1. **State the problem:** We want to maximize the objective function $$f = x_1 + x_2$$ subject to the constraints: $$5x_1 - 2x_2 \leq 7,$$ $$-x_1 + x_2 \leq 5,$$ $$x_1 + x_2 \leq 6,$$ $$x_1, x_2 \geq 0.$$ 2. **Convert inequalities to equalities by adding slack variables:** Let $$s_1, s_2, s_3 \geq 0$$ be slack variables for each constraint: $$5x_1 - 2x_2 + s_1 = 7,$$ $$-x_1 + x_2 + s_2 = 5,$$ $$x_1 + x_2 + s_3 = 6.$$ 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 $$-f = -x_1 - x_2$$. | Basis | $x_1$ | $x_2$ | $s_1$ | $s_2$ | $s_3$ | RHS | |-------|-------|-------|-------|-------|-------|-----| | $s_1$ | 5 | -2 | 1 | 0 | 0 | 7 | | $s_2$ | -1 | 1 | 0 | 1 | 0 | 5 | | $s_3$ | 1 | 1 | 0 | 0 | 1 | 6 | | $-f$ | -1 | -1 | 0 | 0 | 0 | 0 | 4. **Perform simplex iterations:** - Identify entering variable: most negative coefficient in objective row is $$-1$$ for both $$x_1$$ and $$x_2$$; choose $$x_1$$. - Compute ratios for leaving variable: - For $s_1$: $$\frac{7}{5} = 1.4$$ - For $s_2$: $$\frac{5}{-1}$$ (negative, ignore) - For $s_3$: $$\frac{6}{1} = 6$$ - Leaving variable is $s_1$ (smallest positive ratio). 5. **Pivot on $x_1$ in $s_1$ row:** Divide $s_1$ row by 5: $$s_1: x_1 - \frac{2}{5}x_2 + \frac{1}{5}s_1 = \frac{7}{5}$$ Rewrite other rows to eliminate $x_1$: - $s_2$: add $1 \times$ new $s_1$ row: $$s_2: 0x_1 + \left(1 - \frac{2}{5}\right)x_2 + \left(0 + \frac{1}{5}\right)s_1 + 1 s_2 = 5 + \frac{7}{5} = \frac{32}{5}$$ - $s_3$: subtract $1 \times$ new $s_1$ row: $$s_3: 0x_1 + \left(1 + \frac{2}{5}\right)x_2 + \left(0 - \frac{1}{5}\right)s_1 + 1 s_3 = 6 - \frac{7}{5} = \frac{23}{5}$$ - Objective row: add $1 \times$ new $s_1$ row: $$-f: 0x_1 + \left(-1 - \frac{2}{5}\right)x_2 + \frac{1}{5}s_1 = \frac{7}{5}$$ 6. **Repeat simplex steps until no negative coefficients in objective row:** Next entering variable is $x_2$ (coefficient negative in objective row). Calculate ratios for leaving variable: - $s_2$: $$\frac{32/5}{(1 - 2/5)} = \frac{32/5}{3/5} = \frac{32}{5} \times \frac{5}{3} = \frac{32}{3} \approx 10.67$$ - $s_3$: $$\frac{23/5}{(1 + 2/5)} = \frac{23/5}{7/5} = \frac{23}{5} \times \frac{5}{7} = \frac{23}{7} \approx 3.29$$ Leaving variable is $s_3$. 7. **Pivot on $x_2$ in $s_3$ row:** Divide $s_3$ row by $7/5$: $$s_3: 0x_1 + 1 x_2 - \frac{1}{7}s_1 + \frac{5}{7}s_3 = \frac{23}{7}$$ Update other rows to eliminate $x_2$: - $s_1$: add $\frac{2}{5} \times$ new $s_3$ row: $$s_1: x_1 + 0 x_2 + \left(\frac{1}{5} - \frac{2}{5} \times \frac{1}{7}\right)s_1 + \frac{2}{5} \times \frac{5}{7}s_3 = \frac{7}{5} + \frac{2}{5} \times \frac{23}{7}$$ - $s_2$: subtract $(3/5) \times$ new $s_3$ row: $$s_2: 0 x_1 + 0 x_2 + \left(\frac{1}{5} + \frac{3}{5} \times \frac{1}{7}\right)s_1 + \left(1 - \frac{3}{5} \times \frac{5}{7}\right)s_3 = \frac{32}{5} - \frac{3}{5} \times \frac{23}{7}$$ - Objective row: add $(7/5) \times$ new $s_3$ row: $$-f: 0 x_1 + 0 x_2 + \left(\frac{1}{5} - \frac{7}{5} \times \frac{1}{7}\right)s_1 + \frac{7}{5} \times \frac{5}{7}s_3 = \frac{7}{5} + \frac{7}{5} \times \frac{23}{7}$$ 8. **Final solution:** Since all coefficients in objective row are now non-negative, the optimal solution is: $$x_1 = \frac{29}{7} \approx 4.14,$$ $$x_2 = \frac{23}{7} \approx 3.29,$$ with maximum value $$f = x_1 + x_2 = \frac{29}{7} + \frac{23}{7} = \frac{52}{7} \approx 7.43.$$ **Note:** The simplex method steps can be implemented in Excel by setting up the tableau and performing pivot operations iteratively.