1. **State the problem:** We want to maximize the objective function $$z = 3x_1 + 4x_2$$ subject to the constraints: $$x_1 + x_2 \leq 450$$ $$2x_1 + x_2 \leq 600$$ $$x_1, x_2 \geq 0$$ 2. **Graphical method overview:** The feasible region is the area that satisfies all constraints simultaneously, including non-negativity. The maximum value of $$z$$ will occur at a vertex (corner point) of this feasible region. 3. **Find the intercepts of the constraints:** - For $$x_1 + x_2 \leq 450$$: - When $$x_1=0$$, $$x_2=450$$ - When $$x_2=0$$, $$x_1=450$$ - For $$2x_1 + x_2 \leq 600$$: - When $$x_1=0$$, $$x_2=600$$ - When $$x_2=0$$, $$2x_1=600 \Rightarrow x_1=300$$ 4. **Find the intersection point of the two lines:** Solve the system: $$\begin{cases} x_1 + x_2 = 450 \\ 2x_1 + x_2 = 600 \end{cases}$$ Subtract the first from the second: $$2x_1 + x_2 - (x_1 + x_2) = 600 - 450 \Rightarrow x_1 = 150$$ Substitute $$x_1=150$$ into $$x_1 + x_2 = 450$$: $$150 + x_2 = 450 \Rightarrow x_2 = 300$$ 5. **List all corner points of the feasible region:** - $$A = (0,0)$$ - $$B = (0,450)$$ - $$C = (150,300)$$ (intersection) - $$D = (300,0)$$ 6. **Evaluate $$z$$ at each corner point:** - At $$A$$: $$z = 3(0) + 4(0) = 0$$ - At $$B$$: $$z = 3(0) + 4(450) = 1800$$ - At $$C$$: $$z = 3(150) + 4(300) = 450 + 1200 = 1650$$ - At $$D$$: $$z = 3(300) + 4(0) = 900$$ 7. **Conclusion:** The maximum value of $$z$$ is $$1800$$ at the point $$B = (0,450)$$. Hence, the solution is $$x_1=0$$, $$x_2=450$$ with maximum $$z=1800$$.