1. **Stating the problem:** We want to maximize the objective function $$z = 3x + 2y$$ subject to the constraints: $$2x + y \leq 5$$ $$2x + 2y \leq 6$$ $$x, y \geq 0$$ 2. **Understanding the constraints:** These inequalities define a feasible region in the first quadrant (since $x, y \geq 0$). 3. **Find the corner points of the feasible region:** - From $2x + y \leq 5$, when $x=0$, $y \leq 5$; when $y=0$, $x \leq \frac{5}{2}$. - From $2x + 2y \leq 6$, simplify to $x + y \leq 3$; when $x=0$, $y \leq 3$; when $y=0$, $x \leq 3$. 4. **Find intersection of the two lines:** Solve the system: $$\begin{cases} 2x + y = 5 \\ x + y = 3 \end{cases}$$ Subtract second from first: $$2x + y - (x + y) = 5 - 3$$ $$x = 2$$ Substitute $x=2$ into $x + y = 3$: $$2 + y = 3 \Rightarrow y = 1$$ 5. **List corner points:** - Point A: $(0,0)$ - Point B: $(0,3)$ from $x + y = 3$ - Point C: $(2,1)$ intersection - Point D: $(\frac{5}{2},0)$ from $2x + y = 5$ 6. **Evaluate $z$ at each corner:** - At A: $z = 3(0) + 2(0) = 0$ - At B: $z = 3(0) + 2(3) = 6$ - At C: $z = 3(2) + 2(1) = 6 + 2 = 8$ - At D: $z = 3(\frac{5}{2}) + 2(0) = \frac{15}{2} = 7.5$ 7. **Conclusion:** The maximum value of $z$ is $8$ at point $(2,1)$. **Final answer:** $$\boxed{\max z = 8 \text{ at } (x,y) = (2,1)}$$