Which point is not feasible for x+y≤6, x-y≤2, x≥0, and y≥0?

Correct answer: (5,0)

A feasible point must satisfy all four inequalities simultaneously. Test option A, (5,0): x+y=5≤6 is true, x-y=5-0=5≤2 is false, and although non-negativity holds, one failed condition is enough to make the point infeasible. Thus option A is correct. For (2,2), the values are 4≤6, 0≤2, and both coordinates are non-negative, so it is feasible. For (1,4), the values are 5≤6, -3≤2, with x and y non-negative, so it is feasible. For (3,2), the values are 5≤6, 1≤2, and both coordinates are non-negative, so it is also feasible. The key concept is intersection: the feasible region is the common part of all four half-planes, not the region satisfying only the first inequality.