Q.For the linear programming problem Maximize $Z = 3x + 2y$ subject to the constraints $x + y \le 4,\ x \le 3,\ x, y \ge 0$, determine, giving a reason in each case, whether the points $(1,1)$, $(3,3)$ and $(4,0)$ are feasible solutions.

Check $(1,1)$: $x+y=1+1=2\le4$ ✓; $x=1\le3$ ✓; $x=1\ge0,\ y=1\ge0$ ✓. All three constraints hold, so $(1,1)$ is a feasible solution.

Check $(3,3)$: $x+y=3+3=6$, but the constraint requires $x+y\le4$, and $6>4$ — this constraint is violated, so $(3,3)$ is infeasible (it does not matter that $x=3\le3$ is separately satisfied; every constraint must hold at once).

Check $(4,0)$: $x=4$, but the constraint requires $x\le3$, and $4>3$ — violated, so $(4,0)$ is infeasible (note that $x+y=4\le4$ is satisfied here; this point fails only the second constraint, illustrating that a point can be infeasible by breaking just one of several constraints).

Substitute the coordinates of each point into every constraint of the LPP in turn; a point is a feasible solution only if it satisfies ALL constraints simultaneously (including the non-negativity restrictions), and is infeasible if it violates even one.

Concluding a point is feasible after checking only one constraint and finding it satisfied, without checking the remaining constraints; assuming a point that violates one constraint must violate all of them.