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Class 12 Mathematics Chapter 12 of 13

Chapter 12 — Linear Programming

Overview

Chapter 12 — Linear Programming illustration

Chapter: Linear Programming (Class 12 NCERT) — Introduction, importance, key themes, and learning outcomes. Introduction: Linear Programming (LP) studies methods to optimize (maximize or minimize) a linear objective function subject to a set of linear inequalities (constraints). In CBSE Class 12 you learn to translate real-life optimization problems (profit, cost, resource allocation) into mathematical form and solve them graphically when there are two decision variables. Importance: LP provides a systematic way to make optimal decisions in business, economics, engineering and everyday resource allocation. It builds modeling skills, geometric understanding of inequalities, and problem-solving techniques useful for higher studies (operations research, management) and competitive exams. Key themes: formulation of an LPP (decision variables, objective function, constraints, non-negativity conditions); feasible region (intersection of half-planes); graphical solution for two variables; corner-point (extreme point) principle; classification of solutions (unique optimal, multiple optimal, unbounded, infeasible); interpreting and verifying solutions in context. What the student will…

Learning Objectives

  • Define linear programming and related terms such as feasible solution, feasible region, objective function and constraint.
  • Explain the steps for converting real-world optimization situations into a linear programming formulation.
  • Formulate linear inequalities and the corresponding objective function from word problems involving two variables.
  • Graph linear inequalities in two variables and identify the feasible region on the Cartesian plane.
  • Construct the corner (vertex) points of a feasible region by solving pairs of boundary equations.
  • Apply the corner-point (vertex) method to evaluate the objective function and locate the optimal solution.
  • Determine maximum and minimum values of linear objective functions using the graphical method for two-variable problems.
  • Solve linear programming problems subject to non-negativity conditions and interpret the numerical results.

Topics in this chapter

9 topics · tap a topic title to jump straight to it.

⌨️1

Introduction to Linear Programming

What is Linear Programming?
Linear Programming (LP) is a method to achieve the best outcome (maximum profit, minimum cost, etc.) in a mathematical model whose requirements are represented by linear relationships. An LP problem consists of a linear objective function to be maximized or minimized subject to a set of linear constraints (inequalities or equations) and non-negativity conditions on the variables.

Key components

  • Decision variables: unknowns to be determined (e.g., x and y).
  • Objective function: a linear function of decision variables to maximize or minimize, e.g. Z = c1 x + c2 y.
  • Constraints: linear inequalities/equations limiting the variables, e.g. a1 x + b1 y ≤ b.
  • Feasible region: set of all points satisfying the constraints and non-negativity — usually a convex polygon (or unbounded region) in 2D.
  • Optimal solution: a point in the feasible region where the objective function reaches its best value. For LP with linear constraints, an optimum (if exists and is finite) occurs at a vertex (corner point) of the feasible region.

Graphical method (for two variables)

  1. Introduce decision variables x and y (both ≥ 0 unless otherwise stated).
  2. Write the objective function (e.g. Maximize Z = p x + q y).
  3. Plot each linear constraint as a straight line and shade the half-plane satisfying the inequality.
  4. The feasible region is the intersection of all shaded half-planes (and axes for non-negativity).
  5. Find the vertices (corner points) of the feasible region by solving pairs of boundary equations.
  6. Evaluate the objective function at each vertex; the best value gives the optimal solution.

Special cases

  • No feasible solution: constraints are inconsistent (feasible region empty).
  • Unbounded solution: feasible region is unbounded and objective can increase/decrease without bound.
  • Multiple optima: whole edge (segment) of the feasible region gives the same optimal value.

Practical tips: Always check non-negativity conditions, draw constraint lines accurately (use intercepts), label intersection points, and use the corner-point principle to compute objective values only at vertices.

📌 Examples
  • Manufacturing mix: A factory makes tables (x) and chairs (y). Each table needs 3 hours of labor and 2 units of wood; each chair needs 2 hours and 1 unit of wood. Available: 60 labor hours and 30 wood units. Profit: Rs. 50 per table, Rs. 30 per chair. Formulate: Maximize Z = 50x + 30y subject to 3x + 2y ≤ 60, 2x + y ≤ 30, x,y ≥ 0. Solve graphically to get optimal x and y.
  • Diet / cost minimization: Choose quantities x and y of two foods to meet minimum vitamin and calorie requirements at minimum cost. Constraints are linear inequalities for nutrient amounts; objective is Minimize cost = c1 x + c2 y.
  • Transportation / shipping: Minimize total shipping cost from warehouses to stores subject to supply and demand constraints. Formulate as LP with objective summing cost per route times shipment variables and linear supply/demand constraints.
  • Blending problem: Mix two fuels A and B to produce a blend meeting octane and sulfur limits at minimum cost. Variables are amounts of A and B; constraints are linear bounds from blend specifications.
🧮 Formulas
  1. General LP (standard): Maximize or Minimize Z = c1 x1 + c2 x2 + ... + cn xn subject to a11 x1 + a12 x2 + ... + a1n xn (≤, =, or ≥) b1 a21 x1 + a22 x2 + ... + a2n xn (≤, =, or ≥) b2 ... xi ≥ 0 for all i
  2. Two-variable objective: Z = p x + q y (to be maximized or minimized).
  3. Constraint line intercept form (for ax + by = c): x-intercept = (c/a, 0) if a ≠ 0; y-intercept = (0, c/b) if b ≠ 0.
  4. Corner-point principle: If an LP has an optimal solution and the feasible region is a convex polygon, then at least one optimal solution is at a vertex (corner) of the feasible region.
  5. Slack variable conversion (for ≤ constraint): a1 x + b1 y + s1 = c1, with s1 ≥ 0 (s1 is slack variable).
📊 Visual ideas
Feasible region sketch: Draw x and y axes (x ≥ 0, y ≥ 0). For each constraint like ax + by ≤ c, draw the line ax + by = c using intercepts, then shade the side satisfying the inequality. The feasible region is the common shaded area — shade it lightly and outline the polygon.
Corner points and evaluation: Mark and label all intersection points (including axis intercepts) of constraint lines that bound the feasible region. Compute coordinates by solving pairs of linear equations. Put the objective value Z at each point and identify the maximum or minimum.
Objective line technique: Draw a representative objective line p x + q y = k (choose a k), then move this line parallelly towards increasing k (for maximization) or decreasing k (for minimization) until it last touches the feasible region. The touching point is the optimum.
Unbounded/infeasible examples: Show one graph where the feasible region extends infinitely in a direction (indicating possible unbounded objective), and another where half-planes do not intersect (no feasible region).
⌨️2

Formulation of Linear Programming Problems

What is a Linear Programming Problem (LPP)?
A linear programming problem is a mathematical procedure to find the best (maximum or minimum) value of a linear objective function subject to a set of linear constraints (equalities or inequalities) and non-negativity conditions. It models optimization problems where relationships are linear.

Key components

  • Decision variables: unknowns to be determined (x1, x2, ...).
  • Objective function: the linear function to maximize or minimize, e.g., Maximize Z = c1 x1 + c2 x2 + ...
  • Constraints: linear inequalities or equations representing resource limits, e.g., a11 x1 + a12 x2 + ... ≤ b1.
  • Non-negativity conditions: xj ≥ 0 unless negative values are allowed.

