1. **State the problem:** We want to maximize the weekly profit from producing two toy guns: Acer and Bulls-I. 2. **Define variables:** Let $x$ = dozens of Bulls-I produced. Let $y$ = dozens of Acer produced. 3. **Objective function:** Maximize profit $P = 5x + 8y$ (since Bulls-I yields 5 profit per dozen, Acer yields 8). 4. **Constraints:** - Plastic: $2y + 1x \leq 1000$ (pounds of plastic) - Labor: $3y + 4x \leq 2400$ (minutes; 40 hours = 2400 minutes) - Marketing total: $x + y \leq 700$ (dozens) - Marketing difference: $y - x \leq 350$ - Non-negativity: $x \geq 0$, $y \geq 0$ 5. **Rewrite constraints for clarity:** - Plastic: $2y + x \leq 1000$ - Labor: $3y + 4x \leq 2400$ - Total production: $x + y \leq 700$ - Difference: $y \leq x + 350$ 6. **Graphical interpretation:** The feasible region is bounded by these inequalities, forming a polygon. 7. **Find intersection points (corner points) of constraints:** - Intersection of plastic and labor: Solve system: $$\begin{cases} 2y + x = 1000 \\ 3y + 4x = 2400 \end{cases}$$ Multiply first by 4: $$8y + 4x = 4000$$ Subtract second: $$(8y + 4x) - (3y + 4x) = 4000 - 2400 \Rightarrow 5y = 1600 \Rightarrow y = 320$$ Then $x = 1000 - 2y = 1000 - 640 = 360$ - Intersection of plastic and total production: $$\begin{cases} 2y + x = 1000 \\ x + y = 700 \end{cases}$$ From second: $x = 700 - y$ Substitute: $$2y + 700 - y = 1000 \Rightarrow y = 300$$ Then $x = 700 - 300 = 400$ - Intersection of labor and total production: $$\begin{cases} 3y + 4x = 2400 \\ x + y = 700 \end{cases}$$ From second: $y = 700 - x$ Substitute: $$3(700 - x) + 4x = 2400 \Rightarrow 2100 - 3x + 4x = 2400 \Rightarrow x = 300$$ Then $y = 700 - 300 = 400$ - Intersection of difference and total production: $$\begin{cases} y = x + 350 \\ x + y = 700 \end{cases}$$ Substitute first into second: $$x + (x + 350) = 700 \Rightarrow 2x = 350 \Rightarrow x = 175$$ Then $y = 175 + 350 = 525$ - Check if these points satisfy all constraints and non-negativity. 8. **Evaluate profit at each feasible corner point:** - At $(x,y) = (360,320)$: $$P = 5(360) + 8(320) = 1800 + 2560 = 4360$$ - At $(400,300)$: $$P = 5(400) + 8(300) = 2000 + 2400 = 4400$$ - At $(300,400)$: $$P = 5(300) + 8(400) = 1500 + 3200 = 4700$$ - At $(175,525)$: $$P = 5(175) + 8(525) = 875 + 4200 = 5075$$ 9. **Check feasibility of $(175,525)$:** - Plastic: $2(525) + 175 = 1050 + 175 = 1225 > 1000$ (violates plastic constraint) So $(175,525)$ is not feasible. 10. **Check other points for all constraints:** - $(300,400)$: Plastic: $2(400) + 300 = 800 + 300 = 1100 > 1000$ (not feasible) - $(400,300)$: Plastic: $2(300) + 400 = 600 + 400 = 1000$ (feasible) Labor: $3(300) + 4(400) = 900 + 1600 = 2500 > 2400$ (not feasible) - $(360,320)$: Plastic: $2(320) + 360 = 640 + 360 = 1000$ (feasible) Labor: $3(320) + 4(360) = 960 + 1440 = 2400$ (feasible) Total: $360 + 320 = 680 \leq 700$ (feasible) Difference: $320 - 360 = -40 \leq 350$ (feasible) 11. **Conclusion:** The only feasible corner point with maximum profit is $(x,y) = (360,320)$ with profit $P = 4360$. 12. **Compare with current plan:** Current profit is 4100 at $(100,450)$. New plan increases profit to 4360. **Final answer:** Produce 360 dozens of Bulls-I and 320 dozens of Acer to maximize profit at 4360 per week.