Maximize Z = 11 x + 8y subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0.

The correct option is 60 at (4, 2).

To find the maximum value of the objective function, we will determine the feasible region defined by the system of linear inequalities and then evaluate the objective function at each of its corner points (vertices).

Step 1: Identify the constraints and boundary lines
The given constraints are:
1) $x\le 4$
2) $y\le 6$
3) $x+y\le 6$
4) $x\ge 0$, $y\ge 0$

The non-negativity constraints ($x\ge 0$, $y\ge 0$) restrict the feasible region to the first quadrant.

Step 2: Find the corner points (vertices) of the feasible region
The boundary lines are:
- Line 1: $x=4$
- Line 2: $y=6$
- Line 3: $x+y=6$

Let's determine the intersection points of these boundary lines that satisfy all the inequalities:
- The origin $(0,0)$ is a corner point.
- The y-intercept of the line $x+y=6$ (where $x=0$) is $(0,6)$. This satisfies $y\le 6$.
- The x-intercept of the line $x+y=6$ (where $y=0$) would be $(6,0)$, but since $x\le 4$, the boundary is cut off at the line $x=4$. Thus, we look at the intersection of $x=4$ with the x-axis ($y=0$), which gives the point $(4,0)$.
- The intersection of the lines $x=4$ and $x+y=6$ is found by substituting $x=4$ into the equation:
$4+y=6⟹y=2$.
This yields the corner point $(4,2)$.

Therefore, the corner points of the feasible region are: $(0,0)$, $(4,0)$, $(4,2)$, and $(0,6)$.

Step 3: Evaluate the objective function at each corner point
We substitute the coordinates of each corner point into the objective function:
$Z=11x+8y$

1. At $(0,0)$:
$Z=11\left(0\right)+8\left(0\right)=0$

2. At $(4,0)$:
$Z=11\left(4\right)+8\left(0\right)=44$

3. At $(4,2)$:
$Z=11\left(4\right)+8\left(2\right)=44+16=60$

4. At $(0,6)$:
$Z=11\left(0\right)+8\left(6\right)=48$

Comparing the values, the maximum value of $Z$ is 60, which occurs at the coordinate point $(4,2)$.