1. **State the problem:** We want to maximize the objective function $$Q = 3y + 2x$$ subject to the constraints: - $$x \geq 0$$ - $$y \geq 0$$ - $$x + 3y \geq 12$$ - $$x + 2y \leq 10$$ 2. **Understand the constraints:** - The first two constraints restrict $$x$$ and $$y$$ to the first quadrant (including axes). - The third constraint $$x + 3y \geq 12$$ represents the region above or on the line $$x + 3y = 12$$. - The fourth constraint $$x + 2y \leq 10$$ represents the region below or on the line $$x + 2y = 10$$. 3. **Find the feasible region:** The feasible region is where all constraints overlap. 4. **Find intersection points of the boundary lines:** - Intersection of $$x + 3y = 12$$ and $$x + 2y = 10$$: $$\begin{cases} x + 3y = 12 \\ x + 2y = 10 \end{cases}$$ Subtract second from first: $$x + 3y - (x + 2y) = 12 - 10 \Rightarrow y = 2$$ Substitute $$y=2$$ into $$x + 2y = 10$$: $$x + 2(2) = 10 \Rightarrow x + 4 = 10 \Rightarrow x = 6$$ So intersection point is $$(6, 2)$$. - Intersection of $$x + 3y = 12$$ with axes: - When $$x=0$$, $$3y=12 \Rightarrow y=4$$ - When $$y=0$$, $$x=12$$ - Intersection of $$x + 2y = 10$$ with axes: - When $$x=0$$, $$2y=10 \Rightarrow y=5$$ - When $$y=0$$, $$x=10$$ 5. **Check which points satisfy all constraints:** - Check point $$(6, 2)$$: - $$x \geq 0$$ and $$y \geq 0$$ true - $$x + 3y = 6 + 6 = 12 \geq 12$$ true - $$x + 2y = 6 + 4 = 10 \leq 10$$ true - Check point $$(0,4)$$: - $$x + 3y = 0 + 12 = 12 \geq 12$$ true - $$x + 2y = 0 + 8 = 8 \leq 10$$ true - Check point $$(10,0)$$: - $$x + 3y = 10 + 0 = 10 \geq 12$$ false, so not feasible - Check point $$(0,5)$$: - $$x + 3y = 0 + 15 = 15 \geq 12$$ true - $$x + 2y = 0 + 10 = 10 \leq 10$$ true - But $$x + 3y \geq 12$$ is true, but $$x + 2y \leq 10$$ is also true - However, $$x + 3y \geq 12$$ is true, so feasible - Check point $$(12,0)$$: - $$x + 2y = 12 + 0 = 12 \leq 10$$ false, so not feasible 6. **Evaluate objective function $$Q = 3y + 2x$$ at feasible vertices:** - At $$(6, 2)$$: $$Q = 3(2) + 2(6) = 6 + 12 = 18$$ - At $$(0, 4)$$: $$Q = 3(4) + 2(0) = 12 + 0 = 12$$ - At $$(0, 5)$$: $$Q = 3(5) + 2(0) = 15 + 0 = 15$$ 7. **Determine maximum:** The maximum value of $$Q$$ is $$18$$ at $$(6, 2)$$. **Final answer:** $$\boxed{\max Q = 18 \text{ at } (x,y) = (6,2)}$$