Feasible region and feasible solutions

Every linear programming problem eventually asks the same question: out of all the points $(x, y)$ that a word problem allows, which ones are even worth checking? The answer is the feasible region — the patch of the coordinate plane where every constraint of the problem is satisfied at once. Nothing outside it can ever be a candidate answer, no matter how good its objective value looks on paper. This topic sits right after formulating an LPP and right before solving it. A word problem converts into decision variables, an objective function, and a handful of linear inequalities (the constraints). Each inequality carves the plane into a half-plane; the feasible region is what survives when all those half-planes are overlapped. Any point sitting inside is a feasible solution; a point falling outside even one constraint is infeasible. Because CBSE restricts this chapter to two variables, the whole idea can be seen and shaded on ordinary graph paper — geometry doing the work that algebra would otherwise have to do point by point.

The core insight

The feasible region is the intersection of all constraint half-planes (including $x \ge 0$ , $y \ge 0$ ), and every point inside it — a feasible solution — is a legal candidate for the optimal answer, while every point outside violates at least one constraint.

The core idea

A single inequality is a half-plane. A linear inequality in two variables, such as

ax + by \le c,

where: $a, b, c$ are given constants and $(x, y)$ is any point in the plane, describes a half-plane. The boundary line $ax + by = c$ splits the entire plane into two pieces, and the inequality sign tells you which piece (including the line itself, since CBSE constraints are always non-strict, $\le$ or $\ge$ , never $<$ or $>$ ).

Stacking constraints gives the feasible region. A real LPP has several constraints together with the non-negativity conditions $x \ge 0$ , $y \ge 0$ (decision variables such as "number of chairs" cannot be negative). The feasible region is the set of points obeying every one of them simultaneously — the geometric intersection of all the half-planes:

\text{Feasible region} = \{(x, y) : x \ge 0,\ y \ge 0,\ \text{and every constraint inequality holds}\}.

Because of $x \ge 0, y \ge 0$ , the feasible region always lives in the first quadrant.

Feasible vs infeasible solutions. Any point $(x, y)$ lying inside (or on the boundary of) the feasible region is a feasible solution — a legal candidate. A point that breaks even one constraint is an infeasible solution and is never checked further, however attractive its objective value might look.

Bounded and unbounded regions. A feasible region is bounded if it fits inside some sufficiently large circle — it does not run off to infinity in any direction. It is unbounded if it stretches out indefinitely, which happens often with $\ge$ -type constraints, since those push the kept half-plane away from the origin. The distinction matters later: an unbounded feasible region can make a maximisation problem have no finite optimal value, even though minimisation may still work fine.

Corner points (vertices). Wherever two boundary lines of the feasible region cross, that crossing is a corner point (vertex), found by solving the two boundary equations simultaneously. These vertices are not just decoration — the Corner Point Theorem (covered separately) guarantees that if an optimal solution exists, it occurs at one of them, which is what makes graphing the feasible region worth doing in the first place.

Figure

Shaded triangular feasible region in the first quadrant bounded by three lines with its three corner points marked

Derivation AFinding the feasible region and its corner points

The goal is to convert a list of linear constraints into a precisely shaded feasible region and a fully identified list of corner points, using nothing beyond intercepts, a single test-point check per constraint, and solving pairs of linear equations — the exact machinery this CBSE chapter equips a student with.

Step 1. Draw each boundary line by plotting intercepts. For $x + y \le 6$ , set $x=0$ to get $(0,6)$ and $y=0$ to get $(6,0)$ ; join them for the line $x+y=6$ . For $x \ge 2$ , the boundary is simply the vertical line $x = 2$ .

Step 2. Decide the shaded side with a test point. Substitute a convenient point not on the line — the origin $(0,0)$ works unless the line itself passes through it. For $x+y \le 6$ : $0 + 0 = 0 \le 6$ is true, so the side containing the origin is kept. For $x \ge 2$ : the origin gives $0 \ge 2$ , false, so the side away from the origin (to the right of $x=2$ ) is kept.

Step 3. Bring in non-negativity. $y \ge 0$ restricts attention to the upper half-plane; here $x \ge 0$ is automatically satisfied once $x \ge 2$ holds, so it adds nothing new.

Step 4. Shade the common region. Overlay all the kept half-planes — right of $x=2$ , below-left of $x+y=6$ , above the $x$ -axis — and the overlap is the feasible region.

