable:** The most negative value in the objective row is $$-\frac{3}{5}$$ (corresponding to $$x_2$$). So, $$x_2$$ is the entering variable. 4. **Identify the new leaving variable:** Compute the ratios of the right-hand side to the coefficients of $$x_2$$ in the constraints. - Ratio for $$x_1$$: $$\frac{4}{\frac{2}{5}} = 10$$ Since $$x_1$$ has the smallest ratio, it is the leaving variable. #### Pivot to make $$x_2$$ the basic variable in the first row: $$\begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & \text{RHS} \\ \hline x_2 & \frac{5}{2} & 1 & \frac{1}{2} & 10 \\ \hline z & \frac{3}{2} & 0 & \frac{6}{2} & 60 \\ \end{array}$$ Since there are no more negative values in the objective row, the current solution is optimal. ### Solution From the final simplex tableau, we can see: $$x_1 = 0, \quad x_2 = 10, \quad s_1 = 5$$ The optimal value of the objective function is: $$z = 60$$ Thus, the maximum value of $$z = 9x_1 + 3x_2$$ is 60 when $$x_1 = 0$$ and $$x_2 = 10$$. Would you like more details or have any questions? Here are some related questions you might find interesting: 1. How to handle more complex constraints in the simplex method? 2. What are the steps to perform the two-phase simplex method? 3. How to interpret the final simplex tableau? 4. What are the differences between the simplex method and the graphical method for solving LPPs? 5. How to deal with unbounded solutions in the simplex method? Tip: When solving linear programming problems using the simplex method, always ensure to perform the pivoting steps correctly and verify the optimality conditions at each step.