1. **State the problem:** We want to maximize the profit from manufacturing chairs, tables, and bookcases given constraints on cutting, assembly, and finishing hours. 2. **Define variables:** Let $x$ = number of chairs, $y$ = number of tables, $z$ = number of bookcases. 3. **Write the objective function:** Maximize profit: $$P = 20x + 30y + 25z$$ 4. **Write the constraints based on operation hours:** Cutting: $$1x + 1y + 3z \leq 600$$ Assembly: $$1x + 2y + 1z \leq 500$$ Finishing: $$1x + 1y + 1z \leq 300$$ 5. **Non-negativity constraints:** $$x, y, z \geq 0$$ 6. **Convert inequalities to equalities by adding slack variables:** $$x + y + 3z + s_1 = 600$$ $$x + 2y + z + s_2 = 500$$ $$x + y + z + s_3 = 300$$ where $s_1, s_2, s_3 \geq 0$ are slack variables. 7. **Set up the initial simplex tableau:** | Basis | $x$ | $y$ | $z$ | $s_1$ | $s_2$ | $s_3$ | RHS | |-------|-----|-----|-----|-------|-------|-------|-----| | $s_1$ | 1 | 1 | 3 | 1 | 0 | 0 | 600 | | $s_2$ | 1 | 2 | 1 | 0 | 1 | 0 | 500 | | $s_3$ | 1 | 1 | 1 | 0 | 0 | 1 | 300 | | $P$ | -20 | -30 | -25 | 0 | 0 | 0 | 0 | 8. **Perform simplex iterations:** - Identify entering variable: most negative coefficient in $P$ row is $-30$ for $y$. - Identify leaving variable by minimum ratio test: - $s_1$: $600/1=600$ - $s_2$: $500/2=250$ - $s_3$: $300/1=300$ So $s_2$ leaves. 9. **Pivot on $y$ in $s_2$ row:** Divide $s_2$ row by 2: $$[0.5, 1, 0.5, 0, 0.5, 0, 250]$$ 10. **Update other rows to zero out $y$ column:** - $s_1$ new = $s_1$ - $1 \times s_2$ row - $s_3$ new = $s_3$ - $1 \times s_2$ row - $P$ new = $P$ + $30 \times s_2$ row After calculations: $s_1$: $[1 - 0.5, 1 - 1, 3 - 0.5, 1 - 0, 0 - 0.5, 0 - 0, 600 - 250] = [0.5, 0, 2.5, 1, -0.5, 0, 350]$ $s_3$: $[1 - 0.5, 1 - 1, 1 - 0.5, 0 - 0, 0 - 0.5, 1 - 0, 300 - 250] = [0.5, 0, 0.5, 0, -0.5, 1, 50]$ $P$: $[-20 + 30*0.5, -30 + 30*1, -25 + 30*0.5, 0 + 30*0, 0 + 30*0.5, 0 + 30*0, 0 + 30*250] = [-5, 0, -10, 0, 15, 0, 7500]$ 11. **Next iteration:** Entering variable: most negative in $P$ row is $-10$ for $z$. Leaving variable by ratio test: - $s_1$: $350/2.5=140$ - $s_3$: $50/0.5=100$ $s_3$ leaves. 12. **Pivot on $z$ in $s_3$ row:** Divide $s_3$ row by 0.5: $$[1, 0, 1, 0, -1, 2, 100]$$ 13. **Update other rows:** $s_1$ new = $s_1$ - $2.5 \times s_3$ row $P$ new = $P$ + $10 \times s_3$ row $s_1$: $[0.5 - 2.5*1, 0 - 2.5*0, 2.5 - 2.5*1, 1 - 2.5*0, -0.5 - 2.5*(-1), 0 - 2.5*2, 350 - 2.5*100] = [-2, 0, 0, 1, 2, -5, 100]$ $P$: $[-5 + 10*1, 0 + 10*0, -10 + 10*1, 0 + 10*0, 15 + 10*(-1), 0 + 10*2, 7500 + 10*100] = [5, 0, 0, 0, 5, 20, 8500]$ 14. **Check optimality:** All coefficients in $P$ row for variables $x,y,z$ are non-negative, so optimal solution reached. 15. **Read solution:** Basic variables: $y$ and $z$ from previous pivots, and $s_1$. From $s_3$ row: $z = 100$ From $s_2$ row (now $y$ row): $y = 250$ From $s_1$ row: $x = 0$ (since $x$ is non-basic) 16. **Interpretation:** Produce 0 chairs, 250 tables, and 100 bookcases to maximize profit. 17. **Maximum profit:** $$P = 20(0) + 30(250) + 25(100) = 0 + 7500 + 2500 = 10000$$ **Final answer:** Produce 0 chairs, 250 tables, and 100 bookcases for a maximum profit of 10000.