1. **State the problem:** Graph the system of inequalities: $$x_1 + x_2 \leq 7$$ $$x_1 + 6x_2 \leq 12$$ $$x_1, x_2 \geq 0$$ Find the corner points of the feasible region and verify they correspond to basic feasible solutions. 2. **Plot the boundary lines:** - For $$x_1 + x_2 = 7$$, intercepts are $$x_1=7, x_2=7$$. - For $$x_1 + 6x_2 = 12$$, intercepts are $$x_1=12, x_2=2$$. 3. **Find intersection of the two lines:** Solve the system: $$\begin{cases} x_1 + x_2 = 7 \\ x_1 + 6x_2 = 12 \end{cases}$$ Subtract first from second: $$x_1 + 6x_2 - (x_1 + x_2) = 12 - 7$$ $$\cancel{x_1} + 6x_2 - \cancel{x_1} - x_2 = 5$$ $$5x_2 = 5$$ $$x_2 = 1$$ Substitute back: $$x_1 + 1 = 7$$ $$x_1 = 6$$ 4. **List corner points:** - $$ (0,0) $$ from $$x_1, x_2 \geq 0$$ - $$ (7,0) $$ from $$x_1 + x_2 = 7$$ when $$x_2=0$$ - $$ (6,1) $$ intersection point 5. **Verify basic feasible solutions:** Each corner point corresponds to a solution where two constraints are active (equalities), confirming they are basic feasible solutions. **Final answer:** The corner points of the feasible region are $$ (0,0), (7,0), (6,1) $$.