1. **Problem Statement:** We want to maximize the profit from producing tables and chairs given limited hours in carpentry and painting departments. 2. **Define Variables:** Let $x$ = number of tables produced Let $y$ = number of chairs produced 3. **Constraints:** - Carpentry hours: Each table takes 4 hours, each chair takes 3 hours (implied from problem context, but since only tables are given 4 hours and chairs not specified, assume chairs take 3 hours for example; if not given, assume variable $c$ for chairs hours) - Painting hours: Each table takes 1 hour, each chair takes 2 hours (again, assumed 2 hours for chairs as not specified) - Available carpentry hours: 240 - Available painting hours: 100 Since the problem only specifies tables require 4 hours carpentry and 1 hour painting, and chairs require some hours but not specified, we will assume chairs require $a$ hours carpentry and $b$ hours painting. But since problem only gives tables hours, we will only use tables hours and chairs hours as variables. Re-examining problem: It says both require carpentry and painting hours, but only tables hours are given explicitly. Since chairs hours are not given, we cannot form exact constraints for chairs hours. So we will assume chairs require $c$ hours carpentry and $d$ hours painting. But since problem only gives tables hours, we will only use tables hours and chairs hours as variables. Re-examining problem: It says both require carpentry and painting hours, but only tables hours are given explicitly. Since chairs hours are not given, we cannot form exact constraints for chairs hours. So we will assume chairs require $c$ hours carpentry and $d$ hours painting. But problem states: "Each table takes 4 hours of carpentry and 1 hour in painting and varnishing." "Each chair produced is sold for a $5 profit." No hours for chairs given. Assuming chairs take 3 hours carpentry and 2 hours painting (typical assumption for example). So constraints: $$4x + 3y \leq 240$$ $$1x + 2y \leq 100$$ 4. **Objective Function:** Maximize profit: $$P = 7x + 5y$$ 5. **Non-negativity constraints:** $$x \geq 0, y \geq 0$$ 6. **Summary of Linear Programming Model:** Maximize: $$P = 7x + 5y$$ Subject to: $$4x + 3y \leq 240$$ $$x + 2y \leq 100$$ $$x \geq 0, y \geq 0$$ 7. **Graphing:** Plot the constraints on the $xy$-plane: - $4x + 3y = 240$ - $x + 2y = 100$ The feasible region is the area satisfying all inequalities. 8. **Finding corner points:** - Intersection with axes: - For $4x + 3y = 240$: - $x=0 \Rightarrow y=80$ - $y=0 \Rightarrow x=60$ - For $x + 2y = 100$: - $x=0 \Rightarrow y=50$ - $y=0 \Rightarrow x=100$ - Intersection of two lines: Solve system: $$4x + 3y = 240$$ $$x + 2y = 100$$ Multiply second by 3: $$3x + 6y = 300$$ Subtract first: $$(3x + 6y) - (4x + 3y) = 300 - 240$$ $$-x + 3y = 60$$ $$3y = x + 60$$ $$y = \frac{x}{3} + 20$$ Substitute into second: $$x + 2\left(\frac{x}{3} + 20\right) = 100$$ $$x + \frac{2x}{3} + 40 = 100$$ $$\frac{3x}{3} + \frac{2x}{3} = 60$$ $$\frac{5x}{3} = 60$$ $$x = \frac{60 \times 3}{5} = 36$$ Then: $$y = \frac{36}{3} + 20 = 12 + 20 = 32$$ 9. **Evaluate profit at corner points:** - At $(0,0)$: $P=0$ - At $(0,50)$: $P=7(0)+5(50)=250$ - At $(60,0)$: $P=7(60)+5(0)=420$ - At $(36,32)$: $P=7(36)+5(32)=252+160=412$ 10. **Conclusion:** Maximum profit is $420$ at $x=60$ tables and $y=0$ chairs.