1. **State the problem:** We want to maximize the objective function $$Z = 2x_1 + 10x_2$$ subject to the constraints: $$10x_1 + 4x_2 \geq 40$$ $$x_1 + 6x_2 \geq 24$$ $$x_1 + 2x_2 \leq 14$$ with $$x_1, x_2 \geq 0$$. 2. **Rewrite inequalities for clarity:** - Labor: $$10x_1 + 4x_2 \geq 40$$ - Material: $$x_1 + 6x_2 \geq 24$$ - Storage: $$x_1 + 2x_2 \leq 14$$ - Non-negativity: $$x_1 \geq 0, x_2 \geq 0$$ 3. **Convert inequalities to equalities to find boundary lines:** - Labor line: $$10x_1 + 4x_2 = 40$$ - Material line: $$x_1 + 6x_2 = 24$$ - Storage line: $$x_1 + 2x_2 = 14$$ 4. **Find intersection points of constraints to identify feasible region vertices:** - Intersection of Labor and Material: Solve system: $$10x_1 + 4x_2 = 40$$ $$x_1 + 6x_2 = 24$$ From second: $$x_1 = 24 - 6x_2$$ Substitute into first: $$10(24 - 6x_2) + 4x_2 = 40 \Rightarrow 240 - 60x_2 + 4x_2 = 40$$ $$-56x_2 = -200 \Rightarrow x_2 = \frac{200}{56} = \frac{25}{7} \approx 3.571$$ Then $$x_1 = 24 - 6 \times 3.571 = 24 - 21.429 = 2.571$$ - Intersection of Labor and Storage: $$10x_1 + 4x_2 = 40$$ $$x_1 + 2x_2 = 14$$ From second: $$x_1 = 14 - 2x_2$$ Substitute: $$10(14 - 2x_2) + 4x_2 = 40 \Rightarrow 140 - 20x_2 + 4x_2 = 40$$ $$-16x_2 = -100 \Rightarrow x_2 = \frac{100}{16} = 6.25$$ Then $$x_1 = 14 - 2 \times 6.25 = 14 - 12.5 = 1.5$$ - Intersection of Material and Storage: $$x_1 + 6x_2 = 24$$ $$x_1 + 2x_2 = 14$$ Subtract second from first: $$(x_1 + 6x_2) - (x_1 + 2x_2) = 24 - 14 \Rightarrow 4x_2 = 10 \Rightarrow x_2 = 2.5$$ Then $$x_1 = 14 - 2 \times 2.5 = 14 - 5 = 9$$ 5. **Check which intersection points satisfy all constraints:** - Point A: $$(2.571, 3.571)$$ Check Storage: $$2.571 + 2 \times 3.571 = 2.571 + 7.142 = 9.713 \leq 14$$ (valid) - Point B: $$(1.5, 6.25)$$ Check Material: $$1.5 + 6 \times 6.25 = 1.5 + 37.5 = 39 > 24$$ (valid) - Point C: $$(9, 2.5)$$ Check Labor: $$10 \times 9 + 4 \times 2.5 = 90 + 10 = 100 > 40$$ (valid) 6. **Evaluate objective function at each vertex:** - At A: $$Z = 2(2.571) + 10(3.571) = 5.142 + 35.71 = 40.852$$ - At B: $$Z = 2(1.5) + 10(6.25) = 3 + 62.5 = 65.5$$ - At C: $$Z = 2(9) + 10(2.5) = 18 + 25 = 43$$ 7. **Check corner points on axes (if feasible):** - At $$x_1=0$$: Labor: $$4x_2 \geq 40 \Rightarrow x_2 \geq 10$$ Material: $$6x_2 \geq 24 \Rightarrow x_2 \geq 4$$ Storage: $$2x_2 \leq 14 \Rightarrow x_2 \leq 7$$ No $$x_2$$ satisfies both $$x_2 \geq 10$$ and $$x_2 \leq 7$$, so no feasible point at $$x_1=0$$. - At $$x_2=0$$: Labor: $$10x_1 \geq 40 \Rightarrow x_1 \geq 4$$ Material: $$x_1 \geq 24$$ Storage: $$x_1 \leq 14$$ No $$x_1$$ satisfies both $$x_1 \geq 24$$ and $$x_1 \leq 14$$, so no feasible point at $$x_2=0$$. 8. **Conclusion:** The maximum value of $$Z$$ is at point B: $$(1.5, 6.25)$$ with $$Z = 65.5$$. **Final answer:** $$\boxed{\max Z = 65.5 \text{ at } (x_1, x_2) = (1.5, 6.25)}$$