Steps to formulate an LPP

  1. Understand the problem: read context, identify what to optimize (profit, cost, time, etc.).
  2. Define decision variables: assign variables to quantities you can control, with units.
  3. Write the objective function: express the quantity to maximize/minimize in terms of the decision variables.
  4. Write constraints: translate resource limits, requirements and logical conditions into linear inequalities or equations.
  5. Set non-negativity conditions: state xj ≥ 0 if negative values are not meaningful.
  6. Convert to standard form if needed: for algorithmic solution (simplex), convert inequalities to equalities using slack/surplus (and possibly artificial) variables.

Common translations

  • 'At most' or 'not more than' → use ≤
  • 'At least' or 'not less than' → use ≥
  • 'Total available' or 'limited by' usually give ≤ constraints
  • 'Needs to be exactly' → use =

Properties
For problems with two variables, the feasible region (set of points satisfying all constraints) is a convex polygon (possibly unbounded). If an optimal solution exists for a linear objective, at least one optimal solution occurs at a corner (extreme) point of the feasible region (corner-point principle).

Converting inequalities for simplex
To convert ≤ constraints to equalities, add slack variables s ≥ 0: a11 x1 + a12 x2 + ... + s = b. For ≥ constraints, subtract a surplus variable and possibly add an artificial variable when using simplex.

📌 Examples
  • 1) Two-product manufacturing (maximize profit): A company makes product A and B. Profit: Rs 5 per A and Rs 4 per B. Each A uses 2 units of raw material and 1 hour of labour; each B uses 1 unit of raw material and 1 hour of labour. Available: 100 units material, 80 hours labour. Decision variables: x1 = number of A, x2 = number of B. Objective: Maximize Z = 5x1 + 4x2. Constraints: 2x1 + 1x2 ≤ 100 (material), 1x1 + 1x2 ≤ 80 (labour), x1 ≥ 0, x2 ≥ 0.
  • 2) Diet/cost minimization: Choose quantities of food F1 and F2 to meet at least 500 mg vitamin given F1 has 50 mg/serving and F2 30 mg/serving. Costs: Rs 10 and Rs 6 per serving. Decision variables: x1, x2 ≥ 0. Objective: Minimize C = 10x1 + 6x2. Constraint: 50x1 + 30x2 ≥ 500.
  • 3) Mixture/blending problem: Produce 100 kg of a mixture that must contain at least 40% ingredient X. Suppose two sources S1 (60% X) and S2 (20% X) are available. Let x1, x2 be kg from S1 and S2. Objective might be minimize cost or simply find feasible blends. Constraints: x1 + x2 = 100, 0.6x1 + 0.2x2 ≥ 40, x1, x2 ≥ 0.
  • 4) Advertising allocation: A firm has Rs 50,000 to spend on TV and Online ads. Each TV ad costs 2,000 and yields 5000 expected impressions; each Online ad costs 1,000 and yields 2,200 impressions. Maximize impressions: let x1 = number of TV ads, x2 = number of Online ads. Objective: Maximize I = 5000x1 + 2200x2 subject to 2000x1 + 1000x2 ≤ 50000 and x1, x2 ≥ 0.
🧮 Formulas
  1. General objective function (maximize or minimize): Z = c1 x1 + c2 x2 + ... + cn xn
  2. General linear constraint: a1 x1 + a2 x2 + ... + an xn ≤, ≥, or = b
  3. Non-negativity: x1 ≥ 0, x2 ≥ 0, ..., xn ≥ 0
  4. Standard matrix form: Maximize Z = c^T x subject to A x ≤ b, x ≥ 0
  5. Slack variable for ≤ constraint: a1 x1 + ... + an xn + s = b, with s ≥ 0
  6. Surplus for ≥ constraint: a1 x1 + ... + an xn - s = b (s ≥ 0); may need artificial variable for simplex
📊 Visual ideas
For two variables x and y, draw x-axis (x1) and y-axis (x2). For each constraint of form a x + b y ≤ c, draw the line a x + b y = c and shade the side satisfying the inequality; the feasible region is the intersection (typically a convex polygon).
Label and compute intersection points of boundary lines to find corner points. Evaluate the objective function at each corner point; the largest (or smallest) value gives the optimum.
Show iso-profit or iso-cost lines: draw lines c1 x + c2 y = k and slide parallel lines outward (for maximize) until the last line touching the feasible region identifies the optimal corner.
If converting to equalities using slack variables, illustrate how each ≤ constraint becomes a horizontal/vertical/oblique line that bounds the feasible polygon; mark slack variables as distances from the boundary to the axis in simple cases.
📈3

Graphical Method for Solving LPP (Two Variables)

What is a Linear Programming Problem (LPP)?
An LPP is the process of optimizing (maximizing or minimizing) a linear objective function subject to a set of linear constraints (inequalities or equalities). In Class 12 problems the variables are usually two: x and y.

Standard form (two variables)
Objective: Z = ax + by (to be maximized or minimized)
Subject to: linear constraints of the form c1x + d1y (≤, ≥, =) e1, etc., and usually x ≥ 0, y ≥ 0.

Graphical method — idea
When there are two variables, each linear constraint corresponds to a straight line on the xy-plane. The set of points satisfying all constraints is the feasible region (a polygon or unbounded region). The optimal value of Z occurs at one (or more) corner (vertex) of the feasible region (Corner-Point Theorem).

Step-by-step procedure

  1. Write the objective function and all constraints (including x ≥ 0, y ≥ 0 if given).
  2. Convert each constraint into an equality to draw its boundary line: cix + diy = ei.
  3. Plot each boundary line on a graph. Use intercepts (set x=0 to get y-intercept, set y=0 to get x-intercept) or two convenient points to draw the line.
  4. Determine which side of each boundary line satisfies the inequality (test a point, often (0,0) if not on the line).
  5. Shade the common region that satisfies all constraints — this is the feasible region. If no common region exists, the LPP is infeasible.
  6. Identify all corner points (vertices) of the feasible region — intersections of boundary lines and/or axes.
  7. Evaluate the objective function Z = ax + by at each corner point. The largest value (for maximization) or smallest (for minimization) among these is the optimal solution.
  8. Special-case checks: if feasible region is unbounded, check whether objective can increase/decrease without bound; if an objective line coincides with an edge, there are infinitely many optimal solutions.

Iso-profit / Iso-cost line method (alternate visualization)
Write ax + by = Z and regard Z as a parameter. This is a straight line with slope -a/b. For maximization, slide the line parallelly away from the origin until it last touches the feasible region; that touching point is the optimum. For minimization slide toward the origin until first touch.

Corner-Point Theorem
If the feasible region is non-empty and bounded, the maximum or minimum value of a linear objective function occurs at a vertex (corner point) of the feasible region.

Special cases to watch

  • Infeasible: no common feasible region.
  • Unbounded: feasible region extends to infinity; objective may be unbounded (no finite optimum) or may have finite optimum depending on its direction.
  • Multiple optimal solutions: objective line is parallel to a side of feasible region and the entire side yields the same optimal value.
  • Redundant constraint: constraint that does not affect feasible region.

Notes
Graphical method applies only when there are two decision variables. If variables must be integers, graphical method gives fractional solution which may need integer programming techniques.

