1. **State the problem:** We want to maximize the objective function $$f = x_1 + x_2$$ subject to the constraints: $$\begin{cases} 5x_1 - 2x_2 \leq 7 \\ -x_1 + x_2 \leq 5 \\ x_1 + x_2 \leq 6 \\ x_1, x_2 \geq 0 \end{cases}$$ 2. **Convert inequalities to equalities by adding slack variables:** Let $$s_1, s_2, s_3 \geq 0$$ be slack variables for each constraint: $$\begin{cases} 5x_1 - 2x_2 + s_1 = 7 \\ -x_1 + x_2 + s_2 = 5 \\ x_1 + x_2 + s_3 = 6 \end{cases}$$ 3. **Set up the initial simplex tableau:** Variables: $$x_1, x_2, s_1, s_2, s_3$$ Objective function to maximize: $$f = x_1 + x_2$$ or equivalently minimize $$-f = -x_1 - x_2$$ Initial tableau: $$\begin{array}{c|ccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & RHS \\ \hline s_1 & 5 & -2 & 1 & 0 & 0 & 7 \\ s_2 & -1 & 1 & 0 & 1 & 0 & 5 \\ s_3 & 1 & 1 & 0 & 0 & 1 & 6 \\ \hline -z & -1 & -1 & 0 & 0 & 0 & 0 \end{array}$$ 4. **Perform simplex iterations:** - Identify entering variable: most negative coefficient in bottom row is $$-1$$ for both $$x_1$$ and $$x_2$$; choose $$x_1$$. - Compute ratios for leaving variable: $$\frac{7}{5} = 1.4, \quad \frac{5}{-1} = \text{not valid (negative)}, \quad \frac{6}{1} = 6$$ - Minimum positive ratio is 1.4, so $$s_1$$ leaves, $$x_1$$ enters. 5. **Pivot on element in row 1, column 1 (value 5):** Divide row 1 by 5: $$\begin{array}{c|ccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & RHS \\ \hline x_1 & 1 & -\frac{2}{5} & \frac{1}{5} & 0 & 0 & \frac{7}{5} \\ s_2 & -1 & 1 & 0 & 1 & 0 & 5 \\ s_3 & 1 & 1 & 0 & 0 & 1 & 6 \\ \hline -z & -1 & -1 & 0 & 0 & 0 & 0 \end{array}$$ 6. **Eliminate $$x_1$$ from other rows:** - Row 2: add row 1 to row 2: $$-1 + 1 = 0$$ for $$x_1$$ coefficient $$1 + (-\frac{2}{5}) = \frac{3}{5}$$ for $$x_2$$ $$0 + \frac{1}{5} = \frac{1}{5}$$ for $$s_1$$ $$1 + 0 = 1$$ for $$s_2$$ $$0 + 0 = 0$$ for $$s_3$$ $$5 + \frac{7}{5} = \frac{32}{5}$$ for RHS - Row 3: subtract row 1 from row 3: $$1 - 1 = 0$$ for $$x_1$$ $$1 - (-\frac{2}{5}) = \frac{7}{5}$$ for $$x_2$$ $$0 - \frac{1}{5} = -\frac{1}{5}$$ for $$s_1$$ $$0 - 0 = 0$$ for $$s_2$$ $$1 - 0 = 1$$ for $$s_3$$ $$6 - \frac{7}{5} = \frac{23}{5}$$ for RHS - Row -z: add row 1 to row -z: $$-1 + 1 = 0$$ for $$x_1$$ $$-1 + (-\frac{2}{5}) = -\frac{7}{5}$$ for $$x_2$$ $$0 + \frac{1}{5} = \frac{1}{5}$$ for $$s_1$$ $$0 + 0 = 0$$ for $$s_2$$ $$0 + 0 = 0$$ for $$s_3$$ $$0 + \frac{7}{5} = \frac{7}{5}$$ for RHS 7. **New tableau:** $$\begin{array}{c|ccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & RHS \\ \hline x_1 & 1 & -\frac{2}{5} & \frac{1}{5} & 0 & 0 & \frac{7}{5} \\ s_2 & 0 & \frac{3}{5} & \frac{1}{5} & 1 & 0 & \frac{32}{5} \\ s_3 & 0 & \frac{7}{5} & -\frac{1}{5} & 0 & 1 & \frac{23}{5} \\ \hline -z & 0 & -\frac{7}{5} & \frac{1}{5} & 0 & 0 & \frac{7}{5} \end{array}$$ 8. **Next entering variable:** Most negative coefficient in bottom row is $$-\frac{7}{5}$$ for $$x_2$$. 9. **Compute ratios for leaving variable:** $$\frac{32/5}{3/5} = \frac{32}{5} \times \frac{5}{3} = \frac{32}{3} \approx 10.67$$ $$\frac{23/5}{7/5} = \frac{23}{5} \times \frac{5}{7} = \frac{23}{7} \approx 3.29$$ Minimum positive ratio is $$3.29$$, so $$s_3$$ leaves, $$x_2$$ enters. 10. **Pivot on row 3, column 2 (value $$\frac{7}{5}$$):** Divide row 3 by $$\frac{7}{5}$$: $$x_2 = 1, s_1 = -\frac{1}{5} \div \frac{7}{5} = -\frac{1}{7}, s_3 = \frac{1}{\frac{7}{5}} = \frac{5}{7}, RHS = \frac{23}{5} \div \frac{7}{5} = \frac{23}{7}$$ 11. **Update other rows to eliminate $$x_2$$:** - Row 1: add $$\frac{2}{5}$$ times row 3 to row 1: $$x_2$$ coefficient becomes 0 - Row 2: subtract $$\frac{3}{5}$$ times row 3 from row 2 - Row -z: add $$\frac{7}{5}$$ times row 3 to row -z 12. **Final tableau after calculations:** $$\begin{array}{c|ccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & RHS \\ \hline x_1 & 1 & 0 & \frac{1}{7} & 0 & \frac{2}{7} & \frac{41}{7} \\ s_2 & 0 & 0 & \frac{4}{7} & 1 & -\frac{3}{7} & \frac{41}{7} \\ x_2 & 0 & 1 & -\frac{1}{7} & 0 & \frac{5}{7} & \frac{23}{7} \\ \hline -z & 0 & 0 & 0 & 0 & 1 & \frac{72}{7} \end{array}$$ 13. **Interpretation:** - Basic variables: $$x_1 = \frac{41}{7} \approx 5.857, x_2 = \frac{23}{7} \approx 3.286$$ - Slack variables: $$s_1 = 0, s_2 = 0, s_3 = 0$$ (since not basic, their values are zero) - Objective function value: $$f = x_1 + x_2 = \frac{41}{7} + \frac{23}{7} = \frac{64}{7} \approx 9.143$$ **Final answer:** The optimal plan is $$x_1 = \frac{41}{7}, x_2 = \frac{23}{7}$$ with maximum $$f = \frac{64}{7}$$. **Note:** Excel implementation involves setting up the tableau and performing these pivot operations step-by-step using formulas or manual input.