Step 1: Understanding the Concept:
A feasible region is the area on a graph that satisfies all given linear inequalities simultaneously. We can identify the correct region by testing each inequality against the geometric location of the labeled regions relative to the boundary lines.
Step 2: Key Formula or Approach:
For a line $ax+by=c$:
- If $y$-coefficient is positive, $ax+by \ge c$ represents the region above the line, and $\le$ represents the region below.
- For a vertical line $x=k$, $x \le k$ is the left side.
- For a horizontal line $y=k$, $y \ge k$ is the upper side.
Test a sample point in a region to confirm.
Step 3: Detailed Explanation:
Let's analyze the constraints one by one:
1. $x \le 4$: The region must be to the left of the vertical line $x = 4$. Looking at the graph, regions $\text{S}_1, \text{S}_2, \text{S}_4$ are to the left, while $\text{S}_3$ is to the right. This eliminates $\text{S}_3$.
2. $y \ge 3$: The region must be above the horizontal line $y = 3$. Regions $\text{S}_1$ and $\text{S}_3$ are above this line, while $\text{S}_2$ and $\text{S}_4$ are below it. This eliminates $\text{S}_2$ and $\text{S}_4$.
At this point, only $\text{S}_1$ satisfies both simple constraints. Let's verify with the remaining inequalities to be certain.
3. $2x + 3y \ge 12$: This requires the region to be "above" the line $2x+3y=12$ (since the coefficient of $y$ is positive). $\text{S}_1$ is clearly situated above this slanted line.
4. $-x + y \le 3$: Rearranging gives $y \le x + 3$. This requires the region to be "below" the line $-x+y=3$. $\text{S}_1$ is bounded below this line.
Since region $\text{S}_1$ satisfies all four conditions simultaneously, it is the feasible region.
Alternatively, pick a test point clearly inside $\text{S}_1$, roughly at $x=3, y=4$, and check it against all constraints:
- $2(3) + 3(4) = 6 + 12 = 18 \ge 12$ (True)
- $-(3) + 4 = 1 \le 3$ (True)
- $3 \le 4$ (True)
- $4 \ge 3$ (True)
Step 4: Final Answer:
The feasible region is $\text{S}_1$.