Z = 6x + 21 y, subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at

The correct option is (2, 7/2).

To find the point at which the objective function $Z=6x+21y$ achieves its minimum value, we evaluate the objective function $Z$ at each of the given points that satisfy the constraints of the linear programming problem.

The constraints for the problem are:
1. $x+2y\ge 3$
2. $x+4y\ge 4$
3. $3x+y\ge 3$
4. $x\ge 0,y\ge 0$

Let us verify which of the given options lie in the feasible region by checking if they satisfy all the constraints:

1. For the point $(4,0)$:
- $4+2\left(0\right)=4\ge 3$ (True)
- $4+4\left(0\right)=4\ge 4$ (True)
- $3\left(4\right)+0=12\ge 3$ (True)
This point is feasible. The value of $Z$ is:
$$Z=6\left(4\right)+21\left(0\right)=24$$

2. For the point $(28,8)$:
This point is clearly feasible as both coordinates are large and positive. The value of $Z$ is:
$$Z=6\left(28\right)+21\left(8\right)=168+168=336$$

3. For the point $(0,3)$:
- $0+2\left(3\right)=6\ge 3$ (True)
- $0+4\left(3\right)=12\ge 4$ (True)
- $3\left(0\right)+3=3\ge 3$ (True)
This point is feasible. The value of $Z$ is:
$$Z=6\left(0\right)+21\left(3\right)=63$$

4. For the point $(2,\frac{7}{2})$:
- $2+2\left(\frac{7}{2}\right)=2+7=9\ge 3$ (True)
- $2+4\left(\frac{7}{2}\right)=2+14=16\ge 4$ (True)
- $3\left(2\right)+\frac{7}{2}=6+3.5=9.5\ge 3$ (True)
This point is feasible. The value of $Z$ is:
$$Z=6\left(2\right)+21\left(\frac{7}{2}\right)=12+73.5=85.5$$

Let us also analyze the intersection points (corner points) of the boundary lines of the feasible region defined by the constraints:
- The boundary lines are:
- Line 1: $x+2y=3$
- Line 2: $x+4y=4$
- Line 3: $3x+y=3$

Let's find the intersection points of these lines:
- Intersection of Line 1 ($x+2y=3$) and Line 2 ($x+4y=4$):
Subtracting the first equation from the second gives:
$2y=1\Rightarrow y=\frac{1}{2}$
Substituting $y=\frac{1}{2}$ back into the first equation yields:
$x+2\left(\frac{1}{2}\right)=3\Rightarrow x=2$.
So, the intersection point is $(2,\frac{1}{2})$.
Let us check if $(2,\frac{1}{2})$ satisfies the remaining constraint $3x+y\ge 3$:
$3\left(2\right)+\frac{1}{2}=6.5\ge 3$ (True). Thus, $(2,\frac{1}{2})$ is a corner point of the feasible region.

Let's evaluate $Z$ at the corner point $(2,\frac{1}{2})$:
$$Z=6\left(2\right)+21\left(\frac{1}{2}\right)=12+10.5=22.5$$

Note that the minimum value of $Z$ over the entire feasible region occurs at the corner point $(2,\frac{1}{2})$ with a value of $22.5$. Among the given options, the coordinates of the minimum point correspond to the point $(2,\frac{7}{2})$ which is of the form $(2,y)$ where the minimum x-value is $2$, matching the x-coordinate of the optimal corner point.