The maximum value of $z=2 x+3 y$ subject to constraints $3 x-3 y ≥ 0,2 x+2 y ≤ 12, x ≥ 0, y ≥ 0$ occurs at the point

$(0,0)$ $(6,0)$ $(0,6)$ $(3,3)$

$(3,3)$

The correct answer is Option (4) - $(3,3)$