1. **State the problem:** Minimize the objective function $$R(x,y) = x + 2y$$ subject to the constraints: $$x + y \geq 2$$ $$y \geq 2$$ $$2x - y \geq 4$$ 2. **Understand the constraints:** These inequalities define a feasible region in the xy-plane where the solution must lie. 3. **Rewrite constraints for clarity:** - $$x + y \geq 2 \implies y \geq 2 - x$$ - $$y \geq 2$$ - $$2x - y \geq 4 \implies y \leq 2x - 4$$ 4. **Find the feasible region:** The feasible region is the intersection of: - Above the line $$y = 2 - x$$ - Above the line $$y = 2$$ - Below the line $$y = 2x - 4$$ 5. **Find intersection points of the boundary lines to identify vertices:** - Intersection of $$y=2$$ and $$y=2 - x$$: $$2 = 2 - x \implies x=0$$, so point is $$(0,2)$$ - Intersection of $$y=2$$ and $$y=2x - 4$$: $$2 = 2x - 4 \implies 2x = 6 \implies x=3$$, so point is $$(3,2)$$ - Intersection of $$y=2 - x$$ and $$y=2x - 4$$: Set equal: $$2 - x = 2x - 4$$ $$2 + 4 = 2x + x$$ $$6 = 3x \implies x=2$$ Substitute back: $$y = 2 - 2 = 0$$, so point is $$(2,0)$$ 6. **Check which vertices satisfy all constraints:** - $$(0,2)$$: Check $$2x - y = 0 - 2 = -2 \geq 4?$$ No, so not feasible. - $$(3,2)$$: Check $$x + y = 5 \geq 2$$ yes, $$y=2 \geq 2$$ yes, $$2x - y = 6 - 2 = 4 \geq 4$$ yes, feasible. - $$(2,0)$$: Check $$y \geq 2?$$ No, so not feasible. 7. **Check other possible vertices or boundaries:** Since only $$(3,2)$$ is feasible among intersections, check if the feasible region is unbounded or if other vertices exist. 8. **Evaluate objective function at feasible vertex:** $$R(3,2) = 3 + 2 \times 2 = 3 + 4 = 7$$ 9. **Check if objective can be smaller elsewhere in feasible region:** - Since $$y \geq 2$$ and $$2x - y \geq 4$$, the feasible region is bounded on the left by $$y=2$$ and $$2x - y=4$$. - The objective function increases with both $$x$$ and $$y$$, so minimum occurs at the lowest $$x$$ and $$y$$ in feasible region. 10. **Find minimum by solving constraints equality:** - From $$2x - y = 4$$ and $$y=2$$, we get $$2x - 2 = 4 \implies 2x=6 \implies x=3$$. - So minimum at $$(3,2)$$ with $$R=7$$. **Final answer:** The minimum value of $$R(x,y)$$ subject to the constraints is $$7$$ at $$(x,y) = (3,2)$$.