1. **State the problem:** Minimize $$z = 2x + 3y$$ subject to constraints: $$-x + 2y \geq 4$$ $$x + y \geq 6$$ $$x + 3y \geq 9$$ $$x, y \geq 0$$ 2. **Rewrite inequalities to equalities to find boundary lines:** - For $$-x + 2y = 4$$ - For $$x + y = 6$$ - For $$x + 3y = 9$$ 3. **Find intersection points of these lines and axes to identify vertices of feasible region:** - Intersection of $$-x + 2y = 4$$ and $$x + y = 6$$: Add equations: $$(-x + 2y) + (x + y) = 4 + 6 \Rightarrow 3y = 10 \Rightarrow y = \frac{10}{3}$$ Substitute $$y$$ back into $$x + y = 6$$: $$x + \frac{10}{3} = 6 \Rightarrow x = 6 - \frac{10}{3} = \frac{8}{3}$$ So point $$A = \left(\frac{8}{3}, \frac{10}{3}\right)$$ - Intersection of $$-x + 2y = 4$$ and $$x + 3y = 9$$: From $$-x + 2y = 4$$, express $$x = 2y - 4$$ Substitute into $$x + 3y = 9$$: $$(2y - 4) + 3y = 9 \Rightarrow 5y - 4 = 9 \Rightarrow 5y = 13 \Rightarrow y = \frac{13}{5}$$ Then $$x = 2 \times \frac{13}{5} - 4 = \frac{26}{5} - 4 = \frac{6}{5}$$ So point $$B = \left(\frac{6}{5}, \frac{13}{5}\right)$$ - Intersection of $$x + y = 6$$ and $$x + 3y = 9$$: Subtract equations: $$(x + 3y) - (x + y) = 9 - 6 \Rightarrow 2y = 3 \Rightarrow y = \frac{3}{2}$$ Substitute into $$x + y = 6$$: $$x + \frac{3}{2} = 6 \Rightarrow x = \frac{9}{2}$$ So point $$C = \left(\frac{9}{2}, \frac{3}{2}\right)$$ - Check intersections with axes (where $$x=0$$ or $$y=0$$) that satisfy constraints: For $$-x + 2y \geq 4$$ with $$x=0$$: $$2y \geq 4 \Rightarrow y \geq 2$$ For $$x + y \geq 6$$ with $$x=0$$: $$y \geq 6$$ For $$x + 3y \geq 9$$ with $$x=0$$: $$3y \geq 9 \Rightarrow y \geq 3$$ So minimum $$y$$ on $$y$$-axis is $$6$$ to satisfy all constraints. Similarly for $$y=0$$: $$-x + 0 \geq 4 \Rightarrow -x \geq 4 \Rightarrow x \leq -4$$ (not possible since $$x \geq 0$$) So no feasible point on $$x$$-axis. 4. **Evaluate objective function $$z = 2x + 3y$$ at vertices:** - At $$A = \left(\frac{8}{3}, \frac{10}{3}\right)$$: $$z = 2 \times \frac{8}{3} + 3 \times \frac{10}{3} = \frac{16}{3} + 10 = \frac{16}{3} + \frac{30}{3} = \frac{46}{3} \approx 15.33$$ - At $$B = \left(\frac{6}{5}, \frac{13}{5}\right)$$: $$z = 2 \times \frac{6}{5} + 3 \times \frac{13}{5} = \frac{12}{5} + \frac{39}{5} = \frac{51}{5} = 10.2$$ - At $$C = \left(\frac{9}{2}, \frac{3}{2}\right)$$: $$z = 2 \times \frac{9}{2} + 3 \times \frac{3}{2} = 9 + \frac{9}{2} = 9 + 4.5 = 13.5$$ - At $$y=6, x=0$$ (point on $$y$$-axis): $$z = 2 \times 0 + 3 \times 6 = 18$$ 5. **Conclusion:** The minimum value of $$z$$ is $$10.2$$ at point $$\left(\frac{6}{5}, \frac{13}{5}\right)$$. --- **Final answer:** $$\boxed{\min z = 10.2 \text{ at } \left(\frac{6}{5}, \frac{13}{5}\right)}$$