Maximize Z = 6x + 4y, subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, x ≥ 0, y ≥ 0

The correct option is 16 at (2, 1).

To find the maximum value of the objective function $Z=6x+4y$ subject to the given constraints, we can use the graphical method of linear programming. First, we identify the feasible region determined by the following inequalities:
1. $x\le 2$
2. $x+y\ge 0$ (implied by non-negativity) and $x+y\le 3$
3. $-2x+y\le 1$
4. $x\ge 0$, $y\ge 0$

Let us find the boundary lines and their intersection points to determine the corner points (vertices) of the feasible region:
- The line $x=0$ is the y-axis, and $y=0$ is the x-axis.
- The line $x=2$ is a vertical line passing through $(2,0)$.
- The line $x+y=3$ intersects the axes at $(3,0)$ and $(0,3)$.
- The line $-2x+y=1$ intersects the axes at $(-0.5,0)$ and $(0,1)$.

Next, we solve for the intersection points of these boundary lines that lie in the first quadrant ($x\ge 0$, $y\ge 0$):
- Intersection of $x=0$ and $-2x+y=1$ gives the corner point $A(0,1)$.
- Intersection of $-2x+y=1$ and $x+y=3$:
Subtracting the first equation from the second equation:
$(x+y)-(-2x+y)=3-1$
$3x=2\Rightarrow x=\frac{2}{3}$
Substituting $x=\frac{2}{3}$ into $x+y=3$ gives $y=3-\frac{2}{3}=\frac{7}{3}$. This gives the corner point $B(\frac{2}{3},\frac{7}{3})$.
- Intersection of $x+y=3$ and $x=2$:
Substituting $x=2$ into the equation gives $2+y=3\Rightarrow y=1$. This gives the corner point $C(2,1)$.
- Intersection of $x=2$ and the x-axis ($y=0$) gives the corner point $D(2,0)$.
- The origin $O(0,0)$ is also a corner point.

The bounded feasible region is defined by the vertices $O(0,0)$, $A(0,1)$, $B(\frac{2}{3},\frac{7}{3})$, $C(2,1)$, and $D(2,0)$.

Now, we evaluate the objective function $Z=6x+4y$ at each of these corner points:
1. At $O(0,0)$: $Z=6\left(0\right)+4\left(0\right)=0$
2. At $A(0,1)$: $Z=6\left(0\right)+4\left(1\right)=4$
3. At $B(\frac{2}{3},\frac{7}{3})$: $Z=6\left(\frac{2}{3}\right)+4\left(\frac{7}{3}\right)=4+\frac{28}{3}=\frac{40}{3}\approx 13.33$
4. At $C(2,1)$: $Z=6\left(2\right)+4\left(1\right)=12+4=16$
5. At $D(2,0)$: $Z=6\left(2\right)+4\left(0\right)=12$

Comparing these values, the maximum value of $Z$ is $16$, which occurs at the point $(2,1)$.