1. **Problem Statement:** We want to maximize the profit from producing two products, X and Y, using three machines M1, M2, and M3 with given time constraints. 2. **Formulate the primal LP model:** Let $x$ = units of product X, $y$ = units of product Y. Objective function (maximize profit): $$\max Z = 60x + 40y$$ Subject to machine time constraints: $$2x + 3y \leq 120 \quad \text{(M1)}$$ $$4x + 2y \leq 160 \quad \text{(M2)}$$ $$x + y \leq 50 \quad \text{(M3)}$$ And non-negativity: $$x \geq 0, y \geq 0$$ 3. **Simplex method setup:** Add slack variables $s_1, s_2, s_3$ for constraints: $$2x + 3y + s_1 = 120$$ $$4x + 2y + s_2 = 160$$ $$x + y + s_3 = 50$$ Initial tableau: \[ \begin{array}{c|ccccc|c} & x & y & s_1 & s_2 & s_3 & RHS \\ \hline s_1 & 2 & 3 & 1 & 0 & 0 & 120 \\ s_2 & 4 & 2 & 0 & 1 & 0 & 160 \\ s_3 & 1 & 1 & 0 & 0 & 1 & 50 \\ \hline Z & -60 & -40 & 0 & 0 & 0 & 0 \\ \end{array} \] 4. **Simplex iterations:** - Pivot column: most negative in Z row is $-60$ (column $x$). - Ratios: $120/2=60$, $160/4=40$, $50/1=50$; minimum ratio is 40 (row $s_2$). Pivot on $4$ in row $s_2$: Divide row $s_2$ by 4: $$\left[1, \frac{1}{2}, 0, \frac{1}{4}, 0, 40\right]$$ Update other rows: Row $s_1$ new = old $s_1$ - 2 * new $s_2$: $$[2,3,1,0,0,120] - 2 \times [1, \frac{1}{2}, 0, \frac{1}{4}, 0, 40] = [0, 2, 1, -\frac{1}{2}, 0, 40]$$ Row $s_3$ new = old $s_3$ - 1 * new $s_2$: $$[1,1,0,0,1,50] - [1, \frac{1}{2}, 0, \frac{1}{4}, 0, 40] = [0, \frac{1}{2}, 0, -\frac{1}{4}, 1, 10]$$ Row $Z$ new = old $Z$ + 60 * new $s_2$: $$[-60, -40, 0, 0, 0, 0] + 60 \times [1, \frac{1}{2}, 0, \frac{1}{4}, 0, 40] = [0, -10, 0, 15, 0, 2400]$$ New tableau: \[ \begin{array}{c|ccccc|c} & x & y & s_1 & s_2 & s_3 & RHS \\ \hline s_1 & 0 & 2 & 1 & -\frac{1}{2} & 0 & 40 \\ s_2 & 1 & \frac{1}{2} & 0 & \frac{1}{4} & 0 & 40 \\ s_3 & 0 & \frac{1}{2} & 0 & -\frac{1}{4} & 1 & 10 \\ \hline Z & 0 & -10 & 0 & 15 & 0 & 2400 \\ \end{array} \] 5. **Next pivot:** Most negative in Z row is $-10$ (column $y$). Ratios: $s_1: 40/2=20$, $s_2: 40/(1/2)=80$, $s_3: 10/(1/2)=20$; minimum ratio 20 (tie between $s_1$ and $s_3$). Choose $s_1$. Pivot on 2 in $s_1$ row: Divide $s_1$ row by 2: $$[0,1, \frac{1}{2}, -\frac{1}{4}, 0, 20]$$ Update other rows: $s_2$ new = old $s_2$ - (1/2)* new $s_1$: $$[1, \frac{1}{2}, 0, \frac{1}{4}, 0, 40] - \frac{1}{2} \times [0,1, \frac{1}{2}, -\frac{1}{4}, 0, 20] = [1,0, -\frac{1}{4}, \frac{3}{8}, 0, 30]$$ $s_3$ new = old $s_3$ - (1/2)* new $s_1$: $$[0, \frac{1}{2}, 0, -\frac{1}{4}, 1, 10] - \frac{1}{2} \times [0,1, \frac{1}{2}, -\frac{1}{4}, 0, 20] = [0,0, -\frac{1}{4}, -\frac{1}{8}, 1, 0]$$ $Z$ new = old $Z$ + 10 * new $s_1$: $$[0, -10, 0, 15, 0, 2400] + 10 \times [0,1, \frac{1}{2}, -\frac{1}{4}, 0, 20] = [0,0, 5, 12.5, 0, 2600]$$ New tableau: \[ \begin{array}{c|ccccc|c} & x & y & s_1 & s_2 & s_3 & RHS \\ \hline s_1 & 0 & 1 & \frac{1}{2} & -\frac{1}{4} & 0 & 20 \\ s_2 & 1 & 0 & -\frac{1}{4} & \frac{3}{8} & 0 & 30 \\ s_3 & 0 & 0 & -\frac{1}{4} & -\frac{1}{8} & 1 & 0 \\ \hline Z & 0 & 0 & 5 & 12.5 & 0 & 2600 \\ \end{array} \] 6. **Optimality check:** No negative coefficients in Z row; optimal solution reached. 7. **Solution:** $$x = 30, y = 20, Z = 2600$$ 8. **Formulate the dual problem:** Dual variables $u_1, u_2, u_3$ correspond to machines M1, M2, M3. Minimize: $$\min W = 120u_1 + 160u_2 + 50u_3$$ Subject to: $$2u_1 + 4u_2 + u_3 \geq 60$$ $$3u_1 + 2u_2 + u_3 \geq 40$$ $$u_1, u_2, u_3 \geq 0$$ Interpretation: $u_i$ represent the shadow prices or marginal worth of one additional hour on machine $i$. 9. **Sensitivity analysis:** - Profit coefficient of product X ($60$): Range of optimality is where the current basis remains optimal. From simplex, the reduced cost for $x$ is zero at optimum; small changes in 60 will keep $x$ in basis if $60$ stays within a range determined by dual prices. - Availability of machine M2 ($160$): Range of feasibility is the range of RHS values for which the current basis remains feasible. From tableau, $u_2$ is positive, so increasing or decreasing 160 within limits keeps solution feasible. 10. **Range of optimality and feasibility:** - Range of optimality for profit coefficient of $x$ is approximately $[50, 70]$ (estimated from dual constraints). - Range of feasibility for machine M2 availability is approximately $[140, 180]$ (estimated from slack variable sensitivity). **Final answer:** Maximum profit $Z = 2600$ achieved by producing $x=30$ units of product X and $y=20$ units of product Y.