1. **State the problem:** Maximize the objective function $$Z = 7x_1 + 14x_2$$ subject to the constraints: $$3x_1 + 2x_2 \leq 36$$ $$x_1 + 4x_2 \leq 10$$ $$x_1, x_2 \geq 0$$ 2. **Convert inequalities to equations by adding slack variables:** Let $$s_1$$ and $$s_2$$ be slack variables for the two constraints. $$3x_1 + 2x_2 + s_1 = 36$$ $$x_1 + 4x_2 + s_2 = 10$$ with $$s_1, s_2 \geq 0$$. 3. **Set up the initial simplex tableau:** \[ \begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & s_2 & RHS \\ \hline s_1 & 3 & 2 & 1 & 0 & 36 \\ s_2 & 1 & 4 & 0 & 1 & 10 \\ \hline -Z & -7 & -14 & 0 & 0 & 0 \\ \end{array} \] 4. **Identify entering variable:** The most negative coefficient in the bottom row is $$-14$$ for $$x_2$$, so $$x_2$$ enters the basis. 5. **Determine leaving variable:** Calculate ratios of RHS to positive coefficients in $$x_2$$ column: - For $$s_1$$: $$36 / 2 = 18$$ - For $$s_2$$: $$10 / 4 = 2.5$$ Smallest positive ratio is 2.5, so $$s_2$$ leaves the basis. 6. **Pivot on element in row 2, column 2 (value 4):** Divide row 2 by 4: $$x_1/4 + x_2 + s_2/4 = 10/4 = 2.5$$ 7. **Update other rows to make other entries in $$x_2$$ column zero:** - Row 1: $$R1 - 2 \times R2$$ - Row 3: $$R3 + 14 \times R2$$ Calculations: - New row 1: $$3x_1 + 2x_2 + s_1 = 36$$ $$- 2 \times (x_1/4 + x_2 + s_2/4 = 2.5)$$ $$= 3x_1 + 2x_2 + s_1 - \frac{2}{4}x_1 - 2x_2 - \frac{2}{4}s_2 = 36 - 5$$ $$= \left(3 - 0.5\right)x_1 + (2 - 2)x_2 + s_1 - 0.5 s_2 = 31$$ $$= 2.5 x_1 + s_1 - 0.5 s_2 = 31$$ - New row 3: $$-7 x_1 -14 x_2 + 0 s_1 + 0 s_2 = 0$$ $$+ 14 \times (x_1/4 + x_2 + s_2/4 = 2.5)$$ $$= -7 x_1 -14 x_2 + 14 \times \frac{x_1}{4} + 14 x_2 + 14 \times \frac{s_2}{4} = 0 + 35$$ $$= (-7 + 3.5) x_1 + (-14 + 14) x_2 + 0 s_1 + 3.5 s_2 = 35$$ $$= -3.5 x_1 + 3.5 s_2 = 35$$ 8. **New tableau:** \[ \begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & s_2 & RHS \\ \hline s_1 & 2.5 & 0 & 1 & -0.5 & 31 \\ x_2 & 0.25 & 1 & 0 & 0.25 & 2.5 \\ \hline -Z & -3.5 & 0 & 0 & 3.5 & 35 \\ \end{array} \] 9. **Repeat:** The most negative coefficient in the bottom row is $$-3.5$$ for $$x_1$$, so $$x_1$$ enters the basis. 10. **Determine leaving variable:** Ratios for positive $$x_1$$ coefficients: - For $$s_1$$: $$31 / 2.5 = 12.4$$ - For $$x_2$$: $$2.5 / 0.25 = 10$$ Smallest ratio is 10, so $$x_2$$ leaves the basis. 11. **Pivot on element in row 2, column 1 (value 0.25):** Divide row 2 by 0.25: $$x_1 + 4 x_2 + 0 s_1 + s_2 = 10$$ becomes $$x_1 + 4 x_2 + 0 s_1 + s_2 = 10$$ (already the same, but we express in terms of $$x_1$$). Actually, since $$x_2$$ is leaving and $$x_1$$ entering, we solve row 2 for $$x_1$$: $$x_1 = 4 x_2 + 0 s_1 + s_2 = 10$$ But this conflicts with previous step; correct approach is to pivot on 0.25 in row 2, column 1: Divide row 2 by 0.25: $$x_1 + 4 x_2 + 0 s_1 + s_2 = 10$$ becomes $$x_1 + 4 x_2 + 0 s_1 + s_2 = 10$$ Wait, this is inconsistent with previous tableau; re-express row 2: Row 2 is: $$0.25 x_1 + 1 x_2 + 0 s_1 + 0.25 s_2 = 2.5$$ Divide by 0.25: $$x_1 + 4 x_2 + 0 s_1 + s_2 = 10$$ Now express $$x_1$$ in terms of others: $$x_1 = 10 - 4 x_2 - s_2$$ But since $$x_2$$ is leaving, we solve for $$x_2$$: $$x_2 = (2.5 - 0.25 x_1 - 0.25 s_2)$$ This is complex; better to pivot properly: Pivoting on 0.25 in row 2, column 1: - New row 2: divide by 0.25 - Update rows 1 and 3 to zero out $$x_1$$ coefficients. New row 2: $$x_1 + 4 x_2 + 0 s_1 + s_2 = 10$$ Row 1 update: $$R1 - 2.5 \times R2$$ $$2.5 x_1 + s_1 - 0.5 s_2 = 31$$ $$- 2.5 \times (x_1 + 4 x_2 + 0 s_1 + s_2 = 10)$$ $$= 2.5 x_1 + s_1 - 0.5 s_2 - 2.5 x_1 - 10 x_2 - 0 - 2.5 s_2 = 31 - 25$$ $$= s_1 - 3 s_2 - 10 x_2 = 6$$ Row 3 update: $$R3 + 3.5 \times R2$$ $$-3.5 x_1 + 3.5 s_2 = 35$$ $$+ 3.5 \times (x_1 + 4 x_2 + 0 s_1 + s_2 = 10)$$ $$= -3.5 x_1 + 3.5 s_2 + 3.5 x_1 + 14 x_2 + 0 + 3.5 s_2 = 35 + 35$$ $$= 14 x_2 + 7 s_2 = 70$$ 12. **New tableau:** \[ \begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & s_2 & RHS \\ \hline s_1 & 0 & -10 & 1 & -3 & 6 \\ x_1 & 1 & 4 & 0 & 1 & 10 \\ \hline -Z & 0 & 14 & 0 & 7 & 70 \\ \end{array} \] 13. **Check for optimality:** No negative coefficients in the bottom row, so the solution is optimal. 14. **Read solution:** $$x_1 = 10$$ $$x_2 = 0$$ (since $$x_2$$ is not in basis and its coefficient in row 1 is negative, set to zero) $$Z = 7(10) + 14(0) = 70$$ **Final answer:** $$x_1 = 10, x_2 = 0, Z_{max} = 70$$