1. **State the problem:** Graph the system of inequalities: $$x_1 + x_2 \leq 6$$ $$x_1 + 3x_2 \leq 12$$ $$x_1, x_2 \geq 0$$ Find the corner points of the feasible region and verify they correspond to basic feasible solutions. 2. **Understand the constraints:** - Both $x_1$ and $x_2$ must be nonnegative. - The inequalities define a polygonal feasible region bounded by the lines. 3. **Find intercepts for each inequality:** - For $x_1 + x_2 \leq 6$: - When $x_1=0$, $x_2=6$ - When $x_2=0$, $x_1=6$ - For $x_1 + 3x_2 \leq 12$: - When $x_1=0$, $3x_2=12 \Rightarrow x_2=4$ - When $x_2=0$, $x_1=12$ 4. **Find intersection of the two lines:** Solve the system: $$\begin{cases} x_1 + x_2 = 6 \\ x_1 + 3x_2 = 12 \end{cases}$$ Subtract first from second: $$x_1 + 3x_2 - (x_1 + x_2) = 12 - 6$$ $$\cancel{x_1} + 3x_2 - \cancel{x_1} - x_2 = 6$$ $$2x_2 = 6$$ $$x_2 = 3$$ Substitute back: $$x_1 + 3 = 6 \Rightarrow x_1 = 3$$ 5. **List corner points of feasible region:** - $(0,0)$ from $x_1, x_2 \geq 0$ - $(6,0)$ from $x_1 + x_2 = 6$ intercept - $(3,3)$ intersection point - $(0,4)$ from $x_1 + 3x_2 = 12$ intercept 6. **Verify basic feasible solutions:** Basic feasible solutions correspond to corner points where two constraints are active (equalities). Each corner point satisfies two constraints as equalities: - $(0,0)$: $x_1=0$, $x_2=0$ - $(6,0)$: $x_1 + x_2 = 6$, $x_2=0$ - $(3,3)$: $x_1 + x_2 = 6$, $x_1 + 3x_2 = 12$ - $(0,4)$: $x_1=0$, $x_1 + 3x_2 = 12$ All corner points correspond to basic feasible solutions. **Final answer:** The feasible region is bounded by points $(0,0)$, $(6,0)$, $(3,3)$, and $(0,4)$.