📌 Examples
  • Example 1 (Maximization — production problem): Maximize Z = 3x + 5y subject to x + 2y ≤ 8, 3x + y ≤ 9, x ≥ 0, y ≥ 0. Solution outline: plot lines x + 2y = 8 and 3x + y = 9 and the axes, find feasible region (polygon with vertices (0,0), (0,4), (2,3), (3,0)). Evaluate Z at vertices: Z(0,0)=0, Z(0,4)=20, Z(3,0)=9, Z(2,3)=21. Maximum is 21 at (x,y)=(2,3).
  • Example 2 (Minimization): Minimize C = 5x + 4y subject to x + y ≥ 4, 2x + y ≥ 5, x ≥ 0, y ≥ 0. Solution outline: plot lines x + y = 4 and 2x + y = 5; feasible region is the intersection above both lines and in first quadrant. Key vertices include intersections (1,3), (0,5), (4,0). Evaluate C: C(1,3)=17, C(0,5)=20, C(4,0)=20. Minimum is 17 at (1,3).
🧮 Formulas
  1. Objective function: Z = ax + by (maximize or minimize).
  2. Constraint (boundary): c x + d y = e. To plot use intercepts: x-intercept: (e/c, 0) (set y=0), y-intercept: (0, e/d) (set x=0).
  3. Slope of a line c x + d y = e: y = (e/d) - (c/d) x ⇒ slope = -c/d.
  4. Iso-objective line: ax + by = Z ⇒ y = (Z/b) - (a/b)x. Lines for different Z are parallel (slope = -a/b).
  5. Corner-Point Theorem: optimal value (if feasible and bounded) occurs at a vertex of the feasible region.
  6. To find intersection of two lines: solve linear system, e.g., c1 x + d1 y = e1 c2 x + d2 y = e2 solve for x and y (substitution or elimination).
📊 Visual ideas
Graph suggestion for Example 1: Draw x and y axes, mark a suitable scale (e.g., 1 unit = 1). Plot line x+2y=8 using intercepts (8,0) and (0,4). Plot line 3x+y=9 using intercepts (3,0) and (0,9). Shade region satisfying x+2y ≤ 8 (below the line) and 3x+y ≤ 9 (below that line) and x,y ≥ 0. Highlight vertices (0,0), (0,4), (2,3) and (3,0). Mark the optimal vertex (2,3) and annotate Z=21 there. You can also draw an iso-profit line 3x+5y=21 and show it just touches the feasible region at (2,3).
Graph suggestion for Example 2: Use axes with scale 1 unit = 1. Plot x+y=4 (intercepts (4,0),(0,4)) and 2x+y=5 (intercepts (2.5,0),(0,5)). Feasible region is above both lines and in first quadrant. Identify intersections (1,3), (0,5), (4,0). Plot iso-cost lines 5x+4y=C and slide them toward the feasible region; the first contact gives the minimum at (1,3).
Typical annotations to include on all graphs: label axes and units, draw each boundary line solid and mark the inequality direction with a small arrow or shading, clearly mark and label all corner points with coordinates, draw the iso-objective line parallel to itself to show movement, and highlight the optimal point(s).
Special-case graph suggestions: (a) Infeasible: draw two (or more) constraint half-planes that do not overlap — show 'no feasible region' label. (b) Unbounded: draw feasible region extending to infinity and show an objective line whose value can increase without bound in that direction. (c) Multiple optimal: draw an objective line parallel to one edge of the feasible polygon so the whole edge is optimal — shade or bold that edge.
🔢4

Fundamental Concepts and Theorems

Overview

Linear Programming (LP) is the study of optimizing a linear objective function subject to a set of linear constraints (equalities or inequalities). In Class 12, the focus is on two-variable problems solved by the graphical method and on fundamental theorems that guarantee where optima occur.

Key components of an LPP

  • Decision variables: unknowns to be chosen (usually x and y).
  • Objective function: a linear function to be maximized or minimized, e.g. Z = c1 x + c2 y.
  • Constraints: linear inequalities (or equalities) like a11 x + a12 y ≤ b1, together with non‑negativity conditions x ≥ 0, y ≥ 0.
  • Feasible region: set of all points (x,y) satisfying all constraints. For linear constraints this is a convex polygon (possibly unbounded) in the plane.

Why convexity matters

The feasible region formed by linear inequalities is convex: any line segment joining two feasible points lies entirely in the region. This convexity is critical for optimization because it ensures that a global optimum occurs at an extreme point (vertex) of the feasible region.

Fundamental theorem (Corner‑point or Extreme‑point theorem)

If a linear programming problem has an optimal solution and the feasible region is nonempty and bounded, then at least one optimal solution occurs at a vertex (corner point) of the feasible region. Consequently, for two‑variable problems solved graphically, one needs only evaluate the objective function at the vertices to find the optimum.

Special cases

  • No feasible solution: constraints are inconsistent and feasible region is empty.
  • Unbounded solution: feasible region is unbounded and the objective can increase (or decrease) without bound.
  • Multiple (infinitely many) optimal solutions: objective function is parallel to a constraint line and attains the same maximum/minimum along an edge segment of the feasible polygon.

Graphical method summary (2 variables)

  1. Let x ≥ 0, y ≥ 0 (if given) and write all constraints.
  2. Convert each inequality into the boundary line (replace ≤ or ≥ by =).
  3. Plot the boundary lines (use intercepts to draw them quickly) and determine the half‑planes to shade; intersection of all half‑planes is the feasible region.
  4. Identify corner points (vertices) of the feasible region — intersections of pairs of boundary lines or intersections with axes.
  5. Compute the objective value Z at each vertex; choose the vertex giving the largest (or smallest) Z.
  6. Check special cases: if objective lines are parallel to an edge containing vertices with same Z, there are infinitely many optima; if region is unbounded and objective improves indefinitely in an allowed direction, there is no finite optimum.

Practical notion of slack and surplus

For a ≤ constraint, slack = b − (ax + by) ≥ 0. For a ≥ constraint, surplus = (ax + by) − b ≥ 0. These measure unused or excess amounts and are used in algebraic/LP tableau methods.

📌 Examples
  • Manufacturing profit problem: A factory makes two products P1 and P2. Profit per unit: Rs. 40 for P1 and Rs. 30 for P2. Each product uses machine time and raw material: 2 hours and 3 kg for P1, 1 hour and 2 kg for P2. Available: 100 hours and 120 kg. Maximize profit Z = 40x + 30y subject to 2x + 1y ≤ 100, 3x + 2y ≤ 120, x ≥ 0, y ≥ 0. Graph constraints, find vertices, evaluate Z at each vertex to get optimal mix.
  • Diet/blend problem: Choose amounts x and y of two foods to meet at least 50 units of nutrient A and 30 units of nutrient B at minimum cost. If food 1 provides 4A and 1B per unit costing Rs. 5 and food 2 provides 2A and 3B costing Rs. 4, minimize cost C = 5x + 4y subject to 4x + 2y ≥ 50, 1x + 3y ≥ 30, x ≥ 0, y ≥ 0. Use graphical method to find feasible region and minimum cost at a vertex.
  • Workforce scheduling: A store needs at least 8 workers in morning and 5 in evening. Full‑time worker covers both shifts, part‑time covers only one. Minimize cost given wages; formulate linear inequalities and optimize.
  • Mixing/blending (petroleum): Mix two grades of fuel to meet octane requirements at minimum cost. Constraints linear in proportions; optimum at a corner point.
🧮 Formulas
  1. Standard form (maximization): Maximize Z = c1 x + c2 y subject to a11 x + a12 y ≤ b1, a21 x + a22 y ≤ b2, x ≥ 0, y ≥ 0.
  2. Intercepts of line ax + by = c: x‑intercept = (c/a, 0) if a ≠ 0; y‑intercept = (0, c/b) if b ≠ 0.
  3. Solve for intersection (pair of lines): Given a1 x + b1 y = c1 and a2 x + b2 y = c2, solve the 2×2 system (e.g. by substitution or using determinant): x = (c1 b2 − b1 c2)/Δ, y = (a1 c2 − c1 a2)/Δ, where Δ = a1 b2 − b1 a2 (if Δ ≠ 0).
  4. Corner‑point theorem (informal): If an LPP has an optimal solution, there exists an optimal solution at an extreme point (vertex) of the feasible region.
  5. Slack for ≤ constraint: slack = b − (ax + by) ≥ 0; Surplus for ≥ constraint: surplus = (ax + by) − b ≥ 0.
📊 Visual ideas
Feasible region polygon: Plot typical ≤ constraints and non‑negativity axes to show a convex polygon. Mark vertices A, B, C. Evaluate objective Z at each vertex and mark the optimal vertex.
Iso‑profit (objective) lines: Draw a line c1 x + c2 y = k and slide it parallel across the graph until it last touches the feasible region — that touch point is the optimum (show both maximizing and minimizing direction).
Unbounded feasible region: Show constraints giving a region that extends to infinity; draw objective line moving in the improving direction to show no finite optimum.
Infeasible system: Draw two parallel contradictory constraints (e.g. x + y ≤ 1 and x + y ≥ 3) showing no intersection — label 'no feasible solution'.
🧴5

