1. **State the problem:** We want to determine how many chairs ($x$) and tables ($y$) to produce to maximize or minimize an objective (not specified, so we assume maximizing production or profit) subject to machine hour constraints. 2. **Define variables:** Let $x$ = number of chairs produced per week. Let $y$ = number of tables produced per week. 3. **Write constraints from the table:** Machine M1: $3x + 3y \leq 36$ Machine M2: $5x + 2y \leq 50$ Machine M3: $2x + 6y \leq 60$ 4. **Non-negativity constraints:** $x \geq 0$, $y \geq 0$ 5. **Graph the constraints:** - For M1: $3x + 3y = 36 \Rightarrow y = \frac{36 - 3x}{3} = 12 - x$ - For M2: $5x + 2y = 50 \Rightarrow y = \frac{50 - 5x}{2}$ - For M3: $2x + 6y = 60 \Rightarrow y = \frac{60 - 2x}{6} = 10 - \frac{x}{3}$ 6. **Plot these lines and find the feasible region:** The feasible region is the intersection of the half-planes below or on these lines and $x,y \geq 0$. 7. **Find corner points by solving intersections:** - Intersection of M1 and M2: $$\begin{cases} 3x + 3y = 36 \\ 5x + 2y = 50 \end{cases}$$ Multiply first by 2 and second by 3: $$\begin{cases} 6x + 6y = 72 \\ 15x + 6y = 150 \end{cases}$$ Subtract: $$15x + 6y - (6x + 6y) = 150 - 72 \Rightarrow 9x = 78 \Rightarrow x = \frac{78}{9} = \frac{26}{3} \approx 8.67$$ Substitute into M1: $$3 \times \frac{26}{3} + 3y = 36 \Rightarrow 26 + 3y = 36 \Rightarrow 3y = 10 \Rightarrow y = \frac{10}{3} \approx 3.33$$ - Intersection of M1 and M3: $$\begin{cases} 3x + 3y = 36 \\ 2x + 6y = 60 \end{cases}$$ Multiply first by 2: $$6x + 6y = 72$$ Subtract second: $$6x + 6y - (2x + 6y) = 72 - 60 \Rightarrow 4x = 12 \Rightarrow x = 3$$ Substitute into M1: $$3 \times 3 + 3y = 36 \Rightarrow 9 + 3y = 36 \Rightarrow 3y = 27 \Rightarrow y = 9$$ - Intersection of M2 and M3: $$\begin{cases} 5x + 2y = 50 \\ 2x + 6y = 60 \end{cases}$$ Multiply first by 3 and second by 1: $$15x + 6y = 150$$ $$2x + 6y = 60$$ Subtract second from first: $$15x + 6y - (2x + 6y) = 150 - 60 \Rightarrow 13x = 90 \Rightarrow x = \frac{90}{13} \approx 6.92$$ Substitute into M2: $$5 \times 6.92 + 2y = 50 \Rightarrow 34.6 + 2y = 50 \Rightarrow 2y = 15.4 \Rightarrow y = 7.7$$ 8. **Check corner points and axes intercepts:** - Intercepts for M1: $x=0 \Rightarrow y=12$, $y=0 \Rightarrow x=12$ - Intercepts for M2: $x=0 \Rightarrow y=25$, $y=0 \Rightarrow x=10$ - Intercepts for M3: $x=0 \Rightarrow y=10$, $y=0 \Rightarrow x=30$ 9. **Summary of feasible corner points:** - $(0,0)$ - $(0,10)$ from M3 intercept - $(3,9)$ intersection M1 & M3 - $(\frac{26}{3}, \frac{10}{3})$ intersection M1 & M2 - $(\frac{90}{13}, 7.7)$ intersection M2 & M3 - $(0,12)$ from M1 intercept 10. **Objective function:** Since no objective function is given, we cannot find the optimal solution. --- **Desmos graph function:** $$y = \min(12 - x, \frac{50 - 5x}{2}, 10 - \frac{x}{3})$$ This represents the feasible region boundary.