1. **State the problem:** We need to solve the first linear programming problem using the graphical method: Maximize $$Z = 6X_1 + 4X_2$$ Subject to: $$X_1 + 2X_2 \leq 8$$ $$2X_1 + X_2 \leq 10$$ $$X_1, X_2 \geq 0$$ 2. **Plot the constraints:** Rewrite each inequality as an equation to find boundary lines: - $$X_1 + 2X_2 = 8$$ - $$2X_1 + X_2 = 10$$ 3. **Find intercepts for each constraint:** - For $$X_1 + 2X_2 = 8$$: - When $$X_1=0$$, $$2X_2=8 \Rightarrow X_2=4$$ - When $$X_2=0$$, $$X_1=8$$ - For $$2X_1 + X_2 = 10$$: - When $$X_1=0$$, $$X_2=10$$ - When $$X_2=0$$, $$2X_1=10 \Rightarrow X_1=5$$ 4. **Identify feasible region:** The feasible region is bounded by the lines and the axes $$X_1=0$$ and $$X_2=0$$. 5. **Find corner points of the feasible region:** - Intersection of $$X_1 + 2X_2 = 8$$ and $$2X_1 + X_2 = 10$$: Solve the system: $$X_1 + 2X_2 = 8$$ $$2X_1 + X_2 = 10$$ Multiply first equation by 2: $$2X_1 + 4X_2 = 16$$ Subtract second equation: $$2X_1 + 4X_2 - (2X_1 + X_2) = 16 - 10$$ $$3X_2 = 6 \Rightarrow X_2 = 2$$ Substitute $$X_2=2$$ into first equation: $$X_1 + 2(2) = 8 \Rightarrow X_1 + 4 = 8 \Rightarrow X_1 = 4$$ - Other corner points are where constraints meet axes: - $$ (0,0) $$ - $$ (0,4) $$ from $$X_1 + 2X_2 = 8$$ - $$ (5,0) $$ from $$2X_1 + X_2 = 10$$ 6. **Evaluate objective function $$Z$$ at each corner point:** - At $$ (0,0) $$: $$Z = 6(0) + 4(0) = 0$$ - At $$ (0,4) $$: $$Z = 6(0) + 4(4) = 16$$ - At $$ (5,0) $$: $$Z = 6(5) + 4(0) = 30$$ - At $$ (4,2) $$: $$Z = 6(4) + 4(2) = 24 + 8 = 32$$ 7. **Conclusion:** The maximum value of $$Z$$ is $$32$$ at $$ (X_1, X_2) = (4, 2) $$.