Types of Solutions and Special Cases

What is being solved: In linear programming (graphical method) we maximize or minimize a linear objective function Z = ax + by subject to linear inequalities (constraints) and x, y usually ≥ 0. The feasible region is the set of all points (x,y) that satisfy the constraints. The nature of the feasible region and how the objective lines meet it determine the type of solution.

Corner-point (vertex) principle: For a linear objective and a convex polygonal feasible region, an optimal value (maximum or minimum) if it exists occurs at a vertex (corner point) of the feasible region. Thus the graphical method: draw constraints, find vertices (intersections of boundary lines), evaluate Z at each vertex and choose the best value.

Main types of solutions and special cases

  • Unique (single) optimal solution: The objective line touches the feasible region at exactly one vertex. Evaluate Z at all vertices; one vertex gives the best value.
  • Multiple (infinitely many) optimal solutions: The objective line is parallel to a boundary of the feasible region and coincides with it at the optimum. Every point on that edge (line segment) gives the same optimal value — infinitely many optima.
  • Unbounded solution (objective unbounded): The feasible region is unbounded in a direction in which the objective function increases (for maximization) or decreases (for minimization). Then the objective can go to +∞ or −∞ and no finite optimal value exists.
  • No feasible solution (infeasible LP): The constraints have no common point; feasible region is empty. There is no solution that satisfies all constraints simultaneously.
  • Feasible region is a single point or a line segment: If constraints reduce to equalities or force a single intersection, the feasible set may be a single point (one feasible solution) or a line segment (every point on it feasible). If objective is constant on that set, every point is optimal; otherwise pick that point.
  • Redundant constraint: A constraint that does not affect the feasible region (it is implied by others). Removing it does not change the feasible set or optimum.
  • Degeneracy (graphical view): More than two boundary lines meet at a single vertex (e.g., an intersection where three or more constraints are active). Degeneracy can lead to ties in objective values at adjacent vertices or multiple bases representing same vertex.

How to identify these cases (graphical checklist):

  • Plot all constraint lines and shade the feasible side; locate the feasible region (if none, LP is infeasible).
  • Check if the feasible region is bounded (closed polygon) or unbounded (opens to infinity).
  • Draw objective lines ax + by = c and slide them in the direction of increasing (or decreasing) c until the last contact with the feasible region. If the last contact is a vertex — unique optimum; if it coincides with an entire edge — infinitely many optima; if the line can move to infinity without leaving feasible region — unbounded objective.
  • Check constraint redundancy visually: a constraint inside the feasible polygon is redundant.

Typical solution procedure (graphical): 1) Convert inequalities to boundary lines. 2) Draw lines, find intersection points (vertices). 3) List only vertices inside the feasible region. 4) Compute Z = ax + by at each vertex. 5) Select the vertex(ices) giving maximum/minimum. Use the behaviour of objective lines to detect special cases.

📌 Examples
  • Unique optimum: Maximize Z = 3x + 2y subject to x + y ≤ 6, x ≤ 4, y ≤ 3, x, y ≥ 0. The feasible region is a bounded polygon and Z is maximum at one corner (a single solution).
  • Multiple optima: Maximize Z = x + y subject to x + y ≤ 5, x, y ≥ 0. The objective lines x + y = c coincide with the constraint x + y = 5 at optimum, so every point on the segment from (0,5) to (5,0) is optimal (infinitely many solutions).
  • Unbounded objective: Maximize Z = x + y subject to y ≤ x, x, y ≥ 0. The feasible region is unbounded in the direction where x and y grow, so Z can increase without bound — no finite maximum.
  • Infeasible system: Constraints x + y ≤ 1 and x + y ≥ 3 with x, y ≥ 0 have no common point — the feasible region is empty, so no solution exists.
  • Feasible region a single point: Solve x + y = 2 and x − y = 0 together with x, y ≥ 0. Both equalities intersect at a single feasible point (1,1) — that is the only solution.
  • Redundant constraint: For instance, x + y ≤ 5 and 2x + 2y ≤ 10 — the second constraint is redundant (it is just a multiple of the first) and does not change the feasible region.
🧮 Formulas
  1. Standard objective: Z = ax + by (to maximize or minimize).
  2. Typical constraint form: a1 x + b1 y ≤ c1, a2 x + b2 y ≤ c2, ... with x, y ≥ 0.
  3. Corner (intersection) of two lines a1 x + b1 y = c1 and a2 x + b2 y = c2: solve the 2×2 system. Using determinants, x = (c1 b2 − b1 c2) / (a1 b2 − b1 a2), y = (a1 c2 − c1 a2) / (a1 b2 − b1 a2) (denominator ≠ 0).
  4. Corner-point theorem (informal): If the feasible region is nonempty and bounded, an optimum of a linear objective occurs at one (or more) vertices of the feasible region.
  5. Condition for multiple optima: objective vector (a,b) is parallel to an active edge of the feasible polygon; objective line coincides with that edge at the optimum.
  6. Unboundedness check: if feasible region is unbounded in a direction v where dot((a,b), v) > 0 for maximization, then objective is unbounded (can increase indefinitely).
📊 Visual ideas
Unique optimum: Draw a bounded polygon (feasible region), mark vertices, draw several objective lines ax+by=c parallel to each other and show the last one touching exactly one vertex. Label that vertex and report Z there.
Multiple optima: Draw feasible polygon with one edge collinear with an objective line (e.g., x+y=5). Shade the polygon and highlight the entire edge where the objective touches — indicate that every point on this edge gives the same optimal Z.
Unbounded objective: Draw a feasible region that extends to infinity (an open wedge or half-plane). Draw objective lines and show they can be shifted indefinitely in the improving direction without leaving feasible region. Use an arrow to show direction of increase.
Infeasible case: Show two non-overlapping shaded half-planes whose intersection is empty (no common area). Label that the feasible set is empty.
🔢6

Handling Different Types of Constraints

In linear programming (Class 12), constraints are linear equations or inequalities that define the feasible region for the decision variables. Handling different types of constraints correctly is essential to get the correct feasible region and optimum solution. Common constraint types are:

  • ‘≤’ (less-than-or-equal-to) : represents an upper bound. Convert to an equality by adding a slack variable (nonnegative).
  • ‘≥’ (greater-than-or-equal-to) : represents a lower bound. Convert to an equality by subtracting a surplus variable (nonnegative). In algebraic methods you may also need an artificial variable for simplex methods.
  • ‘=’ (equality) : the feasible region is restricted to the line itself (or intersection of lines).
  • Non-negativity constraints (x, y ≥ 0) : restrict variables to the first quadrant. If variables are free (can be negative), replace each free variable by difference of two nonnegative variables.

