Concept guide
Linear programming: the best supplied candidate, and why it is not the continuous optimum
Constraints 3x+2y≤12, x+2y≤6, −x≤0, −y≤0; objective: maximize L(x,y)=x+y. For the seven supplied candidates P1=(0,0), P2=(4,0), P3=(3,1), P4=(2,2), P5=(0,3), P6=(4,1), P7=(3,2), check feasibility constraint by constraint, compare objective values only among feasible candidates, and find the best value among the candidates and all points attaining it; then use the feasible point Q=(3,3/2) and an inequality certificate to prove that the maximum over the continuous feasible region is 9/2, not the candidate value 4.