1. **State the problem:** We need to find the best product mix of products A, B, C, and D to maximize profit given constraints on preparatory and finisher machine hours. 2. **Identify the data:** - Selling price per unit: $P_A=15$, $P_B=18$, $P_C=20$, $P_D=25$ - Variable cost per unit: $VC_A=17$, $VC_B=11$, $VC_C=10$, $VC_D=16$ - Units produced per preparatory machine hour: $U_{prep,A}=3$, $U_{prep,B}=4$, $U_{prep,C}=2$, $U_{prep,D}=3$ - Units produced per finisher machine hour: $U_{fin,A}=2$, $U_{fin,B}=3$, $U_{fin,C}=4$, $U_{fin,D}=2$ - Maximum machine hours: preparatory = 4000, finisher = 4000 3. **Calculate contribution margin per unit:** $$CM_i = P_i - VC_i$$ $$CM_A = 15 - 17 = -2$$ $$CM_B = 18 - 11 = 7$$ $$CM_C = 20 - 10 = 10$$ $$CM_D = 25 - 16 = 9$$ 4. **Calculate contribution margin per machine hour for each machine:** - Preparatory machine hour contribution: $$CM_{prep,i} = CM_i \times U_{prep,i}$$ $$CM_{prep,A} = -2 \times 3 = -6$$ $$CM_{prep,B} = 7 \times 4 = 28$$ $$CM_{prep,C} = 10 \times 2 = 20$$ $$CM_{prep,D} = 9 \times 3 = 27$$ - Finisher machine hour contribution: $$CM_{fin,i} = CM_i \times U_{fin,i}$$ $$CM_{fin,A} = -2 \times 2 = -4$$ $$CM_{fin,B} = 7 \times 3 = 21$$ $$CM_{fin,C} = 10 \times 4 = 40$$ $$CM_{fin,D} = 9 \times 2 = 18$$ 5. **Interpretation:** Product A has negative contribution margin, so it should not be produced. 6. **Formulate constraints:** Let $x_B, x_C, x_D$ be units produced of products B, C, and D. Preparatory machine hours used: $$\frac{x_B}{4} + \frac{x_C}{2} + \frac{x_D}{3} \leq 4000$$ Finisher machine hours used: $$\frac{x_B}{3} + \frac{x_C}{4} + \frac{x_D}{2} \leq 4000$$ 7. **Objective function (maximize profit):** $$Z = 7x_B + 10x_C + 9x_D$$ 8. **Solve the linear programming problem:** To maximize $Z$ subject to the constraints above. 9. **Heuristic approach:** Calculate contribution margin per combined machine hour for each product: - Total machine hours per unit: $$MH_B = \frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}$$ $$CM/MH_B = \frac{7}{7/12} = 12$$ $$MH_C = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$$ $$CM/MH_C = \frac{10}{3/4} = \frac{10 \times 4}{3} = 13.33$$ $$MH_D = \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}$$ $$CM/MH_D = \frac{9}{5/6} = \frac{9 \times 6}{5} = 10.8$$ 10. **Rank products by contribution margin per machine hour:** C (13.33) > B (12) > D (10.8) 11. **Allocate machine hours to product C first:** Max units of C limited by preparatory machine hours: $$x_C \leq 4000 \times 2 = 8000$$ Max units of C limited by finisher machine hours: $$x_C \leq 4000 \times 4 = 16000$$ So max $x_C = 8000$ (preparatory is limiting) 12. **Use machine hours after producing 8000 units of C:** Preparatory hours used: $$\frac{8000}{2} = 4000$$ (all preparatory hours used) Finisher hours used: $$\frac{8000}{4} = 2000$$ No preparatory hours left for B or D, so only product C is produced. 13. **Calculate total profit:** $$Z = 10 \times 8000 = 80000$$ **Final answer:** Produce 8000 units of product C only to maximize profit of 80000.