1. **State the problem:** We want to maximize the profit from producing tables (x) and chairs (y) given constraints on available hours in two departments. 2. **Formulate the objective function:** The profit per table is 30 and per chair is 24. So, the objective function to maximize is: $$Z = 30x + 24y$$ 3. **Formulate the constraints:** Department 1 hours: Producing one table takes 3 hours and one chair takes 6 hours, total hours available is 120: $$3x + 6y \leq 120$$ Department 2 hours: Producing one table takes 6 hours and one chair takes 3 hours, total hours available is 96: $$6x + 3y \leq 96$$ Non-negativity constraints: $$x \geq 0, \quad y \geq 0$$ 4. **Graph the constraints:** - For $$3x + 6y = 120$$, intercepts are: - When $$x=0$$, $$y=20$$ - When $$y=0$$, $$x=40$$ - For $$6x + 3y = 96$$, intercepts are: - When $$x=0$$, $$y=32$$ - When $$y=0$$, $$x=16$$ 5. **Determine the feasible region:** The feasible region is the area where both inequalities hold, including the axes (since $$x,y \geq 0$$). It is bounded by the lines and axes. 6. **Find corner points of the feasible region:** - Intersection with axes: - (0,0) - (0,16) from $$6x+3y=96$$ when $$x=0$$ - (20,0) from $$3x+6y=120$$ when $$y=0$$ - Intersection of the two lines: Solve system: $$3x + 6y = 120$$ $$6x + 3y = 96$$ Multiply first by 1 and second by 2: $$3x + 6y = 120$$ $$12x + 6y = 192$$ Subtract first from second: $$12x + 6y - (3x + 6y) = 192 - 120$$ $$9x = 72 \Rightarrow x = 8$$ Substitute $$x=8$$ into first equation: $$3(8) + 6y = 120 \Rightarrow 24 + 6y = 120 \Rightarrow 6y = 96 \Rightarrow y = 16$$ So intersection point is (8,16). 7. **Evaluate objective function at corner points:** - At (0,0): $$Z=0$$ - At (0,16): $$Z=30(0)+24(16)=384$$ - At (20,0): $$Z=30(20)+24(0)=600$$ - At (8,16): $$Z=30(8)+24(16)=240+384=624$$ 8. **Decision:** Maximum profit is $$624$$ at $$x=8$$ tables and $$y=16$$ chairs. Hence, the optimal production plan is to produce 8 tables and 16 chairs to maximize profit.