1. **Problem Statement:** Angela and Zooey want to decide how many fish dinners ($x$) and beef dinners ($y$) to prepare each night to maximize profit, given constraints on total meals, preparation time, and customer preferences. 2. **Define variables:** - Let $x$ = number of fish dinners - Let $y$ = number of beef dinners 3. **Constraints:** - Maximum meals: $$x + y \leq 60$$ - Preparation time: fish takes 15 min, beef takes 30 min (twice as long), total 20 hours = 1200 minutes $$15x + 30y \leq 1200$$ - Customer ratio: at least 3 fish for every 2 beef dinners $$x \geq \frac{3}{2} y$$ - At least 10% beef dinners: $$y \geq 0.1(x + y) \Rightarrow y \geq 0.1x + 0.1y \Rightarrow 0.9y \geq 0.1x \Rightarrow y \geq \frac{1}{9}x$$ - Non-negativity: $$x \geq 0, \quad y \geq 0$$ 4. **Objective function (profit):** - Profit from fish dinner = 12 - Profit from beef dinner = 16 $$\text{Maximize } P = 12x + 16y$$ 5. **Summary of model:** $$\begin{cases} x + y \leq 60 \\ 15x + 30y \leq 1200 \\ x \geq \frac{3}{2} y \\ y \geq \frac{1}{9} x \\ x,y \geq 0 \end{cases}$$ 6. **Solve graphically:** - Rewrite constraints: - $x + y = 60$ - $15x + 30y = 1200 \Rightarrow x + 2y = 80$ - $x = \frac{3}{2} y$ - $y = \frac{1}{9} x$ - Find feasible region bounded by these lines. - Check corner points: - Intersection of $x + y = 60$ and $x + 2y = 80$: $$x + y = 60 \Rightarrow x = 60 - y$$ Substitute into $x + 2y = 80$: $$60 - y + 2y = 80 \Rightarrow y = 20, x = 40$$ - Intersection of $x = \frac{3}{2} y$ and $x + y = 60$: Substitute $x$: $$\frac{3}{2} y + y = 60 \Rightarrow \frac{5}{2} y = 60 \Rightarrow y = 24, x = 36$$ - Intersection of $x = \frac{3}{2} y$ and $x + 2y = 80$: Substitute $x$: $$\frac{3}{2} y + 2y = 80 \Rightarrow \frac{7}{2} y = 80 \Rightarrow y = \frac{160}{7} \approx 22.86, x = 34.29$$ - Evaluate profit $P$ at these points: - $(40,20)$: $12(40) + 16(20) = 480 + 320 = 800$ - $(36,24)$: $12(36) + 16(24) = 432 + 384 = 816$ - $(34.29,22.86)$: $12(34.29) + 16(22.86) \approx 411.48 + 365.76 = 777.24$ - Maximum profit at $(36,24)$. 7. **Answer for (i):** - Prepare 36 fish dinners and 24 beef dinners each night for maximum profit of 816. --- 8. **(ii) Equal profit for fish and beef dinners:** - New profit per fish dinner = 16 (same as beef) - Objective: $P = 16x + 16y = 16(x + y)$ - Maximize $x + y$ subject to constraints. - Max $x + y$ is 60 (from first constraint). - So any combination with $x + y = 60$ is optimal. - The solution is no longer unique; profit depends only on total meals. --- 9. **(iii) At least 20% beef dinners:** - New constraint: $$y \geq 0.2(x + y) \Rightarrow y \geq 0.2x + 0.2y \Rightarrow 0.8y \geq 0.2x \Rightarrow y \geq \frac{1}{4} x$$ - Replace old $y \geq \frac{1}{9} x$ with $y \geq \frac{1}{4} x$. - Check intersection of $x = \frac{3}{2} y$ and $y = \frac{1}{4} x$: Substitute $x$: $$y = \frac{1}{4} \times \frac{3}{2} y = \frac{3}{8} y$$ This implies $y = \frac{3}{8} y$, only true if $y=0$. - So feasible region shrinks. - Check corner points again with new constraint: - Intersection of $x + y = 60$ and $x + 2y = 80$ remains $(40,20)$. - Check if $(40,20)$ satisfies $y \geq \frac{1}{4} x$: $$20 \geq \frac{1}{4} \times 40 = 10$$ True. - Intersection of $x = \frac{3}{2} y$ and $y = \frac{1}{4} x$: Substitute $x$: $$y = \frac{1}{4} \times \frac{3}{2} y = \frac{3}{8} y$$ Only $y=0$. - So feasible region is smaller, but $(36,24)$ from before satisfies $y \geq \frac{1}{4} x$? $$24 \geq \frac{1}{4} \times 36 = 9$$ True. - So solution remains feasible. - However, the constraint $y \geq \frac{1}{4} x$ is stricter than $y \geq \frac{1}{9} x$, so the feasible region shrinks, possibly reducing maximum profit. - Recalculate profit at $(36,24)$: 816 (still feasible). - Check if other points violate new constraint. - Conclusion: The stricter beef minimum may reduce flexibility but does not change the optimal solution here. --- **Final answers:** - (i) Prepare 36 fish and 24 beef dinners for max profit 816. - (ii) Equal profit makes any combination with total 60 meals optimal. - (iii) Increasing beef minimum to 20% does not change optimal solution but reduces feasible region.