Standard graphical method steps when constraints involve two variables:

  1. Write each constraint in the form ax + by ≤ c or ax + by ≥ c (or = c).
  2. Draw the boundary line ax + by = c for each constraint. Use intercepts or solve two points to plot the line.
  3. Determine the half-plane that satisfies the inequality by testing a point (usually origin if not on the line).
  4. The feasible region is the intersection of all half-planes and non-negativity conditions. It will be a polygon (possibly unbounded), a line segment, a single point, empty (infeasible), or unbounded.
  5. Use the corner-point (vertex) method: compute all corner (intersection) points of the feasible polygon and evaluate the objective function Z = ax + by at each vertex. The maximum or minimum occurs at one (or more) vertices.

Special cases to recognize:

  • Redundant constraint : does not change the feasible region (lies outside or inside polygon).
  • Infeasible system : no point satisfies all constraints.
  • Unbounded solution : feasible region is unbounded and objective function can increase/decrease indefinitely (no finite optimum).
  • Multiple optimal solutions : objective function is parallel to a constraint edge; every point on that edge gives the same optimal value.
  • Binding vs non-binding : at optimum, a constraint is binding if it holds as equality at the optimum vertex; otherwise, it is non-binding.

Algebraic conversions often used:

  • For ax + by ≤ c: add slack s ≥ 0 so ax + by + s = c.
  • For ax + by ≥ c: subtract surplus s ≥ 0 so ax + by - s = c (and possibly add artificial variable if using simplex).
  • For a variable free in sign: x = x' - x'' with x', x'' ≥ 0.

Handling these conversions carefully ensures correct plotting, correct vertex calculation, and correct identification of optimal values.

📌 Examples
  • Factory production: Two products P1 and P2 require machine hours and raw material. Machine hours constraint might be 3x + 2y ≤ 120 (≤ type, add slack). Raw material might require at least 20 units of P1: x ≥ 20 (≥ type, lower bound). Non-negativity x,y ≥ 0.
  • Workforce scheduling: At least 10 workers must be on shift: w ≥ 10 (≥). Shift capacity limits total workers: w ≤ 30 (≤). If a worker count could be any integer but not negative, include non-negativity and integrality (LP model ignores integrality).
  • Diet problem: Nutrient equality: protein constraint 10x + 15y = 200 (equality). Vitamin minimum: 2x + 3y ≥ 50 (≥ type). This restricts feasible mixes to intersection of half-planes and possibly a line.
  • Investment with bounds: Invest between 2000 and 10000 in plan A: 2000 ≤ x ≤ 10000 (both ≥ and ≤ constraints). Other constraints limit total investment = x + y ≤ 20000.
🧮 Formulas
  1. Objective function: Z = px + qy (maximize or minimize). Evaluate Z at each vertex of feasible region.
  2. Convert ≤ to equality by adding slack: ax + by ≤ c => ax + by + s = c, s ≥ 0.
  3. Convert ≥ to equality by subtracting surplus: ax + by ≥ c => ax + by - s = c, s ≥ 0 (artificial variable may be needed for algebraic methods).
  4. Free variable conversion: If x is unrestricted, write x = x1 - x2 with x1, x2 ≥ 0.
  5. Intersection (Cramer's rule) for two lines: a1 x + b1 y = c1 and a2 x + b2 y = c2 => x = (c1 b2 - b1 c2) / (a1 b2 - b1 a2), y = (a1 c2 - c1 a2) / (a1 b2 - b1 a2).
  6. Test half-plane: pick a test point (often (0,0)) and check inequality; if true, shade that side of the boundary line.
📊 Visual ideas
Basic plot: Draw axes x and y. For each constraint ax + by ≤ c draw line ax + by = c (use intercepts x=c/a when y=0 and y=c/b when x=0). Shade the half-plane satisfying the inequality. Intersection of all shaded areas is feasible region; mark and label vertices.
Handling ≥: same as ≤ but shade the opposite side of the line. Show one ≥ and one ≤ constraint to illustrate flipping shading.
Equality constraint: draw the line ax + by = c and show feasible region restricted to that line (e.g., intersection with x,y ≥ 0 becomes a segment or point).
Unbounded feasible region: draw constraints that allow region to extend indefinitely (e.g., x ≥ 0, y ≥ 0, x - y ≥ 0). Show objective line moving outward to indicate no finite maximum if it keeps increasing.
🧴7

Solution Strategy and Shortcuts

What it is: In Linear Programming (graphical method for two variables) you form an objective function (to maximize or minimize) and linear constraints. The solution strategy finds the feasible region (set of all points satisfying the constraints) and uses the Corner‑Point (Extreme Point) principle: the optimum value of the objective occurs at one (or more) vertices of the feasible region.

Step‑by‑step solution strategy:

  • 1. Define variables: assign x and y to measurable quantities (non‑negative if stated).
  • 2. Form the objective function: Z = ax + by (either maximize or minimize).
  • 3. Write constraints as linear inequalities (include x ≥ 0, y ≥ 0 if required).
  • 4. Convert inequalities to equalities to draw boundary lines: ax + by = c.
  • 5. Plot boundary lines on the xy‑plane (use intercepts for quick sketch).
  • 6. Determine feasible region by testing a convenient point (e.g., (0,0)) or using the direction of inequalities; shade it.
  • 7. Find corner (vertex) points — intersections of pairs of boundary lines and intersections with axes.
  • 8. Evaluate objective Z at each vertex. The maximum/minimum among these is the required optimum.
  • 9. Check feasibility/special cases: no feasible region (infeasible), unbounded region (may have no finite optimum), or infinitely many optima (objective parallel to a constraint along an edge).

Common shortcuts and tips:

  • Plot lines fast using intercepts: for ax + by = c, x‑intercept = c/a (set y=0), y‑intercept = c/b (set x=0).
  • To find intersection of two lines, solve the 2×2 linear system. Use substitution, elimination, or Cramer’s rule if you prefer.
  • Test (0,0) to quickly determine which side of a line to shade when 0 satisfies all nonstrict inequalities.
  • Use the slope of objective line (-a/b) to compare with constraint slopes: if objective is parallel to a bounding edge and you can move the objective line along that direction inside the feasible region, you may have infinitely many optimal solutions along that edge.
  • If feasible region is unbounded, compare objective slope with feasible directions. For maximization, if you can move objective line indefinitely in increasing direction without leaving feasible region, no finite maximum exists (similarly for minimization).
  • Eliminate redundant constraints (those that do not affect the feasible region) to simplify graphing.
  • Compute slack for a constraint: slack = RHS − (ax + by). If slack = 0 at optimum, the constraint is binding.

