1. **State the problem:** We want to maximize the objective function $$f = x_1 + x_2$$ subject to the constraints: $$5x_1 - 2x_2 \leq 7,$$ $$-x_1 + x_2 \leq 5,$$ $$x_1 + x_2 \leq 6,$$ $$x_1 \geq 0, x_2 \geq 0.$$ 2. **Convert inequalities to equalities by adding slack variables:** Let $$s_1, s_2, s_3 \geq 0$$ be slack variables for each inequality respectively. The system becomes: $$5x_1 - 2x_2 + s_1 = 7,$$ $$-x_1 + x_2 + s_2 = 5,$$ $$x_1 + x_2 + s_3 = 6.$$ 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: \[ \begin{array}{c|ccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & RHS \\ \hline s_1 & 5 & -2 & 1 & 0 & 0 & 7 \\ s_2 & -1 & 1 & 0 & 1 & 0 & 5 \\ s_3 & 1 & 1 & 0 & 0 & 1 & 6 \\ \hline -f & -1 & -1 & 0 & 0 & 0 & 0 \end{array} \] 4. **Identify entering variable:** Look at the bottom row (objective function coefficients). The most negative coefficient is $$-1$$ for both $$x_1$$ and $$x_2$$. Choose $$x_1$$ to enter. 5. **Determine leaving variable:** Calculate ratios $$\frac{RHS}{x_1}$$ for positive $$x_1$$ coefficients: - For $$s_1$$: $$\frac{7}{5} = 1.4$$ - For $$s_2$$: coefficient of $$x_1$$ is $$-1$$ (negative), ignore - For $$s_3$$: $$\frac{6}{1} = 6$$ Minimum positive ratio is 1.4, so $$s_1$$ leaves. 6. **Pivot on $$x_1$$ in row $$s_1$$:** Divide row $$s_1$$ by 5: $$s_1: x_1 - \frac{2}{5}x_2 + \frac{1}{5}s_1 = \frac{7}{5}$$ 7. **Eliminate $$x_1$$ from other rows:** - Row $$s_2$$: add row $$s_1$$ multiplied by 1 to row $$s_2$$: $$s_2 + 1 \times s_1: (-1 + 1)x_1 + (1 - \frac{2}{5})x_2 + (0 + \frac{1}{5})s_1 + 1 s_2 = 5 + \frac{7}{5}$$ Simplify: $$0 x_1 + \frac{3}{5} x_2 + \frac{1}{5} s_1 + s_2 = \frac{32}{5}$$ - Row $$s_3$$: subtract row $$s_1$$ multiplied by 1: $$s_3 - 1 \times s_1: (1 - 1)x_1 + (1 + \frac{2}{5})x_2 + (0 - \frac{1}{5})s_1 + s_3 = 6 - \frac{7}{5}$$ Simplify: $$0 x_1 + \frac{7}{5} x_2 - \frac{1}{5} s_1 + s_3 = \frac{23}{5}$$ - Row $$-f$$: add row $$s_1$$ multiplied by 1: $$-f + 1 \times s_1: (-1 + 1)x_1 + (-1 - \frac{2}{5})x_2 + 0 + \frac{1}{5} s_1 = 0 + \frac{7}{5}$$ Simplify: $$0 x_1 - \frac{7}{5} x_2 + \frac{1}{5} s_1 = \frac{7}{5}$$ 8. **New tableau:** \[ \begin{array}{c|ccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & RHS \\ \hline x_1 & 1 & -\frac{2}{5} & \frac{1}{5} & 0 & 0 & \frac{7}{5} \\ s_2 & 0 & \frac{3}{5} & \frac{1}{5} & 1 & 0 & \frac{32}{5} \\ s_3 & 0 & \frac{7}{5} & -\frac{1}{5} & 0 & 1 & \frac{23}{5} \\ \hline -f & 0 & -\frac{7}{5} & \frac{1}{5} & 0 & 0 & \frac{7}{5} \end{array} \] 9. **Repeat:** Entering variable: $$x_2$$ (coefficient $$-\frac{7}{5}$$ in objective row). Leaving variable: compute ratios for positive $$x_2$$ coefficients: - $$s_2: \frac{32/5}{3/5} = \frac{32}{5} \times \frac{5}{3} = \frac{32}{3} \approx 10.67$$ - $$s_3: \frac{23/5}{7/5} = \frac{23}{5} \times \frac{5}{7} = \frac{23}{7} \approx 3.29$$ Minimum ratio is $$3.29$$, so $$s_3$$ leaves. 