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 and Model II. 2. **Define variables:** Let $x$ = number of Model I screens made. Let $y$ = number of Model II screens made. 3. **Constraints:** - Metal constraint: Model I requires 2 units, Model II requires 1 unit, total metal available is 6 units. $$2x + y \leq 6$$ - Time constraint: Model I requires 7 hours, Model II requires 8 hours, total time available is 28 hours. $$7x + 8y \leq 28$$ - Non-negativity constraints: $$x \geq 0, \quad y \geq 0$$ 4. **Objective function:** Maximize sales revenue: $$Z = 120x + 80y$$ 5. **Graphical method:** Plot the constraints: - From $2x + y = 6$, intercepts are $x=3$ (when $y=0$) and $y=6$ (when $x=0$). - From $7x + 8y = 28$, intercepts are $x=4$ (when $y=0$) and $y=3.5$ (when $x=0$). 6. **Find corner points of feasible region:** - Point A: $(0,0)$ - Point B: $(3,0)$ from metal constraint - Point C: Intersection of $2x + y = 6$ and $7x + 8y = 28$ - Point D: $(0,3.5)$ from time constraint 7. **Calculate intersection point C:** 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, $$y = 6 - 2 \times \frac{20}{9} = 6 - \frac{40}{9} = \frac{54}{9} - \frac{40}{9} = \frac{14}{9} \approx 1.56$$ 8. **Evaluate objective function at corner points:** - At A $(0,0)$: $Z = 120(0) + 80(0) = 0$ - At B $(3,0)$: $Z = 120(3) + 80(0) = 360$ - At C $(\frac{20}{9}, \frac{14}{9})$: $$Z = 120 \times \frac{20}{9} + 80 \times \frac{14}{9} = \frac{2400}{9} + \frac{1120}{9} = \frac{3520}{9} \approx 391.11$$ - At D $(0,3.5)$: $Z = 120(0) + 80(3.5) = 280$ 9. **Conclusion:** The maximum revenue is approximately $391.11$ when the machinist makes about 2.22 Model I screens and 1.56 Model II screens. Since the number of screens must be whole numbers, the machinist should consider nearby integer points within constraints for practical purposes.