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. **Graphical method:** Plot the constraints and find feasible region. 6. **Find intercepts:** - For metal: when $x=0$, $y=6$; when $y=0$, $x=3$ - For time: when $x=0$, $y=3.5$; when $y=0$, $x=4$ 7. **Find corner points of feasible region:** - $(0,0)$ - $(0,3.5)$ - $(3,0)$ - Intersection of $2x + y = 6$ and $7x + 8y = 28$ 8. **Solve intersection:** From $2x + y = 6$, $y = 6 - 2x$ Substitute into $7x + 8y = 28$: $$7x + 8(6 - 2x) = 28$$ $$7x + 48 - 16x = 28$$ $$-9x = -20$$ $$x = \frac{20}{9} \approx 2.22$$ Then, $$y = 6 - 2 \times \frac{20}{9} = 6 - \frac{40}{9} = \frac{54}{9} - \frac{40}{9} = \frac{14}{9} \approx 1.56$$ 9. **Evaluate revenue at corner points:** - 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 $(2.22,1.56)$: $R=120(2.22)+80(1.56)=266.4+124.8=391.2$ 10. **Conclusion:** Maximum revenue is approximately 391.2 when making about 2 Model I and 1.56 Model II screens. Since partial screens may not be possible, the machinist should consider making 2 Model I and 1 or 2 Model II screens depending on practical constraints.