1. **Problem Statement:** We want to maximize the profit from producing Alloy Gamma (x tons) and Alloy Delta (y tons) given constraints on raw materials Copper, Zinc, and Nickel. 2. **Variables:** Let: - $x$ = tons of Alloy Gamma produced - $y$ = tons of Alloy Delta produced - $c, z, N$ = slack variables for unused Copper, Zinc, Nickel respectively 3. **Objective Function:** Maximize profit: $$\text{Maximize } P = 500x + 450y$$ 4. **Constraints:** From the raw material limits: - Copper: $$0.2x + 0.4y + c = 1200$$ - Zinc: $$0.3x + 0.5y + z = 1500$$ - Nickel: $$0.4x + 0.2y + N = 1600$$ with $c, z, N \geq 0$. 5. **Initial Tableau:** Given in the problem. 6. **First Pivot:** Pivot column is $x$ (since $c_j - z_j$ is max for $x$), pivot row is Nickel (N) because $\frac{1600}{0.4} = 4000$ is minimum positive ratio. 7. **Second Tableau:** Calculate new basic variables after pivoting on $x$ in Nickel row: - New basic variable: $x$ replaces $N$. - Update quantities and coefficients: Nickel row divided by pivot element 0.4: $$x + 0.5y + 2.5N = 4000$$ Update other rows: Copper row: subtract $0.2 \times$ new Nickel row from Copper row: $$c - 0.2x - 0.4y = 1200$$ becomes $$c - 0.2(x + 0.5y + 2.5N) - 0.4y = 1200 - 0.2 \times 1600$$ Simplify: $$c - 0.2x - 0.1y - 0.5N - 0.4y = 1200 - 320 = 880$$ $$c - 0.2x - 0.5y - 0.5N = 880$$ Replace $x$ with Nickel row expression: $$c - 0.2(4000 - 0.5y - 2.5N) - 0.5y - 0.5N = 880$$ Simplify: $$c - 800 + 0.1y + 0.5N - 0.5y - 0.5N = 880$$ $$c - 800 - 0.4y = 880$$ $$c - 0.4y = 1680$$ Similarly for Zinc row: $$z - 0.3x - 0.5y = 1500$$ Substitute $x$: $$z - 0.3(4000 - 0.5y - 2.5N) - 0.5y = 1500$$ $$z - 1200 + 0.15y + 0.75N - 0.5y = 1500$$ $$z - 1200 - 0.35y + 0.75N = 1500$$ $$z - 0.35y + 0.75N = 2700$$ Objective row update: $$z_j = 500 \times 1 + 450 \times 0 = 500$$ $$c_j - z_j$$ for $y$: $$450 - (500 \times 0.5) = 450 - 250 = 200 > 0$$ So $y$ can enter. Second Tableau: Basic | Cj | 500 | 450 | 0 | 0 | 0 | Quantity ---|---|---|---|---|---|---|--- $x$ | 500 | 1 | 0.5 | 0 | 0 | 2.5 | 4000 $c$ | 0 | 0 | -0.4 | 1 | 0 | 0 | 1680 $z$ | 0 | 0 | -0.35 | 0 | 1 | 0.75 | 2700 Gross | zj | 500 | 250 | 0 | 0 | 1250 | 0 Net | cj - zj | 0 | 200 | 0 | 0 | -1250 | Total Profit Max+9 | Yes | No | Yes | No | No | No 8. **Second Pivot:** Pivot column is $y$ (highest positive net profit 200), pivot row is $c$ because $\frac{1680}{0.4} = 4200$ is less than $\frac{2700}{0.35} \approx 7714$. 9. **Third Tableau:** Pivot on $y$ in $c$ row: Divide $c$ row by $-0.4$: $$y - 2.5c = -4200$$ Update other rows: $x$ row: $$x + 0.5y + 2.5N = 4000$$ Substitute $y$: $$x + 0.5(-2.5c - 4200) + 2.5N = 4000$$ $$x - 1.25c - 2100 + 2.5N = 4000$$ $$x - 1.25c + 2.5N = 6100$$ $z$ row: $$z - 0.35y + 0.75N = 2700$$ Substitute $y$: $$z - 0.35(-2.5c - 4200) + 0.75N = 2700$$ $$z + 0.875c + 1470 + 0.75N = 2700$$ $$z + 0.875c + 0.75N = 1230$$ Objective row: $$z_j = 500 + 450 \times 0.5 = 500 + 225 = 725$$ Net profit for $c$: $$0 - 450 \times (-2.5) = 1125 > 0$$ $c$ can enter. Third Tableau: Basic | Cj | 500 | 450 | 0 | 0 | 0 | Quantity ---|---|---|---|---|---|---|--- $x$ | 500 | 1 | 0 | -1.25 | 0 | 2.5 | 6100 $y$ | 450 | 0 | 1 | 2.5 | 0 | 0 | 4200 $z$ | 0 | 0 | 0 | 0.875 | 1 | 0.75 | 1230 Gross | zj | 500 | 450 | 0 | 0 | 1250 | 0 Net | cj - zj | 0 | 0 | 1125 | 0 | -1250 | Total Profit Max+9 | Yes | Yes | Yes | No | No | No 10. **Solution:** - Produce $x = 6100$ tons of Alloy Gamma - Produce $y = 4200$ tons of Alloy Delta - Total profit: $$P = 500 \times 6100 + 450 \times 4200 = 3,050,000 + 1,890,000 = 4,940,000$$ Slack variables $c, z, N$ are non-basic and zero, meaning all raw materials are fully used. **Final answer:** Produce 6100 tons of Alloy Gamma and 4200 tons of Alloy Delta weekly to maximize profit of 4,940,000.