Plotting Linear Inequalities

Plot the region defined by

First boundary. $x_1+2x_2=8$ has intercepts $(8,0)$ and $(0,4)$. Test the origin: $0\le 8$ holds, so the allowed side is the side containing the origin, below and left of this line.

Second boundary. $3x_1+x_2=9$ has intercepts $(3,0)$ and $(0,9)$. Test the origin: $0\le 9$ holds, so again the origin's side is allowed.

Sign restrictions. $x_1\ge 0$ is the region right of the vertical axis; $x_2\ge 0$ is the region above the horizontal axis. Together they confine everything to the first quadrant.

The intersection. The surviving region is a quadrilateral with corners $(0,0)$, $(3,0)$, $(2,3)$ and $(0,4)$. The corner $(2,3)$ is where the two slanted boundaries meet, found by solving $x_1+2x_2=8$ with $3x_1+x_2=9$ simultaneously.

Verification. Take the interior point $(1,1)$: $1+2=3\le 8$, $3+1=4\le 9$, and both coordinates are nonnegative. All four constraints hold, confirming the shaded side of every boundary.

Not a rule: 'less than shades below'. For $x_1+2x_2\le 8$ the origin's side is indeed below the line. For $-x_1-2x_2\le -8$, the same boundary, the origin fails the test, since $0\le -8$ is false, so the allowed side is above. Identical line, identical inequality sign, opposite side.

Not a valid test point: the origin on the boundary. For $2x_1-3x_2\le 0$ the line passes through the origin and substituting gives $0\le 0$, which is true but says nothing about a side. Test $(1,0)$ instead: $2\le 0$ is false, so the allowed side is the other one.

Not the intersection: shading each constraint separately and taking whatever is covered. A point satisfying one constraint but not another is infeasible. The region is where all of the shadings overlap, not where any of them reaches.

Not optional: the sign restrictions. Omitting $x_1\ge 0$ and $x_2\ge 0$ from $x_1+2x_2\le 8$ leaves a half-plane extending infinitely into three other quadrants, with no corners at all. The region and the problem are both different.

Not a special case: a missing variable. $x_1\le 4$ still has a boundary, the vertical line $x_1=4$, and two sides, tested exactly as any other constraint is.

Two errors are possible when plotting a constraint, and they are not equally visible.

A misplaced boundary is usually visible. Compute an intercept wrongly and the line sits somewhere the constraint cannot justify, cutting an axis beyond the bound it came from, or crossing another boundary far outside the quadrant. The drawing contradicts the arithmetic that produced it.

A wrong side does not. Consider $3x_1+x_2\ge 9$ with sign restrictions. Suppose the origin test is skipped and the side chosen by the feel of the inequality. Testing properly: $0\ge 9$ is false, so the allowed side is away from the origin, and the region is unbounded. Shade the origin's side instead and you get a neat bounded triangle with corners $(0,0)$, $(3,0)$, $(0,9)$. A perfectly respectable-looking region belonging to a different problem.

The consequence propagates silently. That triangle has corners, those corners get evaluated, one of them wins, and a specific optimum with a specific value is reported. Every step after the shading is performed correctly on incorrect input. Nothing downstream can detect it, because nothing downstream ever sees the original inequality again.

The safeguard is cheap and must be habitual. Test a point and write the verdict down, for every constraint, before shading. Then verify the finished region with one interior point checked against all the original inequalities. Two arithmetic checks, and together they close the only failure mode in this procedure that the picture cannot reveal.

The side of a boundary line allowed by an inequality can be decided by looking at the inequality sign, so a less-than constraint always shades below or towards the origin.