Maximize $Z=4x+6y$ subject to $3x+2y\le 12$, $x+y\ge 2$, $x,y\ge 0$.

Constraints:$3x+2y\le 12$ Points: $(4,0)$ and $(0,6)$.$x+y\ge 2$ Points: $(2,0)$ and $(0,2)$.Corner points of the feasible region (intersection of half-planes):Point 1: $(0,2)$ Point 2: $(0,6)$ Point 3: $(4,0)$ Point 4: $(2,0)$ Evaluate $Z=4x+6y$:$Z(0,2)=4(0)+6(2)=12$$Z(0,6)=4(0)+6(6)=36$$Z(4,0)=4(4)+6(0)=16$$Z(2,0)=4(2)+6(0)=8$ Maximum value is $36$ at $(0,6)$.