Feasible and Infeasible Regions: Explained with Examples

In linear programming, every decision has to respect real limits: factory hours, budgets, raw materials, storage space. Each limit is written as a mathematical inequality involving the decision variables. The feasible region is the set of all points (x, y) that satisfy every single constraint at the same time.

This matters because any acceptable plan must lie inside the feasible region — the best answer can never be outside it. If no point satisfies all the constraints together, the region is empty and the problem is called infeasible: it has no solution, which usually means the model itself needs fixing.

Key Concepts

• Constraint: an inequality such as 2x + 3y ≤ 60 that a solution must obey.
• Feasible region: the intersection (common part) of all the half-planes given by the constraints; every point in it is a valid plan.
• Infeasible: the constraints contradict each other, so no point satisfies all of them; the feasible region is empty.
• Convex region: a feasible region of a linear program is always convex — the straight line joining any two feasible points stays inside it.
• Boundary line: the equality form of a constraint, such as 2x + 3y = 60; points on the line satisfy the constraint with equality.

Formulas

• Objective function: Z = ax + by, the quantity to maximise or minimise (profit, cost, time).
• Constraint (general form): a₁x + b₁y ≤ c₁ or a₁x + b₁y ≥ c₁, where a₁, b₁, c₁ are constants.
• Non-negativity restrictions: x ≥ 0, y ≥ 0 — you cannot make a negative number of chairs.
• Boundary line: a₁x + b₁y = c₁, found by replacing the inequality sign with =.
• Feasibility test: a point (x₀, y₀) is feasible if it satisfies every inequality; test each one by substitution.

Worked Example

Maximise Z = 3x + 5y subject to 2x + y ≤ 12, x + 2y ≤ 12, x ≥ 0, y ≥ 0. The corner points of the feasible region are the origin O(0, 0), A(6, 0) from 2x + y = 12 with y = 0, and C(0, 6) from x + 2y = 12 with x = 0. The two lines meet at B: solving 2x + y = 12 and x + 2y = 12 together, multiply the second equation by 2 to get 2x + 4y = 24, then subtract the first: 3y = 12, so y = 4 and x = 4, giving B(4, 4). Now test the point (6, 6): substituting gives 2(6) + 6 = 18, which is greater than 12, so (6, 6) violates the first constraint and is infeasible — it lies outside the region. Meanwhile B(4, 4) satisfies 2(4) + 4 = 12 ≤ 12 and 4 + 2(4) = 12 ≤ 12, so it is feasible and, as the corner-point method will show, the optimal corner.

Common Misconceptions

• Points on the boundary are infeasible. Wrong — with ≤ or ≥ the boundary line is included, so boundary points are feasible. Only strict inequalities exclude the boundary.
• Infeasible means the answer is zero. No — infeasible means there is no answer at all, because the demands contradict each other (for example, x ≤ 3 and x ≥ 5 simultaneously).
• The region can lie anywhere on the graph. With x ≥ 0, y ≥ 0 the feasible region always stays in the first quadrant; shading other quadrants is a common exam mistake.

Real-World Applications

• Factories plan production mixes that respect machine hours, labour and raw material limits.
• Diet planners choose food quantities that meet nutrition needs at minimum cost.
• Transport companies route goods from warehouses to shops without exceeding vehicle capacity.
• Farmers allocate land between crops under water and fertiliser limits.

The feasible region is the set of all plans that respect every constraint, and an empty region means the constraints contradict each other. Mastering how to read and draw it is the foundation of every linear programming solution that follows.