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Linear Programming Notes, Formulas, Traps & PYQs | ExamiFly | Examifly

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1 Question | 4 Marks in JEE Main (Guaranteed Score Booster)

Mastering Linear Programming: Formulas, Traps & Exam Strategy | ExamiFly

Q: How do I quickly determine which side of a boundary line $ax + by \le c$ to shade under exam time pressure?

Substitute the test point $(0,0)$ into $ax + by \le c$. If $a(0) + b(0) \le c$ holds true, shade the half-plane containing $(0,0)$. If false, shade the opposite half-plane. If the line passes through origin, test $(1,0)$ instead.

Prerequisites:Straight Lines, Systems of Linear Inequalities, Coordinate Geometry Basics

Linear Programming ExamiFly Revision Notes and Formulas

Chapter Overview

Think of Linear Programming ( $LPP$ ) as the mathematical art of resource allocation under tight real-world constraints. Whether a manufacturing plant optimizes profit or a transport network minimizes fuel consumption, $LPP$ translates these physical limitations—like time, capital, and raw materials—into a system of linear inequalities. At its core, the Convex Polygon Theorem guarantees that the optimal value (maximum or minimum) of a linear objective function $Z = ax + by$ always occurs at one of the extreme vertices (corner points) of the convex feasible region formed by the intersection of these linear constraints. In JEE Main, $LPP$ represents a guaranteed $4$ marks with near-zero computational risk if you master boundary conditions and unbounded region tests. The key to eliminating errors lies in precision graphing, identifying non-negativity constraints $x \ge 0, y \ge 0$ , and rigorously testing open half-planes when the feasible region extends to infinity.

Core Subtopics & Syllabus Breakdown

Mathematical Formulation of Linear Programming Problems

Graphical Method of Solution for Systems of Linear Inequalities

Bounded Feasible Regions and Corner Point Method

Unbounded Feasible Regions and Open Half-Plane Tests

Optimization of Objective Function and Edge Case Analysis

Key Formulas & High-Yield Equations

Objective Function

$Z = ax + by$

Linear function to be maximized or minimized, where

a, b \in \mathbb{R}

are constant coefficients and

x, y \ge 0

are decision variables.

Non-Negativity Constraints

$x \ge 0, \quad y \ge 0$

Strict mathematical boundary defining that real-world physical quantities cannot be negative, restricting the feasible region to Quadrant

I

Corner Point Theorem

$Z_{opt} = \max_{(x_i, y_i) \in V} (ax_i + by_i)$

Applies strictly when the feasible region

R

is a bounded convex polygon; the optimal value occurs at one of the vertices

V

R

Unbounded Maximization Test

$ax + by > M$

M

is the maximum value at a corner point. Valid only if the open half-plane

ax + by > M

has no common points with the feasible region.

Unbounded Minimization Test

$ax + by < m$

m

is the minimum value at a corner point. Valid only if the open half-plane

ax + by < m

has no common points with the feasible region.

Common Mistakes & Traps to Avoid

High Error Zones

Assuming the corner point value $M$ is the maximum in an unbounded feasible region without performing the open half-plane test $ax + by > M$ .

Correct Approach:Graph the open half-plane

ax + by > M

. If it shares even a single point with the feasible region, conclude that

Z

has no maximum value.

Omitting explicit non-negativity constraints $x \ge 0$ and $y \ge 0$ , leading to extending the feasible region into Quadrants $II$ , $III$ , or $IV$ .

Correct Approach:Always draw axis boundary lines

x = 0

and

y = 0

first, and shade strictly within Quadrant

I

unless specified otherwise.

Misidentifying the correct shaded region for inequalities like $ax + by \ge c$ vs $ax + by \le c$ , leading to incorrect corner point coordinates.

Correct Approach:Substitute

(0,0)

into

ax + by \le c

. If true, shade towards the origin; if false, shade away from the origin.

Past Year Question (PYQ) Trends

2024Easy

Questions Asked: 1

Topics: Bounded Feasible Regions, Corner Point Method

2023Easy-Medium

Questions Asked: 1

Topics: Optimization of Objective Function, Multiple Constraints

2022Medium

Questions Asked: 1

Topics: Unbounded Region Iso-Cost Method, Non-Feasible Solutions

Solved High-Yield Sample Questions

Practice Mode

Question #1JEE Main 2024

Maximize the objective function $Z = 3x + 4y$ subject to constraints $x + y \le 4$ , $x \ge 0$ , $y \ge 0$ . What is the maximum value of $Z$ ?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Identify the corner points of the bounded feasible region defined by $x + y \le 4$ , $x \ge 0$ , and $y \ge 0$ , which are $A(0,0)$ , $B(4,0)$ , and $C(0,4)$ .

Step 2

Evaluate the objective function $Z = 3x + 4y$ at each vertex: $Z(0,0) = 0$ , $Z(4,0) = 3(4) + 4(0) = 12$ , and $Z(0,4) = 3(0) + 4(4) = 16$ .

Step 3

Comparing these values, the maximum value of $Z$ is $16$ , occurring at the vertex $C(0,4)$ .

Select an option & click Check Answer to reveal solution

Question #2JEE Main 2023

Minimize $Z = 200x + 500y$ subject to $x + 2y \ge 10$ , $3x + 4y \le 24$ , $x \ge 0$ , $y \ge 0$ . The minimum value of $Z$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Find the vertices of the bounded feasible region by solving boundary lines $x + 2y = 10$ and $3x + 4y = 24$ , giving intersection $C(4,3)$ . Vertices on the $y$ -axis are $A(0,5)$ and $B(0,6)$ .

Step 2

Evaluate $Z = 200x + 500y$ at all vertices: $Z(0,5) = 2500$ , $Z(0,6) = 3000$ , and $Z(4,3) = 200(4) + 500(3) = 2300$ .

Step 3

Comparing all vertex values, the minimum value of $Z$ is $2300$ at point $(4,3)$ .

Select an option & click Check Answer to reveal solution

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Frequently Asked Questions

Can an objective function

Z = ax + by

achieve its maximum value at infinitely many points?

Yes. If two adjacent corner points yield the exact same maximum value $Z_{max}$ , then every single point lying on the line segment connecting these two vertices also gives the same maximum value $Z_{max}$ .

What happens if the feasible region is unbounded and

ax + by > M

intersects the region?

If the open half-plane $ax + by > M$ shares any point with the unbounded feasible region, then $M$ is not the maximum value, and the objective function $Z$ has no maximum value in that region.

How do I quickly determine which side of a boundary line

ax + by \le c

to shade under exam time pressure?

Substitute the test point $(0,0)$ into $ax + by \le c$ . If $a(0) + b(0) \le c$ holds true, shade the half-plane containing $(0,0)$ . If false, shade the opposite half-plane. If the line passes through origin, test $(1,0)$ instead.

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