Linear Programming Including Inequalities

The feasible region is defined as the set of all points $(x,y)$ that satisfy every constraint in the problem simultaneously.

Every single constraint must be satisfied at the same time — not just one or some of them.

The correct answer is 'The set of all points that satisfy every constraint simultaneously' — this matches the definition exactly.

Satisfying only one constraint is not enough — a feasible solution must meet all restrictions, so this option is wrong.

The feasible region is an area (often a polygon), not a single point. The optimal point is found after identifying the region.

The boundary lines are just the edges of the region, not the region itself. The feasible region includes all interior points too.

By elimination, 'The set of all points that satisfy every constraint simultaneously' is the only correct definition.

The inequality uses $\ge$, which means 'greater than or equal to'.

A solid line is used when the inequality includes equality ($\le$ or $\ge$), meaning points on the boundary line are valid solutions.

Since $\ge$ includes the boundary, draw a solid line through $y=3x-4$.

Dashed lines are used for strict inequalities ($<$ or $>$), where the boundary is NOT included. Here we have $\ge$, so this is wrong.

The claim that $\ge$ means 'greater than only' is false — $\ge$ explicitly includes equality.

A boundary line is always required to show where the inequality transitions from true to false.

A solid line is correct because $\ge$ includes the boundary.