Question

Solve the following inequations graphically and write the corner points of the feasible region: \( 2x + 3y \leq 6 \), \( x + y \geq 2 \), \( x \geq 0 \), \( y \geq 0 \).

Hint:

Graph inequalities by finding intercepts and intersections; feasible region is where all conditions overlap.

Answer 1

Plot the lines:
1. \( 2x + 3y = 6 \): Intercepts (3, 0), (0, 2). Shade below (\( \leq \)).
2. \( x + y = 2 \): Intercepts (2, 0), (0, 2). Shade above (\( \geq \)).
3. \( x \geq 0 \), \( y \geq 0 \): First quadrant.
Intersection points:
- \( 2x + 3y = 6 \) and \( x + y = 2 \): Solve: \[ 2x + 3y = 6, \quad x + y = 2 \Rightarrow y = 2 - x. \] \[ 2x + 3(2 - x) = 6 \Rightarrow 2x + 6 - 3x = 6 \Rightarrow -x = 0 \Rightarrow x = 0, \quad y = 2. \] Point: (0, 2).
- \( 2x + 3y = 6 \) and \( y = 0 \): \( 2x = 6 \Rightarrow x = 3 \). Point: (3, 0).
- \( x + y = 2 \) and \( x = 0 \): \( y = 2 \). Point: (0, 2).
- \( x + y = 2 \) and \( y = 0 \): \( x = 2 \). Point: (2, 0).
Feasible region corner points: (0, 2), (2, 0), (3, 0).