Consider the constraint:
The solution space of this constraint is one of the closed half-planes defined by the equation: . To show this, let us consider a point which satsifies Equation 10 as equality, and another point for which Equation 10 is also valid. For any such pair of points, it holds that:
Interpreting the left side of Eq. 11 as the inner (dot) product of the two vectors and , and recognizing that , it follows that line , itself, can be defined by point and the set of points such that vector is at right angles with vector . Furthermore, the set of points that satisfy the > (<) part of Equation 11 have the vector forming an acute (obtuse) angle with vector , and therefore, they are ``above'' (``below'') the line. Hence, the set of points satisfying each of the two inequalities implied by Equation 10 is given by one of the two half-planes the boundary of which is defined by the corresponding equality constraint. Figure 1 summarizes the above discussion.
Figure 1: Half-planes: the feasible region of a linear inequality
An easy way to determine the half-plane depicting the solution space of a linear inequality, is to draw the line depicting the solution space of the corresponding equality constraint, and then test whether the point (0,0) satisfies the inequality. In case of a positive answer, the solution space is the half-space containing the origin, otherwise, it is the other one.
From the above discussion, it follows that the feasible region for the prototype LP of Equation 5 is the shaded area in the following figure:
Figure 2: The feasible region of the prototype example LP