Special cases (recognize quickly):

  • No feasible region → no solution (constraints inconsistent).
  • Unbounded feasible region with objective improving without bound → no finite optimum.
  • Objective line parallel to a constraint edge that forms a face of the feasible region → infinitely many optima along that edge.
📌 Examples
  • Example 1 (Maximization, worked): Maximize Z = 3x + 4y subject to x + 2y ≤ 8, 3x + y ≤ 9, x ≥ 0, y ≥ 0. Strategy: plot boundaries using intercepts (x+2y=8 → (8,0),(0,4); 3x+y=9 → (3,0),(0,9)); determine feasible region (test (0,0) satisfies both); find vertices: (0,0), (3,0), intersection of the two lines (solve x+2y=8 and 3x+y=9 → x=2,y=3), and (0,4). Evaluate Z: Z(0,0)=0, Z(3,0)=9, Z(2,3)=18, Z(0,4)=16. Maximum Z = 18 at (2,3).
  • Example 2 (Minimization, brief): Minimize C = 5x + 3y subject to x + y ≥ 4, x + 2y ≥ 5, x ≥ 0, y ≥ 0. Convert to feasible region (intersection above lines), find corner points (intersections and axes if applicable), evaluate C at each corner, choose smallest value. (Procedure identical; inequalities flip shading direction.)
  • Real‑life example (Manufacturing): A factory makes products A and B. Profit per unit: A = ₹50, B = ₹70. Machine1 available 40 hours with A taking 2 h and B 1 h; Machine2 available 30 hours with A taking 1 h and B 2 h. Let x = units of A, y = units of B. Maximize Z = 50x + 70y subject to 2x + y ≤ 40, x + 2y ≤ 30, x,y ≥ 0. Solve graphically to decide production mix that maximizes profit under machine limits.
  • Real‑life example (Diet/minimization): Minimize cost of food combination subject to nutritional requirements. x and y represent quantities of two foods; constraints represent minimum nutrients; objective is cost = c1 x + c2 y. Graphical method finds cheapest feasible combination (or indicates no feasible combination).
🧮 Formulas
  1. Objective function: Z = ax + by (maximize or minimize).
  2. Constraint (line form): ax + by ≤ c or ax + by ≥ c (boundary: ax + by = c).
  3. Intercepts: x‑intercept = c/a (set y = 0), y‑intercept = c/b (set x = 0).
  4. Intersection of two lines (2×2 system): Given a1 x + b1 y = c1 and a2 x + b2 y = c2, use elimination/substitution or Cramer's rule: x = det([[c1,b1],[c2,b2]]) / det([[a1,b1],[a2,b2]]), y = det([[a1,c1],[a2,c2]]) / det([[a1,b1],[a2,b2]]).
  5. Slack for a constraint (≤ type): slack = RHS − (ax + by); slack = 0 means binding constraint.
  6. Corner‑Point Principle: For a linear objective and convex feasible region, an optimal solution (if finite) occurs at a vertex (corner) of the feasible region.
📊 Visual ideas
Basic sketch for a 2‑constraint problem: draw x and y axes, plot both boundary lines using intercepts, shade the feasible side for each inequality, mark the polygonal feasible region and label vertices (A,B,C,...). Evaluate Z at each vertex and annotate values beside points.
Show objective line technique: draw one line ax + by = k (for some k), then draw a parallel line moving it toward increasing Z (for maximization) until it last touches the feasible region — the touching point(s) are optimum. Indicate parallel movement with arrows.
Unbounded region example: sketch feasible region extending to infinity in one direction; draw objective line whose slope allows indefinite increase; mark that no finite maximum exists.
Multiple optimal solutions: draw a feasible polygon where one edge is parallel to the objective line; highlight the entire edge as the set of optimal points.
🔢8

Examples and Practice Problems

Linear Programming Problems (LPP) are optimisation problems in two (or more) variables where you maximise or minimise a linear objective function subject to linear inequality constraints. The Chapter's "Examples and Practice Problems" develop skill in modelling real situations as LPP, drawing the feasible region, and using the graphical (corner-point) method to find the optimum.

Standard procedure (graphical method for two variables):

  • Step 1 — Define variables: choose x, y to represent decision quantities clearly.
  • Step 2 — Form the objective function: Z = px + qy (to be maximised or minimised).
  • Step 3 — Write constraints as linear inequalities (including non-negativity x ≥ 0, y ≥ 0).
  • Step 4 — Graph each constraint as a straight line; determine which side satisfies the inequality and shade it. The intersection of all shaded half-planes is the feasible region.
  • Step 5 — Identify corner (extreme) points of the feasible region — these are intersection points of pairs of boundary lines (or intercepts with axes).
  • Step 6 — Evaluate Z at each corner point. For a bounded feasible region, the maximum or minimum occurs at one of the corner points (corner-point theorem).
  • Step 7 — Interpret the result in the original problem context and check feasibility (including integer requirements if any).

Important observations:

  • If the feasible region is empty → LPP is infeasible.
  • If the feasible region is unbounded and the objective function increases without bound along that region → objective is unbounded (no finite optimum).
  • Sometimes constraints are redundant (do not affect the feasible region); identify them by graphing or algebraically.
  • To find optimum quickly on graph, use isoprofit/isocost lines: draw the line px + qy = constant and slide it parallelly until the last point it touches the feasible region.
📌 Examples
  • Example 1 (Maximisation — production problem): Problem: A factory makes chairs (x) and tables (y). Profit per chair = Rs 20, per table = Rs 30. Each chair uses 2 labour-hours and 3 kg material; each table uses 3 labour-hours and 4 kg material. Available: 100 labour-hours and 120 kg material. Formulate and solve. Formulation: Maximise Z = 20x + 30y subject to 2x + 3y ≤ 100, 3x + 4y ≤ 120, x ≥ 0, y ≥ 0. Graph and corners: x-intercepts: (50,0) from first, (40,0) from second → feasible x ≤ 40. y-intercepts: (0,33.33) and (0,30) → feasible y ≤ 30. Corner points: (0,0), (40,0), (0,30). Evaluate Z: (0,0)=0, (40,0)=800, (0,30)=900. Optimum: max Z = 900 at (x,y) = (0,30) — produce 30 tables, 0 chairs. Remark: Intersection of the two constraint lines lies outside first quadrant (gives x<0), so not a vertex of feasible region.
  • Example 2 (Minimisation — mixing problem): Problem: Two raw materials A and B cost Rs 3/kg and Rs 5/kg respectively. Each kg of A supplies 4 units of a nutrient, each kg of B supplies 7 units. A product must supply at least 100 units of the nutrient. Find the cheapest mix. Formulation: Let x, y be kg of A and B. Minimise C = 3x + 5y subject to 4x + 7y ≥ 100, x ≥ 0, y ≥ 0. Graph and corners: Nutrient line intersects axes at (25,0) and (0,100/7 ≈ 14.285). Feasible region is the half-plane above that line. Candidate minimal cost is at boundary where the constraint meets axes; evaluate costs: at (25,0): C = 75; at (0,14.285): C ≈ 71.43. Optimum: min C ≈ 71.43 at x = 0, y ≈ 14.285 (use only B). Remark: Because region is unbounded above, the minimum occurs on the boundary nearest the origin in cost-direction.
  • Example 3 (Infeasible and Unbounded cases): (a) Infeasible: Constraints x + y ≤ 2 and x + y ≥ 5 with x,y ≥ 0 give no common solution → no feasible region. (b) Unbounded objective: Maximise Z = x + y subject to x - y ≥ 1, x ≥ 0, y ≥ 0. The feasible region extends infinitely and Z can be made arbitrarily large along feasible directions (unbounded), so there is no finite maximum. These illustrate how to detect infeasibility and unboundedness from inequalities/graph.
🧮 Formulas
  1. General objective function: Z = p x + q y (p, q constants).
  2. Constraints: a1 x + b1 y ≤ c1, a2 x + b2 y ≤ c2, ... and x ≥ 0, y ≥ 0 (non-negativity).
  3. To find intersection of two lines: Solve linear system a1 x + b1 y = c1 and a2 x + b2 y = c2 (use substitution or elimination).
  4. Slope of objective (isoprofit/isocost) line px + qy = k is: y = (-p/q) x + k/q → slope = -p/q. Shift this line parallel to find optimum.
  5. Corner-point theorem: If the feasible region is a convex polygon (closed and bounded) and the objective is linear, then a maximum or minimum (if it exists) occurs at a vertex (corner) of the feasible region.
  6. To convert a ≥ inequality to ≤: multiply both sides by −1 (careful to reverse the inequality).
📊 Visual ideas
General graph recipe: Draw x and y axes. For each constraint a x + b y ≤ c: draw the line a x + b y = c (find x- and y-intercepts), then shade the half-plane satisfying the inequality. The feasible region is the intersection of all shaded half-planes; mark it (prefer a different color).
For Example 1: Plot lines 2x + 3y = 100 and 3x + 4y = 120. Mark feasible region in first quadrant (x ≥ 0, y ≥ 0). Identify vertices (0,0), (40,0), (0,30). Plot and label Z lines like 20x + 30y = k; slide parallel lines outward to see last touch at (0,30).
For Example 2: Plot the nutrient line 4x + 7y = 100. Shade the region above the line (since ≥). Mark intercepts (25,0) and (0,100/7 ≈14.285). Plot cost lines 3x + 5y = k and slide to find the smallest k that meets the feasible region (touch at intercept with smaller cost).
To show infeasible/unbounded: (a) For infeasible, sketch two parallel contradictory strips (no overlap), e.g. x + y ≤ 2 and x + y ≥ 5 — highlight absence of intersection. (b) For unbounded, sketch feasible region extending to infinity and show objective lines moving parallelly to larger values without bound.
⚖️9

