Question

Solve the following inequalities graphically and write the corner points of the feasible region: \[ 2x + 3y \leq 6, \quad x + y \geq 2, \quad x \geq 0, \quad y \geq 0 \]

Hint:

To solve a system of inequalities graphically, first graph the boundary lines, then shade the feasible region where all inequalities are satisfied.

Answer 1

We need to solve the system of inequalities graphically. First, graph each inequality and find the feasible region. Step 1: Graph the inequality \( 2x + 3y \leq 6 \). The boundary line is \( 2x + 3y = 6 \), which can be rewritten as \( y = -\frac{2}{3}x + 2 \). Plot this line and shade the region below it. Step 2: Graph the inequality \( x + y \geq 2 \). The boundary line is \( x + y = 2 \), which can be rewritten as \( y = 2 - x \). Plot this line and shade the region above it. Step 3: Graph the inequalities \( x \geq 0 \) and \( y \geq 0 \), which represent the first quadrant. Step 4: Find the corner points of the feasible region by solving the system of equations formed by the boundary lines. The intersection points are: - Intersection of \( 2x + 3y = 6 \) and \( x + y = 2 \): \[ x = 0, y = 2 \quad \text{(Point A: (0, 2))} \] - Intersection of \( 2x + 3y = 6 \) and \( x = 0 \): \[ x = 0, y = 2 \quad \text{(Point B: (0, 2))} \] - Intersection of \( x + y = 2 \) and \( y = 0 \): \[ x = 2, y = 0 \quad \text{(Point C: (2, 0))} \] The corner points of the feasible region are \( (0, 2) \), \( (2, 0) \).