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ALGEBRA HIGHER MATHEMATICS

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Chapter 12
LINEAR PROGRAMMING
12.1 Overview
12.1.1 An Optimisation Problem A problem which seeks to maximise or minimise a
function is called an optimisation problem. An optimisation problem may
involve maximisation of profit, production etc or minimisation of cost, from available
resources etc.

12.1.2 A Linnear Programming Problem (LPP)
A linear programming problem deals with the optimisation (maximisation/
minimisation) of a linear function of two variables (say x and y) known as objective
function subject to the conditions that the variables are non-negative and satisfy a set
of linear inequalities (called linear constraints). A linear programming problem is a
special type of optimisation problem.
12.1.3 Objective Function Linear function Z = ax + by, where a and b are constants,
which has to be maximised or minimised is called a linear objective function.
12.1.4 Decision Variables In the objective function Z = ax + by, x and y are called
decision variables.
12.1.5 Constraints The linear inequalities or restrictions on the variables of an LPP
are called constraints. The conditions x ≥ 0, y ≥ 0 are called non-negative constraints.
12.1.6 Feasible Region The common region determined by all the constraints including
non-negative constraints x ≥ 0, y ≥ 0 of an LPP is called the feasible region for the
problem.
12.1.7 Feasible Solutions Points within and on the boundary of the feasible region
for an LPP represent feasible solutions.
12.1.8 Infeasible Solutions Any Point outside feasible region is called an infeasible
solution.
12.1.9 Optimal (feasible) Solution Any point in the feasible region that gives the
optimal value (maximum or minimum) of the objective function is called an optimal
solution.
Following theorems are fundamental in solving LPPs.




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12.1.10 Theorem 1 Let R be the feasible region (convex polygon) for an LPP and let
Z = ax + by be the objective function. When Z has an optimal value (maximum or
minimum), where x and y are subject to constraints described by linear inequalities,
this optimal value must occur at a corner point (vertex) of the feasible region.
Theorem 2 Let R be the feasible region for a LPP and let Z = ax + by be the objective
function. If R is bounded, then the objective function Z has both a maximum and a
minimum value on R and each of these occur at a corner point of R.
If the feasible region R is unbounded, then a maximum or a minimum value
of the objective function may or may not exist. However, if it exits, it must occur at a
corner point of R.
12.1.11 Corner point method for solving a LPP
The method comprises of the following steps :
(1) Find the feasible region of the LPP and determine its corner points (vertices)
either by inspection or by solving the two equations of the lines intersecting at
that point.
(2) Evaluate the objective function Z = ax + by at each corner point.
Let M and m, respectively denote the largest and the smallest values of Z.
(3) (i) When the feasible region is bounded, M and m are, respectively, the
maximum and minimum values of Z.
(ii) In case, the feasible region is unbounded.
(a) M is the maximum value of Z, if the open half plane determined by
ax + by > M has no point in common with the feasible region. Otherwise, Z has
no maximum value.
(b) Similarly, m is the minimum of Z, if the open half plane determined by
ax + by < m has no point in common with the feasible region. Otherwise, Z has
no minimum value.
12.1.12 Multiple optimal points If two corner points of the feasible region are optimal
solutions of the same type, i.e., both produce the same maximum or minimum, then
any point on the line segment joining these two points is also an optimal solution of
the same type.

12.2 Solved Examples
Short Answer (S.A.)
Example 1 Determine the maximum value of Z = 4x + 3y if the feasible region for an
LPP is shown in Fig. 12.1.




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, LINEAR PROGRAMMING 243



Solution The feasible region is bounded. Therefore, maximum of Z must occur at the
corner point of the feasible region (Fig. 12.1).

Corner Point Value of Z
O, (0, 0) 4 (0) + 3 (0) = 0
A (25, 0) 4 (25) + 3 (0) = 100
B (16, 16) 4 (16) + 3 (16) = 112 ← (Maximum)
C (0, 24) 4 (0) + 3 (24) = 72
Hence, the maximum value of Z is 112.




Fig.12.1

Example 2 Determine the minimum value of Z = 3x + 2y (if any), if the feasible
region for an LPP is shown in Fig.12.2.
Solution The feasible region (R) is unbounded. Therefore minimum of Z may or may
not exist. If it exists, it will be at the corner point (Fig.12.2).

Corner Point Value of Z
A, (12, 0) 3 (12) + 2 (0) = 36
B (4, 2) 3 (4) + 2 (2) = 16
C (1, 5) 3 (1) + 2 (5) = 13 ← (smallest)
D (0, 10) 3 (0) + 2 (10) = 20




20/04/2018

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