Consider the LPP: Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. The optimal feasible solution occurs at

The correct option is Both (6, 0) and (0, 3).

To understand why this is the correct answer, let's analyze the given Linear Programming Problem (LPP) step-by-step.
We are asked to minimize the objective function:
$$Z=x+2y$$
subject to the following constraints:
1) $2x+y\ge 3$
2) $x+2y\ge 6$
3) $x,y\ge 0$

Let's find the feasible region defined by these inequalities in the first quadrant ($x\ge 0$, $y\ge 0$).
First, let's plot the boundary lines:
Line 1: $2x+y=3$. The intercepts are $(1.5,0)$ and $(0,3)$.
Line 2: $x+2y=6$. The intercepts are $(6,0)$ and $(0,3)$.

Now, let's determine the corner points of the unbounded feasible region:
- Since the inequalities are both of the "$\ge$" type, the feasible region lies above and to the right of both lines in the first quadrant.
- Let's check the intersection of the two boundary lines. Setting them equal or solving simultaneously:
From Line 1, $y=3-2x$. Substituting this into Line 2:
$$x+2(3-2x)=6$$
$$x+6-4x=6$$
$$-3x=0\Rightarrow x=0$$
If $x=0$, then $y=3$.
So, the lines intersect at the point $(0,3)$, which is also a corner point.
The corner points of the feasible region are $(6,0)$ and $(0,3)$.

Let's evaluate the objective function $Z=x+2y$ at these corner points:
1. At the point $(6,0)$:
$$Z=6+2\left(0\right)=6$$
2. At the point $(0,3)$:
$$Z=0+2\left(3\right)=6$$

Since both corner points yield the minimum value of $Z=6$, the LPP has multiple optimal solutions. In fact, any point on the line segment joining the points $(6,0)$ and $(0,3)$ is an optimal solution. Therefore, the optimal feasible solution occurs at both $(6,0)$ and $(0,3)$.