It may appear logical to assume that the optimum solution will require the first (second)
set of inequalities to be replaced with equations if .The counterexample
in Table 5.30 shows that this assumption is not correct.
Show that the application of the suggested procedure yields the solution = 2,
= 3, = 4, and = 2, with z = $27, which is worse than the feasible solution
= 2, = 7, and = 6, with z = $15