Step 5. Find corner points by solving pairs of boundary equations. Solving $x=2$ with $y=0$ gives $(2,0)$ ; solving $x=2$ with $x+y=6$ gives $(2,4)$ ; solving $x+y=6$ with $y=0$ gives $(6,0)$ . These three points are the vertices of the feasible region. Result: the feasible region is the triangle with vertices $(2,0)$ , $(2,4)$ , $(6,0)$ — closed and bounded, since it fits inside a circle of radius $6$ .

Worked example

Find and describe the feasible region for the constraints $x + y \le 6$ , $x \ge 2$ , $y \ge 0$ , $x \ge 0$ , and classify the points $(3,2)$ and $(1,1)$ .

Step 1. Plot the boundary $x+y=6$ through $(6,0)$ and $(0,6)$ ; testing the origin gives $0 \le 6$ , true, so the half-plane containing the origin is kept.

Step 2. Plot the boundary $x=2$ ; testing the origin gives $0 \ge 2$ , false, so the half-plane to the right of $x=2$ is kept.

Step 3. Add $y \ge 0$ (upper half-plane); $x \ge 0$ is already implied by $x \ge 2$ .

Step 4. Solve boundary pairs for corners: $x=2,\ y=0 \Rightarrow (2,0)$ ; $x=2,\ x+y=6 \Rightarrow (2,4)$ ; $x+y=6,\ y=0 \Rightarrow (6,0)$ .

Step 5. Check the two test points against every constraint. For $(3,2)$ : $3+2=5\le 6$ ✓, $3\ge 2$ ✓, $2\ge 0$ ✓ — all satisfied, so it is feasible. For $(1,1)$ : $1+1=2\le 6$ ✓, but $1 \ge 2$ is false — it fails one constraint, so it is infeasible.

The feasible region is the bounded triangle with vertices $(2,0)$ , $(2,4)$ , $(6,0)$ ; $(3,2)$ is a feasible solution and $(1,1)$ is infeasible.

Competitive stretch

For $2x+y \ge 10$ , $x+3y \ge 15$ , $x\ge 0$ , $y\ge 0$ : both boundary lines fail the origin test ( $0 \ge 10$ and $0\ge 15$ are both false), so both half-planes are kept away from the origin. Solving $2x+y=10$ and $x+3y=15$ together gives $x=3,\ y=4$ . The corner points come out to $(15,0)$ , $(3,4)$ , $(0,10)$ , and the region extends infinitely upward and to the right — this feasible region is unbounded.

Common traps

Shading the wrong side of a line. Never guess from the inequality symbol alone — always substitute a real test point (the origin, if it is not on the line) and check the arithmetic.
Forgetting the boundary is included. Since CBSE constraints use $\le$ / $\ge$ , the boundary line itself belongs to the feasible region (drawn solid, not dashed).
Assuming every feasible region is bounded. Regions built from several $\ge$ constraints are frequently unbounded — always check whether the shaded area closes up or runs off to infinity.
Mixing up "feasible" with "optimal." Every point inside the region is a feasible solution, but only certain corner points ever give the optimal value of the objective function $Z$ .

Ask the tutor: Not sure which side of a boundary line to shade, or whether your region is bounded or unbounded — ask the AI tutor to check your test-point arithmetic and corner-point calculations step by step.

Why do we even bother finding the feasible region before solving the whole problem?

Because it instantly rules out every point that could never be an answer. Instead of testing every possible $(x,y)$ pair for the objective function, you narrow the search to one bounded or unbounded patch of the plane, and — as the Corner Point Method shows next — only its vertices ever need checking. The feasible region is what turns an infinite search into a handful of arithmetic checks.

How do I quickly tell which side of a boundary line to shade without a graph in front of me?

Pick any point not on the line — the origin $(0,0)$ is easiest unless the line passes through it — and substitute its coordinates into the inequality. If the inequality comes out true, shade the side containing that point; if false, shade the opposite side. This test-point method works for every linear inequality and removes any need to "just look" at the sign.

Can a feasible region have no points at all?

Yes — if the constraints contradict each other (for example $x \ge 5$ together with $x \le 2$ ), no point can satisfy both, and the feasible region is empty. When that happens the LPP has no feasible solution at all, so it certainly has no optimal one either; this is checked before attempting the Corner Point Method.

Feasible region and feasible solutions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads