Chapter 12 ยท Introduction

Chapter 12 โ€” Linear Programming

A workshop has only so many hours, a farmer only so much money, a dietitian only so many ingredients. Linear programming turns every such limit into a straight line, finds the region of all possible plans, and picks the best plan from a handful of corner points.

A small furniture workshop in Jaipur makes tables and chairs. A table earns more than a chair, so why not make only tables? Because every table needs three hours at the carpentry bench, and the workshop has just thirty bench-hours a week; every piece also needs time at the polishing station, which is open only twenty hours. Make too many tables and the carpenters run out of time; make too many chairs and the polishers do. Somewhere among all the possible weekly plans is the one that earns the most โ€” and finding it by trial and error, plan after plan, is slow and never tells you that you have really found the best.

In this chapter you learn a method that settles such questions completely. Each limit โ€” โ€œat most 30 hoursโ€, โ€œat least 8 kg of nitrogenโ€, โ€œno negative chairsโ€ โ€” becomes a linear inequality in two variables, whose graph is a half-plane. All the limits together cut out a feasible region: the set of every plan that is actually possible. The quantity to be made as large or as small as possible โ€” profit, cost, time โ€” is a linear objective function $Z=ax+by$, and the lines $Z=k$ are parallel. Slide such a line across the region and you will see that the best value is always reached at a corner point. So instead of infinitely many plans you test only a few corners. The same idea, scaled up to thousands of variables, schedules airline crews, blends petrol in refineries and plans power-grid supply every day.

Linear inequalitiesโ†’Half-planesโ†’Formulating an LPPโ†’Feasible regionโ†’Corner pointsโ†’Optimal solutionโ†’Unbounded & empty regions
๐ŸŽฏ By the end of this chapter you canโ€ฆ
  • draw the half-plane of a linear inequality $ax+by\le c$ (or $\lt,\ \ge,\ \gt$) using the boundary line and a test point;
  • translate a word problem into a linear programming problem: name the decision variables, write the objective function, the constraints and the non-negative constraints;
  • shade the feasible region of a system of linear inequalities and find all its corner points by intersecting the boundary lines;
  • solve a linear programming problem by the corner point method, and explain the result with the sliding line $Z=k$;
  • decide whether an unbounded feasible region has a maximum or minimum by testing an open half-plane, and recognise problems with no feasible solution or with many optimal solutions;
  • answer the Board's 1-mark, 3-mark, 5-mark and case-study questions on linear programming.
Teaser โ€” Slide the Profit Line

The blue region holds every weekly plan the workshop can carry out ($x$ tables, $y$ chairs; illustrative numbers: $3x+y\le30$ bench-hours, $x+2y\le20$ polishing hours). Profit is $Z=600x+300y$ rupees (illustrative). The red line joins all plans with the same profit $Z=k$. Press Play or drag $k$: as the profit rises, the line slides away from the origin, always parallel to itself. Watch the green part โ€” the plans on the line that are possible โ€” shrink to a single corner just before the line leaves the region.

profit lineโ€”
possible plans on itโ€”
best cornerโ€”

Sections in this chapter

โœ… Before you start
You need three things from earlier classes. (1) Solving a pair of linear equations in two variables, by substitution or elimination (Class 10 Mathematics, Ch. 3 โ€” Pair of Linear Equations in Two Variables) โ€” every corner point of a feasible region is found this way. (2) Drawing a straight line from its intercepts and reading its slope (Class 11 Mathematics, Ch. 9 โ€” Straight Lines). (3) The rules for handling inequalities โ€” adding the same number to both sides, and reversing the sign when you multiply or divide by a negative number (Class 11 Mathematics, Ch. 5 โ€” Linear Inequalities). The graphs of inequalities in two variables are not assumed: Section 12.1 builds them from scratch. The idea of an optimisation problem from Chapter 6 returns here, but without derivatives โ€” the objective function is linear, so calculus cannot find its largest value, and geometry does instead. Quick self-test: solve $x+2y=20,\ 3x+y=30$, and find where $2x+3y=12$ meets the axes. (Answers: $x=8,\ y=6$; at $(6,0)$ and $(0,4)$.)