1. **Problem Statement:** Formulate and solve the linear programming (LP) model to maximize profit from chairs and tables made with limited resources. 2. **Variables:** Let $x$ = number of chairs, $y$ = number of tables. 3. **Objective Function:** Maximize profit: $$Z = 20x + 16y$$ 4. **Constraints:** - Cutting time: $$3x + y \leq 24$$ (hours) - Assembly time: $$4x + 2y \leq 48$$ (hours) - Hockey sticks: $$2x + 5y \leq 10$$ (sticks) - At least 2 chairs per table: $$x \geq 2y$$ - Non-negativity: $$x \geq 0, y \geq 0$$ --- 5. **Plotting the constraints:** - Lines for each inequality boundary: - Cutting time: $y = 24 - 3x$ - Assembly time: $y = 24 - 2x$ - Hockey sticks: $y = \frac{10 - 2x}{5}$ - Chairs to tables ratio: $x = 2y$ 6. **Corner points (intersections of constraints):** - Intersect $x=2y$ and $3x+y=24$: $$3(2y)+y=24 \Rightarrow 6y + y =24 \Rightarrow 7y=24 \Rightarrow y=\frac{24}{7} \approx 3.43, x=2y=6.86$$ - Intersect $x=2y$ and $4x + 2y=48$: $$4(2y) + 2y = 48 \Rightarrow 8y + 2y=48 \Rightarrow 10y=48 \Rightarrow y=4.8, x=9.6$$ - Intersect $x=2y$ and $2x+5y=10$: $$2(2y)+5y=10 \Rightarrow 4y+5y=10 \Rightarrow 9y=10 \Rightarrow y=\frac{10}{9} \approx 1.11, x=2.22$$ - Intersect $3x + y=24$ and $4x + 2y=48$: Multiply first by 2: $$6x + 2y = 48$$ Subtract second: $$6x+2y - (4x+2y) = 48 -48 \Rightarrow 2x = 0 \Rightarrow x=0$$ Substitute into $3x + y=24$: $$3(0) + y = 24 \Rightarrow y=24$$ - Intersect $3x + y=24$ and $2x + 5y=10$: Multiply first by 5: $$15x + 5y = 120$$ Subtract second: $$15x + 5y - (2x + 5y) = 120 - 10 \Rightarrow 13x =110 \Rightarrow x= \frac{110}{13} \approx 8.46$$ Substitute in $3x + y=24$: $$3(8.46) + y = 24 \Rightarrow 25.38 + y=24 \Rightarrow y=-1.38$$ (Not feasible since $y\geq0$) - Intersect $4x + 2y=48$ and $2x + 5y=10$: Multiply second by 2: $$4x + 10y = 20$$ Subtract first: $$ (4x + 10y) - (4x + 2y) = 20 - 48 \Rightarrow 8y = -28 \Rightarrow y = -3.5$$ (Not feasible) 7. **Feasible corner points (non-negative and satisfy all constraints):** - $(0,0)$ - $(0,1.11)$ from $2x + 5y=10$ and $x=0$ - $(2.22,1.11)$ from $x = 2y$ and $2x + 5y=10$ - $(6.86,3.43)$ from $x=2y$ and $3x + y=24$ - $(9.6,4.8)$ from $x=2y$ and $4x + 2y=48$ 8. **Evaluate objective at corner points:** - At $(0,0)$: $Z=0$ - At $(0,1.11)$: $Z=20(0)+16(1.11)=17.78$ - At $(2.22,1.11)$: $Z=20(2.22)+16(1.11)=44.4+17.78=62.18$ - At $(6.86,3.43)$: $Z=20(6.86)+16(3.43)=137.2+54.88=192.08$ - At $(9.6,4.8)$: $Z=20(9.6)+16(4.8)=192+76.8=268.8$ 9. **Check feasibility of $(9.6,4.8)$ with hockey sticks constraint:** $$2(9.6) + 5(4.8) = 19.2 + 24 = 43.2 > 10$$ Not feasible. 10. **Check hockey sticks constraint at $(6.86,3.43)$:** $$2(6.86) + 5(3.43) = 13.72 + 17.15 = 30.87 > 10$$ Not feasible. 11. **Out of feasible points, the binding hockey sticks constraint reduces $y$ drastically; compute intersection between $2x + 5y=10$ and $x=2y$ (already given): $(2.22,1.11)$. This is hence likely the optimal solution. **Final answer:** Max profit is $$Z = 62.18$$ at approximately $$x=2.22$$ chairs and $$y=1.11$$ tables.