Interpretation of Results and Practical Considerations

What to interpret from a linear programming (LP) solution

After solving an LP (graphically or by simplex), you must interpret the solution in terms of the original application. Key points to read from the solution are:

  • Feasibility: Is there any point that satisfies all constraints? If not, the problem is infeasible — no practical solution meets the requirements.
  • Optimality: If feasible, is there a point that maximizes/minimizes the objective? For a bounded feasible region, an optimal value exists and occurs at one (or more) corner(s) of the feasible region.
  • Uniqueness vs. Multiplicity: The solution may be unique (one corner) or multiple (all points on an edge give the same objective value). Multiple optima happen when the objective line is parallel to a constraint boundary that forms the active edge.
  • Unboundedness: If the objective can increase (or decrease) indefinitely while remaining feasible, the problem is unbounded — no finite optimum.
  • Slack and Surplus: For ≤ constraints, slack = (RHS − LHS) at the solution; it measures unused resource. For ≥ constraints, surplus = (LHS − RHS) measures excess above the minimum requirement.

How to read the numeric solution

  1. Identify the feasible region (intersection of all constraints including non-negativity if present).
  2. List corner points (vertices) of the region and compute the objective function value at each corner.
  3. The corner with the largest (for maximization) or smallest (for minimization) objective value is optimal (if region is bounded).
  4. Compute slack/surplus for each constraint at the optimal point to see which resources are binding (zero slack) and which are not.

Practical considerations (modelling and real-world use)

  • LP assumes linear relationships and continuous variables. If variables must be integers (e.g., number of machines, people), use integer programming or round carefully — naive rounding can violate constraints.
  • Data uncertainty: coefficients and RHS values (profits, costs, capacities) may change. Check sensitivity: small changes in coefficients may change which corner is optimal.
  • Multiple optima give flexibility: choose the solution that best fits secondary criteria (e.g., ease of implementation) while keeping the same objective value.
  • Unbounded solutions usually indicate a missing realistic constraint (e.g., market limit, capacity limit). Re-examine model assumptions and add constraints as needed.
  • Rounding: when presenting final values, round in a way that keeps feasibility and re-check constraints, or reformulate as integer LP if necessary.
  • Scalability: Graphical methods apply only to two-variable LPs. For larger problems use algebraic simplex or solver software and then interpret results similarly (binding constraints, slack, shadow prices).

Summary checklist when you interpret results

  • Confirm feasibility.
  • Identify whether optimal solution is unique, multiple, unbounded, or infeasible.
  • Report optimal variable values and objective value.
  • Report slack/surplus for each constraint and note binding constraints.
  • Consider real-world limits, integer needs, and sensitivity to data changes before acting on the numerical solution.
📌 Examples
  • 1) Unique optimum — Manufacturing example: Maximize profit Z = 40x + 30y subject to 2x + y ≤ 100 (machine hours), x + y ≤ 60 (labor), x ≥ 0, y ≥ 0. Find corner points (0,0),(0,60),(20,40),(50,0). Evaluate Z: 0, 1800, 2000, 2000. Here Z = 2000 at two corners (20,40) and (50,0) — multiple optima. If one corner strictly best, that would be unique.
  • 2) Multiple optimal solutions: Maximize Z = 3x + 6y subject to x + 2y ≤ 8, x ≥ 0, y ≥ 0. The objective line 3x + 6y = constant is parallel to the constraint boundary x + 2y = 8 (they have same slope), so every point on the segment between the intercepts that lies on x + 2y = 8 gives the same maximum Z. Choose any point on that segment and check practical considerations.
  • 3) Unbounded solution: Maximize Z = x + y subject to y ≤ x, x ≥ 0, y ≥ 0. The feasible region is the wedge y ≤ x in the first quadrant and extends to infinity; Z can grow without bound (pick x large), so no finite maximum exists. This signals a missing real-world constraint (e.g., production capacity).
  • 4) Infeasible example: Minimize Z = x + y subject to x + y ≤ 1 and x + y ≥ 3 with x, y ≥ 0. No (x,y) satisfies both constraints simultaneously, so the problem is infeasible — requires revising constraints or data.
🧮 Formulas
  1. Objective function: Z = ax + by (for two variables).
  2. General linear constraint: a1x + b1y ≤ c1 (or ≥, =).
  3. Slack (for ≤): slack = RHS − LHS = c − (a x + b y). Slack = 0 means the constraint is binding.
  4. Surplus (for ≥): surplus = LHS − RHS = (a x + b y) − c. Surplus = 0 means the constraint is binding.
  5. Corner-point principle: If the feasible region is non-empty and bounded, an optimal solution (max or min) occurs at one (or more) vertices of the feasible polygon.
  6. Multiple optimality condition (graphical): If the objective line is parallel to a constraint edge that forms part of the feasible boundary, all points on that edge are optimal.
📊 Visual ideas
Unique optimum: Draw x and y axes, plot linear constraints to form a bounded polygon (feasible region shaded). Mark all vertices, draw an objective line (e.g., ax + by = k) and slide it parallel until it last touches the polygon at a single corner. Label the corner coordinates and the optimal Z.
Multiple optima: Draw feasible polygon where one edge is parallel to the objective line. Shade feasible region and highlight the whole edge as optimal. Label endpoints and show the objective line coincident with the edge.
Unbounded case: Draw feasible region that extends to infinity (a wedge or unbounded polygon). Draw objective lines moving in the improving direction and show they never stop — objective keeps increasing. Note missing bounding constraint.
Infeasible case: Draw constraints that do not intersect (no common region). Show each half-plane and indicate there's no shaded common area. Label this as infeasible.

