Skip to content
🚨 This site is a work in progress. Exciting updates are coming soon!

5.2 Q-1

Question Statement

Graph the feasible region of the following systems of linear inequalities and find the corner points.

System 1:

  1. 2xβˆ’3y≀62x - 3y \leq 6
  2. 2x+3y≀122x + 3y \leq 12
  3. xβ‰₯0x \geq 0
  4. yβ‰₯0y \geq 0

Background and Explanation

To graph a system of linear inequalities, we:

  • Convert each inequality to an equation (boundary line).
  • Plot the boundary lines on the graph.
  • Test points to determine which side of each line is part of the feasible region.
  • The feasible region is the area that satisfies all inequalities, and the corner points are the intersection points of the boundary lines.

Solution

Step 1: Graph the boundary lines

For the first inequality 2xβˆ’3y≀62x - 3y \leq 6:

  • Convert to the equation 2xβˆ’3y=62x - 3y = 6.
  • Find the x-intercept (set y=0y = 0):
    2x=6β‡’x=32x = 6 \Rightarrow x = 3. Point: (3,0)(3, 0).
  • Find the y-intercept (set x=0x = 0):
    βˆ’3y=6β‡’y=βˆ’2-3y = 6 \Rightarrow y = -2. Point: (0,βˆ’2)(0, -2).

However, since yβ‰₯0y \geq 0, we disregard the point (0,βˆ’2)(0, -2), as it lies outside the feasible region.

For the second inequality 2x+3y≀122x + 3y \leq 12:

  • Convert to the equation 2x+3y=122x + 3y = 12.
  • Find the x-intercept (set y=0y = 0):
    2x=12β‡’x=62x = 12 \Rightarrow x = 6. Point: (6,0)(6, 0).
  • Find the y-intercept (set x=0x = 0):
    3y=12β‡’y=43y = 12 \Rightarrow y = 4. Point: (0,4)(0, 4).

Step 2: Check the feasibility of the points

  • (0,0)(0, 0) is valid for all inequalities.
  • (3,0)(3, 0), (6,0)(6, 0), and (0,4)(0, 4) are valid as they satisfy all the inequalities.

Step 3: Find the intersection points (corner points)

  • Solve 2xβˆ’3y=62x - 3y = 6 and 2x+3y=122x + 3y = 12 simultaneously to find the intersection point:
    • Adding both equations:
      4x=18β‡’x=924x = 18 \Rightarrow x = \frac{9}{2}.
    • Substitute x=92x = \frac{9}{2} into 2xβˆ’3y=62x - 3y = 6:
      2Γ—92βˆ’3y=6β‡’9βˆ’3y=6β‡’y=12 \times \frac{9}{2} - 3y = 6 \Rightarrow 9 - 3y = 6 \Rightarrow y = 1.

Thus, the intersection point is (92,1)\left(\frac{9}{2}, 1\right).

Step 4: Identify the corner points

The corner points are (0,0)(0, 0), (3,0)(3, 0), (6,0)(6, 0), and (92,1)\left(\frac{9}{2}, 1\right).


Key Formulas or Methods Used

  • The method of graphing inequalities involves:
    1. Converting inequalities to equations.
    2. Plotting boundary lines.
    3. Checking which side of the line satisfies the inequality.
    4. Finding intersection points of boundary lines to determine corner points.

Summary of Steps

  1. Convert inequalities to equations.
  2. Graph the boundary lines using intercepts.
  3. Check feasibility of points (ensuring they satisfy all inequalities).
  4. Solve for intersection points of boundary lines to find corner points.