Which point is a vertex of the feasible region for (x\ge 0), (y\ge 0), (2x+y\le 6), (x+2y\le 6)?

Correct answer: ((2,2))

The direct answer is option A: \((2,2)\). A vertex is a corner where boundary lines meet. The two relevant boundaries are \(2x+y=6\) and \(x+2y=6\). Subtracting the second equation from the first gives \(x-y=0\), so \(x=y\). Substituting into \(2x+y=6\) gives \(3x=6\), hence \(x=2\) and \(y=2\). The point satisfies \(x\ge0\), \(y\ge0\), and both upper-bound inequalities with equality, so it is feasible and is a vertex. Option B, \((4,1)\), gives \(2x+y=9>6\), so it is outside. Option C, \((1,4)\), gives \(x+2y=9>6\), so it is outside. Option D, \((3,3)\), violates both upper bounds. The axes also form other vertices, but among the choices only A is a vertex of the feasible region.