Systems of Inequalities

A single inequality carves out a half-plane. When we combine several inequalities and demand that all of them hold at once, their overlapping region can form a polygon, an unbounded strip, or even be empty. Understanding these systems is the gateway to optimization and linear programming.

Feasible Region

📐Definition — System of Inequalities and Feasible Region

A system of inequalities is a collection of two or more inequalities in the same variables. A point $(x, y)$ is a solution if it satisfies every inequality in the system simultaneously.

The feasible region (or solution set) is the intersection of all individual half-planes — the set of all points that satisfy every inequality at once.

The feasible region can take three forms:

A bounded polygon (a triangle, quadrilateral, etc.) — enclosed on all sides.
An unbounded region — extending infinitely in some direction.
The empty set — when the inequalities are contradictory and no point satisfies all of them.

Method for Graphing Systems

⚡Theorem — Graphing a System of Inequalities

Graph each inequality individually: draw its boundary line and shade the appropriate half-plane.
Identify the overlap — the feasible region is where all shaded regions intersect.
If the feasible region is bounded, find the corner points (vertices) by solving pairs of boundary equations simultaneously.

Worked Example

✎Example — Finding the Feasible Region

Find and describe the feasible region of the system:

$\begin{cases} x + y \leq 4 \\ x - y \geq -2 \\ x \geq 0 \end{cases}$

Solution.

Step 1. Graph each inequality:

$x + y \leq 4$ : boundary $x + y = 4$ (solid). Test $(0,0)$ : $0 \leq 4$ ✓. Shade below-left.
$x - y \geq -2$ , equivalently $y \leq x + 2$ : boundary $x - y = -2$ (solid). Test $(0,0)$ : $0 \geq -2$ ✓. Shade below.
$x \geq 0$ : shade to the right of the $y$ -axis.

Step 2. The feasible region is where all three shaded areas overlap.

Step 3. Find corner points by solving pairs of boundary lines:

$x + y = 4$ and $x - y = -2$ : add the equations to get $2x = 2$ , so $x = 1$ , $y = 3$ . Corner: $(1, 3)$ .
$x + y = 4$ and $x = 0$ : gives $y = 4$ . Corner: $(0, 4)$ .
$x - y = -2$ and $x = 0$ : gives $y = 2$ . Corner: $(0, 2)$ .

The feasible region is the triangle with vertices $(0, 2)$ , $(0, 4)$ , and $(1, 3)$ .

Corner Points and Linear Programming

⚡Theorem — Corner Point Theorem

If a linear objective function $z = ax + by$ is to be maximized or minimized over a bounded feasible region (a polygon), then the optimal value occurs at a vertex (corner point) of the feasible region.

This is the foundation of linear programming. Instead of testing infinitely many points, you only need to evaluate the objective function at each corner point and pick the best one.

For the triangle above, to maximize $z = 2x + y$ :

At $(0, 2)$ : $z = 0 + 2 = 2$
At $(0, 4)$ : $z = 0 + 4 = 4$
At $(1, 3)$ : $z = 2 + 3 = 5$ ← maximum

The maximum value of $z = 2x + y$ on this feasible region is $5$ , achieved at $(1, 3)$ .

Practice

✏Exercise — Find a Feasible Region

Find the feasible region and its corner points for the system:

$\begin{cases} 2x + y \leq 6 \\ x + 2y \leq 6 \\ x \geq 0 \\ y \geq 0 \end{cases}$

Then find the maximum of $z = 3x + 2y$ over this region.