Class XII
Linear programming: Tutorial 1
Solving linear inequalities
Introduction: These tutorials in mathematics are aimed at facilitating learning and
understanding of the class 12 syllabus even when the school is not working.
We start with chapter 12 on ‘Linear Programming’.
This short tutorial is a recall of the topic ‘Linear Inequalities’ which you have already done
in class 11.
I shall explain the graphical solution of linear inequalities with the help of examples.
Example 1: Solve graphically the linear inequality 𝒙 + 𝒚 ≤ 𝟒, 𝒙 ≥ 𝟎 & 𝒚 ≥ 𝟎
Solution: Since the values of x and y are non-negative, the graph lies in the first quadrant.
Step 1: Plot the graph of the linear equation x + y = 4.
A(0,4)
Feasible or required region
O (0,0) B(4,0)
It is a straight line passing through the points (0,4) and (4,0).
Step 2: the line x + y = 4 divides the cartesian plane into two half planes.
Step 3: Choose any one point in any one half plane say (0,0).
Step4: Substitute (0,0) in the given inequality 𝟎 + 𝟎 ≤ 𝟒 𝒊𝒔 𝒕𝒓𝒖𝒆
Thus, the region or half plane containing the point (0,0) is the required feasible solution of
the given inequality.
Note that:
• If the inequality is strict inequality, that is, there is no equality sign, then the
line is not a part of the feasible region
