Practice: Plotting Linear Inequalities

Constraint 1: $2x_1+x_2\le 10$. Boundary $2x_1+x_2=10$, intercepts $(5,0)$ and $(0,10)$. Test the origin: $0\le 10$ is true, so the allowed side contains the origin, below and left of the line.

Constraint 2: $-x_1+2x_2\le 4$. Boundary $-x_1+2x_2=4$, intercepts $(-4,0)$ and $(0,2)$. Test the origin: $0\le 4$ is true, so again the origin's side is allowed, below and right of this line.

Sign restrictions. $x_1\ge 0$ and $x_2\ge 0$ confine everything to the first quadrant.

Corners. The origin $(0,0)$; along the $x_1$-axis to $(5,0)$ where constraint 1 binds; the intersection of the two slanted boundaries; and $(0,2)$ where constraint 2 meets the $x_2$-axis.

For the intersection, solve $2x_1+x_2=10$ with $-x_1+2x_2=4$. From the first, $x_2=10-2x_1$. Substituting: $-x_1+2(10-2x_1)=4$, so $-x_1+20-4x_1=4$, giving $-5x_1=-16$ and $x_1=16/5=3.2$, then $x_2=10-6.4=3.6$.

So the corners are $(0,0)$, $(5,0)$, $(3.2,3.6)$ and $(0,2)$.

Verification. Take the interior point $(1,1)$: $2+1=3\le 10$ holds; $-1+2=1\le 4$ holds; both coordinates are nonnegative. All four constraints are satisfied, confirming every shaded side.

Note on the second constraint. Its $x_1$-intercept is negative, at $(-4,0)$, which lies outside the first quadrant. That is perfectly normal. The boundary line extends across the whole plane, and only the part bounding the feasible region is visible in the quadrant.

What is correct. The boundary is right: $3x_1+x_2=9$ does pass through $(3,0)$ and $(0,9)$.

The error. The side was chosen from the inequality symbol rather than by testing a point. Testing the origin: $3(0)+0=0$, and $0\ge 9$ is false. So the origin's side is not allowed, and the correct region is the far side of the line, away from the origin.

The correct region. The part of the first quadrant on or beyond the line $3x_1+x_2=9$. It is unbounded: it contains $(t,9)$ for every $t\ge 0$, and indeed every point far enough out in the quadrant. Its corners are $(3,0)$ and $(0,9)$, and it has no third corner.

Why 'greater-than means the larger area' fails. The words suggest size, but the inequality is about the value of $3x_1+x_2$, not about area. The region where that expression is large is the one far from the origin, which here happens to be the unbounded side. There is no general correspondence between the direction of the symbol and the visual size of a region.

Why this is hard to catch. The output is entirely plausible. A triangle with corners $(0,0)$, $(3,0)$, $(0,9)$ is a perfectly ordinary feasible region, and nothing about it looks wrong. Every later step would then be executed correctly on it: corners enumerated, objective evaluated, an optimum reported, all of it valid work on the wrong region, and none of it capable of detecting the error, because nothing downstream ever revisits the original inequality.

The check that would have caught it. Test a point and write the verdict down before shading, for every constraint. Then verify the finished region with one interior point: taking $(1,1)$ from the student's triangle gives $3+1=4$, and $4\ge 9$ is false, that single check exposes the whole error in one line.

Deciding a plan. Evaluate the left-hand side at the plan and compare it with $60$. For $x=(4,5,2,6)$: $20+10+14+6=50\le 60$, so the plan satisfies the constraint with $10$ units to spare.

The analogue of the test point. It is the same thing, unchanged, any specific point whose satisfaction you evaluate directly. In two variables a single test point settles a whole side because the expression cannot cross from below the bound to above it without passing through equality; that argument never mentioned the dimension, so it holds here identically.

The analogue of the boundary. The hyperplane $5x_1+2x_2+7x_3+x_4=60$. It is three-dimensional inside four-dimensional space and cannot be pictured, but it does exactly what a line did: divide the space into two half-spaces, one allowed and one not.

A plan exactly on the boundary. The constraint holds with equality. The resource is exactly exhausted, with nothing to spare. The constraint is then binding (or active) at that plan. This matters well beyond drawing: optima in linear programming sit where constraints bind, and which constraints are tight at a point is what identifies it as a corner.

What transfers and what does not. The drawing does not, and it was never the substance. What transfers is the structure: a boundary where the expression equals the bound, two sides, and a direct evaluation to decide which side a point is on. That is why the plotting skill matters as a skill rather than as a picture. The picture stops at two variables, and the reasoning does not.