Linear Inequalities

Concept:Find the corner points of the feasible region defined by the linear inequalities.Explanation:The feasible region is bounded by the lines $5x+y=100$, $x+y=60$, $x=0$, and $y=0$.Find intersections of boundary lines:• Intersection of $x=0$ and $y=0$: $(0,0)$.• Intersection of $5x+y=100$ with $y=0$: $x=20$, so $(20,0)$.• Intersection of $5x+y=100$ with $x+y=60$: subtract equations to get $4x=40$, $x=10$, then $y=50$, so $(10,50)$.• Intersection of $x+y=60$ with $x=0$: $y=60$, so $(0,60)$.Check all points satisfy all inequalities: $(0,0)$, $(20,0)$, $(10,50)$, $(0,60)$ are feasible.Points like $(60,0)$ and $(0,100)$ violate one inequality.Answer:The corner points are $(0,0)$, $(20,0)$, $(10,50)$, and $(0,60)$, which matches option A.