10. **Pivot on $$x_2$$ in row $$s_3$$:** Divide row $$s_3$$ by $$\frac{7}{5}$$: $$x_2 + \left(-\frac{1}{5} \div \frac{7}{5}\right) s_1 + \left(0 \div \frac{7}{5}\right) s_2 + \left(\frac{1}{\frac{7}{5}}\right) s_3 = \frac{23/5}{7/5}$$ Simplify: $$x_2 - \frac{1}{7} s_1 + 0 s_2 + \frac{5}{7} s_3 = \frac{23}{7}$$ 11. **Eliminate $$x_2$$ from other rows:** - Row $$x_1$$: add $$\frac{2}{5}$$ times row $$s_3$$: $$x_1 + 0 x_2 + \left(\frac{1}{5} - \frac{2}{5} \times \frac{1}{7}\right) s_1 + 0 s_2 + 0 + \frac{2}{5} \times \frac{5}{7} s_3 = \frac{7}{5} + \frac{2}{5} \times \frac{23}{7}$$ Simplify: $$x_1 + \frac{3}{35} s_1 + \frac{2}{7} s_3 = \frac{7}{5} + \frac{46}{35} = \frac{49}{35} + \frac{46}{35} = \frac{95}{35} = \frac{19}{7}$$ - Row $$s_2$$: subtract $$\frac{3}{5}$$ times row $$s_3$$: $$0 x_1 + 0 x_2 + \left(\frac{1}{5} - \frac{3}{5} \times -\frac{1}{7}\right) s_1 + 1 s_2 + 0 - \frac{3}{5} \times \frac{5}{7} s_3 = \frac{32}{5} - \frac{3}{5} \times \frac{23}{7}$$ Simplify: $$s_1 \text{ coeff} = \frac{1}{5} + \frac{3}{35} = \frac{7}{35} + \frac{3}{35} = \frac{10}{35} = \frac{2}{7}$$ $$s_3 \text{ coeff} = 0 - \frac{3}{7} = -\frac{3}{7}$$ $$RHS = \frac{32}{5} - \frac{69}{35} = \frac{224}{35} - \frac{69}{35} = \frac{155}{35} = \frac{31}{7}$$ - Row $$-f$$: add $$\frac{7}{5}$$ times row $$s_3$$: $$0 x_1 + 0 x_2 + \left(\frac{1}{5} - \frac{7}{5} \times -\frac{1}{7}\right) s_1 + 0 + 0 + \frac{7}{5} \times \frac{5}{7} s_3 = \frac{7}{5} + \frac{7}{5} \times \frac{23}{7}$$ Simplify: $$s_1 \text{ coeff} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$$ $$s_3 \text{ coeff} = 0 + 1 = 1$$ $$RHS = \frac{7}{5} + \frac{23}{5} = \frac{30}{5} = 6$$ 12. **Final tableau:** \[ \begin{array}{c|ccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & RHS \\ \hline x_1 & 1 & 0 & \frac{3}{35} & 0 & \frac{2}{7} & \frac{19}{7} \\ x_2 & 0 & 1 & -\frac{1}{7} & 0 & \frac{5}{7} & \frac{23}{7} \\ s_2 & 0 & 0 & \frac{2}{7} & 1 & -\frac{3}{7} & \frac{31}{7} \\ \hline -f & 0 & 0 & \frac{2}{5} & 0 & 1 & 6 \end{array} \] 13. **Interpretation:** All coefficients in the objective row for non-basic variables are non-negative, so the current solution is optimal. 14. **Optimal solution:** $$x_1 = \frac{19}{7} \approx 2.71,$$ $$x_2 = \frac{23}{7} \approx 3.29,$$ $$f_{max} = 6.$$