Key Concepts

Linear Programming (LP)
A mathematical method to maximize or minimize a linear objective function subject to a set of linear constraints (equalities or inequalities).
Objective Function
The linear function in LP that is to be maximized or minimized (e.g., profit or cost).
Decision Variables
Unknown quantities in an LP model whose values are to be determined (commonly x, y, ...).
Constraints
Linear equations or inequalities that restrict the values of decision variables (resource limits, demands, etc.).
Feasible Region
The set of all points (values of variables) that satisfy all constraints including non-negativity; usually a polygon in 2D.
Feasible Solution
A specific set of values for decision variables that satisfies all constraints.
Optimal Solution
A feasible solution that gives the best (maximum or minimum) value of the objective function.
Corner Point / Vertex
A point in the feasible region where two or more constraint boundaries meet; candidates for optimality in LP.
Corner Point Theorem
In LP with a convex feasible region, an optimal solution (if it exists) occurs at a corner point (vertex).
Bounded Solution
An LP problem is bounded if the feasible region is closed and the objective function attains a finite optimum value.
Unbounded Solution
When the objective function can increase (or decrease) without bound over the feasible region, so no finite optimum exists.
Infeasible
An LP problem with no point satisfying all constraints simultaneously; feasible region is empty.
Slack Variable
A nonnegative variable added to a '≤' constraint to convert it into an equality for solution methods.
Surplus Variable
A nonnegative variable subtracted from a '≥' constraint to convert it into an equality.
Artificial Variable
A temporary variable introduced to obtain an initial feasible solution when converting equalities or '≥' constraints; used in simplex or Big M method.
Basic Feasible Solution
A feasible solution obtained by setting non-basic variables to zero and solving for the basic variables (used in simplex and corner-point analysis).
Graphical Method
A technique to solve LP problems with two decision variables by graphing constraints, identifying the feasible region, and checking corner points.
Simplex Method
An algorithmic procedure to solve LP problems (especially with many variables) by moving between basic feasible solutions to improve the objective.
Duality (Dual Problem)
Every LP (primal) has an associated dual LP; solutions are related and provide bounds and economic interpretation of constraints.
Iso-profit / Iso-cost Line
A line representing all points giving the same objective value (Z = constant); used to visualize improvement direction in graphical method.

End-of-Chapter Trial Paper & Test Questions

Topic-wise questions to test your understanding of every concept in this chapter.

  1. Define linear programming and name its four essential components. / रैखिक प्रोग्रामन को परिभाषित कीजिए तथा इसके चार आवश्यक घटकों के नाम लिखिए।
    Show answer

    LP is a method to maximize or minimize a linear objective function subject to linear constraints; components are decision variables, objective function, constraints, and non-negativity conditions. / LP एक विधि है जिसमें रैखिक उद्देश्य फलन को रैखिक प्रतिबंधों के अधीन अधिकतम या न्यूनतम किया जाता है; घटक हैं निर्णय चर, उद्देश्य फलन, प्रतिबंध, तथा अऋणात्मकता शर्तें।

  2. State the Corner-Point (Extreme Point) Theorem. / कोणीय-बिंदु (चरम बिंदु) प्रमेय लिखिए।
    Show answer

    If an LPP has an optimal solution and the feasible region is non-empty and bounded, then at least one optimal solution occurs at a vertex (corner point) of the feasible region. / यदि किसी LPP का इष्टतम हल हो तथा सुसंगत क्षेत्र अरिक्त एवं परिबद्ध हो, तो कम से कम एक इष्टतम हल सुसंगत क्षेत्र के एक शीर्ष (कोणीय बिंदु) पर होता है।

  3. Maximize Z = 3x + 5y subject to x + 2y ≤ 8, 3x + y ≤ 9, x,y ≥ 0. Find the optimal value. / x + 2y ≤ 8, 3x + y ≤ 9, x,y ≥ 0 के अधीन Z = 3x + 5y को अधिकतम कीजिए। इष्टतम मान ज्ञात कीजिए।
    Show answer

    Vertices (0,0),(0,4),(2,3),(3,0); Z = 0,20,21,9 respectively; maximum Z = 21 at (2,3). / शीर्ष (0,0),(0,4),(2,3),(3,0); Z क्रमशः = 0,20,21,9; अधिकतम Z = 21, (2,3) पर।

  4. Distinguish between a feasible solution and an optimal solution. / सुसंगत हल तथा इष्टतम हल में अंतर बताइए।
    Show answer

    A feasible solution is any point satisfying all constraints; an optimal solution is a feasible solution that gives the best (maximum or minimum) value of the objective function. / सुसंगत हल वह बिंदु है जो सभी प्रतिबंधों को संतुष्ट करता है; इष्टतम हल वह सुसंगत हल है जो उद्देश्य फलन का सर्वोत्तम (अधिकतम या न्यूनतम) मान देता है।

  5. When does an LPP have multiple (infinitely many) optimal solutions? / किसी LPP के अनेक (अनंत) इष्टतम हल कब होते हैं?
    Show answer

    When the objective line is parallel to a bounding edge of the feasible region, so every point on that edge gives the same optimal value. / जब उद्देश्य रेखा सुसंगत क्षेत्र की किसी सीमांत भुजा के समांतर हो, तब उस भुजा के प्रत्येक बिंदु पर समान इष्टतम मान मिलता है।

  6. Formulate the LPP: A factory makes tables (x) and chairs (y); a table needs 3 labour-hours and 2 wood units, a chair 2 hours and 1 unit; 60 hours and 30 units are available; profit ₹50/table, ₹30/chair. / LPP बनाइए: एक कारखाना मेज (x) तथा कुर्सियाँ (y) बनाता है; मेज में 3 श्रम-घंटे व 2 लकड़ी इकाई, कुर्सी में 2 घंटे व 1 इकाई; 60 घंटे व 30 इकाई उपलब्ध; लाभ ₹50/मेज, ₹30/कुर्सी।
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    Maximize Z = 50x + 30y subject to 3x + 2y ≤ 60, 2x + y ≤ 30, x,y ≥ 0. / Z = 50x + 30y को 3x + 2y ≤ 60, 2x + y ≤ 30, x,y ≥ 0 के अधीन अधिकतम कीजिए।

  7. What is meant by an unbounded solution and an infeasible LPP? / असीमित हल तथा असंगत LPP से क्या तात्पर्य है?
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    Unbounded: the feasible region extends infinitely so the objective can increase/decrease without bound (no finite optimum); Infeasible: constraints have no common point, so the feasible region is empty. / असीमित: सुसंगत क्षेत्र अनंत तक फैला हो जिससे उद्देश्य बिना सीमा बढ़/घट सके (कोई परिमित इष्टतम नहीं); असंगत: प्रतिबंधों का कोई उभयनिष्ठ बिंदु नहीं, अतः सुसंगत क्षेत्र रिक्त।

  8. Minimize C = 5x + 4y subject to x + y ≥ 4, 2x + y ≥ 5, x,y ≥ 0. Find the minimum. / x + y ≥ 4, 2x + y ≥ 5, x,y ≥ 0 के अधीन C = 5x + 4y को न्यूनतम कीजिए।
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    Vertices (0,5),(1,3),(4,0); C = 20,17,20; minimum C = 17 at (1,3). / शीर्ष (0,5),(1,3),(4,0); C = 20,17,20; न्यूनतम C = 17, (1,3) पर।

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