1. **Problem Statement:** Solve the Linear Programming Problem (L.P.P.) by the Simplex method: Maximize $$Z = 6x + 11y$$ Subject to constraints: $$2x + y \leq 104$$ $$x + 2y \leq 76$$ $$x, y \geq 0$$ 2. **Formulating the problem:** We want to maximize the objective function $$Z = 6x + 11y$$ under the given constraints. 3. **Convert inequalities to equalities by adding slack variables:** Let $$s_1$$ and $$s_2$$ be slack variables for the two constraints. $$2x + y + s_1 = 104$$ $$x + 2y + s_2 = 76$$ with $$x, y, s_1, s_2 \geq 0$$. 4. **Initial Simplex Tableau:** \[ \begin{array}{c|cccc|c} & x & y & s_1 & s_2 & RHS \\ \hline s_1 & 2 & 1 & 1 & 0 & 104 \\ s_2 & 1 & 2 & 0 & 1 & 76 \\ Z & -6 & -11 & 0 & 0 & 0 \\ \end{array} \] 5. **Simplex Method Steps:** - Identify entering variable: most negative coefficient in Z-row is -11 (for y), so y enters. - Determine leaving variable by minimum ratio test: - For s_1: $$\frac{104}{1} = 104$$ - For s_2: $$\frac{76}{2} = 38$$ Minimum ratio is 38, so s_2 leaves. 6. **Pivot on y in s_2 row:** Divide s_2 row by 2: $$x + 2y + s_2 = 76 \Rightarrow \frac{1}{2}x + y + \frac{1}{2}s_2 = 38$$ Rewrite s_2 row: $$x + 2y + s_2 = 76 \Rightarrow y = 38 - \frac{1}{2}x - \frac{1}{2}s_2$$ 7. **Update s_1 and Z rows to eliminate y:** - For s_1 row: $$2x + y + s_1 = 104$$ Substitute $$y$$: $$2x + (38 - \frac{1}{2}x - \frac{1}{2}s_2) + s_1 = 104$$ Simplify: $$2x - \frac{1}{2}x + s_1 - \frac{1}{2}s_2 = 104 - 38$$ $$\frac{3}{2}x + s_1 - \frac{1}{2}s_2 = 66$$ - For Z row: $$Z = 6x + 11y$$ Substitute $$y$$: $$Z = 6x + 11(38 - \frac{1}{2}x - \frac{1}{2}s_2)$$ $$Z = 6x + 418 - \frac{11}{2}x - \frac{11}{2}s_2$$ $$Z = 418 + (6 - 5.5)x - 5.5 s_2$$ $$Z = 418 + 0.5x - 5.5 s_2$$ 8. **New tableau:** \[ \begin{array}{c|cccc|c} & x & y & s_1 & s_2 & RHS \\ \hline s_1 & \frac{3}{2} & 0 & 1 & -\frac{1}{2} & 66 \\ y & \frac{1}{2} & 1 & 0 & \frac{1}{2} & 38 \\ Z & 0.5 & 0 & 0 & -5.5 & 418 \\ \end{array} \] 9. **Next iteration:** - Entering variable: x (coefficient 0.5 in Z-row) - Leaving variable by minimum ratio test: - For s_1: $$\frac{66}{\frac{3}{2}} = 44$$ - For y: $$\frac{38}{\frac{1}{2}} = 76$$ Minimum ratio is 44, so s_1 leaves. 10. **Pivot on x in s_1 row:** Divide s_1 row by $$\frac{3}{2}$$: $$x + \frac{2}{3}s_1 - \frac{1}{3}s_2 = 44$$ 11. **Update y and Z rows to eliminate x:** - For y row: $$y = 38 - \frac{1}{2}x - \frac{1}{2}s_2$$ Substitute $$x$$: $$y = 38 - \frac{1}{2}(44 - \frac{2}{3}s_1 + \frac{1}{3}s_2) - \frac{1}{2}s_2$$ Simplify: $$y = 38 - 22 + \frac{1}{3}s_1 - \frac{1}{6}s_2 - \frac{1}{2}s_2$$ $$y = 16 + \frac{1}{3}s_1 - \frac{2}{3}s_2$$ - For Z row: $$Z = 418 + 0.5x - 5.5 s_2$$ Substitute $$x$$: $$Z = 418 + 0.5(44 - \frac{2}{3}s_1 + \frac{1}{3}s_2) - 5.5 s_2$$ Simplify: $$Z = 418 + 22 - \frac{1}{3}s_1 + \frac{1}{6}s_2 - 5.5 s_2$$ $$Z = 440 - \frac{1}{3}s_1 - \frac{16}{3}s_2$$ 12. **Final tableau:** \[ \begin{array}{c|cccc|c} & x & y & s_1 & s_2 & RHS \\ \hline x & 1 & 0 & \frac{2}{3} & -\frac{1}{3} & 44 \\ y & 0 & 1 & \frac{1}{3} & -\frac{2}{3} & 16 \\ Z & 0 & 0 & -\frac{1}{3} & -\frac{16}{3} & 440 \\ \end{array} \] 13. **Optimality check:** All coefficients in Z-row for non-basic variables (s_1, s_2) are negative, so the current solution is optimal. 14. **Solution:** $$x = 44, y = 16$$ Maximum value of $$Z = 6(44) + 11(16) = 264 + 176 = 440$$ **Answer:** The maximum value of $$Z$$ is $$440$$ at $$x = 44$$ and $$y = 16$$.