1. **State the problem:** A machinist has 6 units of metal and 28 hours of free time to make two models of decorative computer screens. Model I requires 2 units of metal and 7 hours, Model II requires 1 unit of metal and 8 hours. Prices are 120 for Model I and 80 for Model II. We want to maximize sales revenue. 2. **Define variables:** Let $x$ = number of Model I screens, $y$ = number of Model II screens. 3. **Write constraints:** Metal: $2x + y \leq 6$ Time: $7x + 8y \leq 28$ Non-negativity: $x \geq 0$, $y \geq 0$ 4. **Objective function:** Maximize revenue $R = 120x + 80y$ 5. **Find feasible region vertices by solving constraints:** - Intersection of $2x + y = 6$ and $7x + 8y = 28$: Multiply first by 8: $16x + 8y = 48$ Subtract second: $(16x + 8y) - (7x + 8y) = 48 - 28 \Rightarrow 9x = 20 \Rightarrow x = \frac{20}{9} \approx 2.22$ Substitute into $2x + y = 6$: $2(\frac{20}{9}) + y = 6 \Rightarrow \frac{40}{9} + y = 6 \Rightarrow y = 6 - \frac{40}{9} = \frac{54}{9} - \frac{40}{9} = \frac{14}{9} \approx 1.56$ - Intersection with axes: $x=0$: from $2(0) + y \leq 6 \Rightarrow y \leq 6$ From $7(0) + 8y \leq 28 \Rightarrow y \leq 3.5$ So max $y$ at $x=0$ is 3.5 $y=0$: from $2x + 0 \leq 6 \Rightarrow x \leq 3$ From $7x + 0 \leq 28 \Rightarrow x \leq 4$ So max $x$ at $y=0$ is 3 6. **Evaluate objective function at vertices:** - At $(0,0)$: $R=0$ - At $(0,3.5)$: $R=120(0)+80(3.5)=280$ - At $(3,0)$: $R=120(3)+80(0)=360$ - At $(\frac{20}{9}, \frac{14}{9}) \approx (2.22,1.56)$: $R=120(\frac{20}{9}) + 80(\frac{14}{9}) = \frac{2400}{9} + \frac{1120}{9} = \frac{3520}{9} \approx 391.11$ 7. **Conclusion:** Maximum revenue is approximately 391.11 when producing about 2 Model I and 1.56 Model II screens. Since partial screens may not be practical, the machinist should consider producing 2 Model I and 1 or 2 Model II screens depending on feasibility.