Step 1: Understanding the Concept:
We need to find the region in the first quadrant that satisfies all given linear inequalities.
Let's rewrite the inequalities in a standard format \( y \le mx + c \) or \( y \ge mx + c \) to easily identify the region relative to the boundary lines. Step 2: Key Formula or Approach:
1. \( x - 2 \le y \implies y \ge x - 2 \) (Region above the line \( y = x - 2 \))
2. \( x \ge y - 1 \implies y \le x + 1 \) (Region below the line \( y = x + 1 \))
3. \( x \ge 2 \) (Region to the right of vertical line \( x = 2 \))
4. \( y \le 4 \) (Region below horizontal line \( y = 4 \))
5. \( x, y \ge 0 \) (First quadrant) Step 3: Detailed Explanation:
Let's find the vertices of this bounded region by finding intersection points of the boundary lines.
- Intersection of left boundary \( x=2 \) and bottom boundary \( y=x-2 \):
\( y = 2 - 2 = 0 \). Point: \( (2, 0) \).
- Intersection of left boundary \( x=2 \) and top boundary \( y=x+1 \):
\( y = 2 + 1 = 3 \). Point: \( (2, 3) \).
- Intersection of top boundary \( y=4 \) and \( y=x+1 \):
\( 4 = x + 1 \implies x = 3 \). Point: \( (3, 4) \).
- Intersection of top boundary \( y=4 \) and bottom boundary \( y=x-2 \):
\( 4 = x - 2 \implies x = 6 \). Point: \( (6, 4) \).
The feasible region is a polygon with vertices \( (2, 0), (6, 4), (3, 4), (2, 3) \).
Let's analyze the given graphs:
- Graph A: The region is bounded by \( y \)-axis. It does not satisfy \( x \ge 2 \).
- Graph B: The region has a vertical left edge starting at \( x=2 \). It has a horizontal top edge at \( y=4 \). The slanted edges correspond to \( y=x+1 \) and \( y=x-2 \). This perfectly matches our analysis.
- Graph C: The region includes areas where \( x<2 \).
- Graph D: Displays two separate regions, which is incorrect for this set of inequalities. Step 4: Final Answer:
Graph (B) correctly represents the feasible region.
