Maximize Z = 4x + 6y, subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0

The correct option is 36 at (0, 6).

Let us analyze the given linear programming problem and explain step-by-step why this answer is correct.
We are required to maximize the objective function:
$$Z=4x+6y$$
subject to the following constraints:
1) $3x+2y\le 12$
2) $x+y\ge 4$
3) $x\ge 0,y\ge 0$

Step 1: Identify the boundary lines and find their intercepts.
For the first constraint line, $3x+2y=12$:
If $x=0$, then $2y=12\Rightarrow y=6$. This gives the point $(0,6)$.
If $y=0$, then $3x=12\Rightarrow x=4$. This gives the point $(4,0)$.

For the second constraint line, $x+y=4$:
If $x=0$, then $y=4$. This gives the point $(0,4)$.
If $y=0$, then $x=4$. This gives the point $(4,0)$.

Step 2: Determine the feasible region.
Since $x,y\ge 0$, the feasible region lies entirely within the first quadrant.
- The inequality $3x+2y\le 12$ represents the region containing the origin $(0,0)$ because $3(0)+2(0)=0\le 12$ is true.
- The inequality $x+y\ge 4$ represents the region away from the origin because $0+0=0\ge 4$ is false.
Combining these, the bounded feasible region has the corner points:
- $A(4,0)$
- $B(0,4)$
- $C(0,6)$

Step 3: Evaluate the objective function Z at each corner point.
Let us calculate the value of $Z=4x+6y$ at each vertex:
1) At point $A(4,0)$:
$$Z=4(4)+6(0)=16$$
2) At point $B(0,4)$:
$$Z=4(0)+6(4)=24$$
3) At point $C(0,6)$:
$$Z=4(0)+6(6)=36$$

Conclusion:
Comparing the values, the maximum value of $Z$ is $36$, which occurs at the corner point $(0,6)$.
Thus, the correct answer is indeed 36 at (0, 6).