1. **Stating the problem:** We are given a system of inequalities: $$x + y \geq 180$$ $$2x + y \geq 200$$ $$5x + 4y \geq 800$$ with constraints $$x \geq 0$$ and $$y \geq 0$$. The objective function to maximize is: $$f(x,y) = 8{,}000{,}000x + 6{,}000{,}000y$$ 2. **Identify the constraints and feasible region:** The inequalities define a feasible region in the first quadrant bounded by lines: - $$x + y = 180$$ - $$2x + y = 200$$ - $$5x + 4y = 800$$ along with $$x \geq 0$$ and $$y \geq 0$$. 3. **Find vertices of the feasible region:** Calculate intersections of each pair of lines that satisfy all constraints: - Intersection of $$x + y = 180$$ and $$2x + y = 200$$: Substract first from second: $$ (2x + y) - (x + y) = 200 - 180 \implies x = 20 $$ Plug $$x=20$$ into $$x + y = 180$$: $$ 20 + y = 180 \implies y = 160 $$ - Intersection of $$x + y = 180$$ and $$5x + 4y = 800$$: Express $$y = 180 - x$$, substitute: $$5x + 4(180 - x) = 800 \implies 5x + 720 - 4x = 800 \implies x = 80$$ Then, $$y = 180 - 80 = 100$$ - Intersection of $$2x + y = 200$$ and $$5x + 4y = 800$$: Express $$y = 200 - 2x$$, substitute: $$5x + 4(200 - 2x) = 800 \implies 5x + 800 - 8x = 800 \implies -3x = 0 \implies x = 0$$ Then, $$y = 200 - 2(0) = 200$$ 4. **Check vertices against constraints:** All vertices: (20, 160), (80, 100), (0, 200) satisfy $$x \geq 0$$, $$y \geq 0$$ and the inequalities. 5. **Evaluate the objective function at each vertex:** - At (20, 160): $$f = 8{,}000{,}000 \times 20 + 6{,}000{,}000 \times 160 = 160{,}000{,}000 + 960{,}000{,}000 = 1{,}120{,}000{,}000$$ - At (80, 100): $$f = 8{,}000{,}000 \times 80 + 6{,}000{,}000 \times 100 = 640{,}000{,}000 + 600{,}000{,}000 = 1{,}240{,}000{,}000$$ - At (0, 200): $$f = 8{,}000{,}000 \times 0 + 6{,}000{,}000 \times 200 = 0 + 1{,}200{,}000{,}000 = 1{,}200{,}000{,}000$$ 6. **Determine maximum value and solution:** The maximum value of $$f$$ is $$1{,}240{,}000{,}000$$ at $$(80, 100)$$. **Final answer:** $$\boxed{\max f = 1{,}240{,}000{,}000 \text{ at } (x,y) = (80, 100)}$$