Consider a LPP given by

Minimum Z = 6x + 10y


Subjected to x ≥ 6; y ≥ 2; 2x + y ≥ 10; x, y ≥ 0


Redundant constraints in this LPP are


Given


Objective Function is Z = 6x + 10y


Constraints are:


x ≥ 6


y ≥ 2


2x + y ≥ 10


x, y ≥ 0


A redundant constraint is that, which doesn’t intersect with the feasible region of the out non-redundant constraints.


Here, the problem is a minimization problem and as per the constraints x≥ 0 and y ≥ 0 the feasible solution is located in the 1st quadrant.


Now, if we map all the three inequalities in a graph, we have



From the graph, it is very clear that, the graph of the inequality 2x + y ≥ 10 is not intersecting the feasible region formed by the constraints x ≥ 6; y ≥ 2; x≥ 0 and y ≥ 0.


Hence the inequality 2x + y ≥ 10 is not really making any difference to the feasible region from by x ≥ 6; y ≥ 2; x≥ 0 and y ≥ 0.


Therefore inequality 2x + y ≥ 10 remains redundant.


The answer of the question is option C.

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