Valencian Community PAU Applied Mathematics for Social Sciences II study materials
Valencian Community PAU Applied Mathematics for Social Sciences II study resources

Valencian Community PAU Applied Mathematics for Social Sciences II: Linear programming in two variables

Valencian Community PAU Applied Mathematics for Social Sciences II · Work through your exam, one topic at a time.

All guide sections

Linear programming in two variables

This linear-programming practice chapter develops translation into inequalities, graphical feasible regions, objective evaluation and elementary sensitivity ideas. These are shared mathematical foundations; no particular regional examination format or required question is claimed.

Core rules

  • Decision variables with x ≥ 0, y ≥ 0; each limited resource gives a ≤ inequality, each minimum requirement a ≥ inequality.
  • The objective z = ax + by is linear, so its level curves are parallel lines and when a nonconstant linear objective has a finite optimum over a full-dimensional polygon, some boundary point attains it.
  • Fundamental theorem: if the feasible region is non-empty and bounded, the optimum is attained at a vertex. If it is unbounded, test whether z grows without limit along an open direction.
  • To draw a constraint, plot its boundary as an equation and test a point (usually the origin) to choose the half-plane. A vertex is the intersection of two boundary lines that satisfies every constraint.
  • Duality: max c^T x with Ax ≤ b, x ≥ 0 has dual min b^T y with A^T y ≥ c, y ≥ 0. When a feasible primal problem has a finite optimum, strong duality gives a dual optimum of the same value. An optimal dual variable can describe a marginal resource value within the applicable sensitivity range.

Worked example

Maximise z = 3x + 2y subject to x + y ≤ 4, x + 2y ≤ 6, x ≥ 0, y ≥ 0.

Boundaries: x + y = 4 cuts the axes at (4,0) and (0,4); x + 2y = 6 cuts them at (6,0) and (0,3). The lines meet where x + y = 4 and x + 2y = 6, that is y = 2, x = 2. Vertices of the region: (0,0), (4,0), (2,2), (0,3). Objective values: 0, 12, 10, 6. The maximum is z = 12 at (4,0). If the objective were z = x + y, the values would be 0, 4, 4, 3, and every point of the edge joining (4,0) and (2,2) would be optimal. This is the tied-edge case.

Traps that cost marks

  • Omitting x, y ≥ 0 and losing two edges of the region.
  • Checking only the intersection of the two main lines.
  • Multiplying an inequality by a negative number without reversing the sign.
  • Giving a fractional vertex as the answer when the variables count whole units.
  • Declaring a maximum without checking that the region is bounded in that direction.