Math Problem Statement

identify corner points: Solve the system​ graphically, and indicate whether the solution region is bounded or unbounded. Find the coordinates of each corner point.

3 x plus y3x+y

greater than or equals≥

1515

x plus 2 yx+2y

greater than or equals≥

10

x

greater than or equals≥

0

y

greater than or equals≥

0

Solution

To solve the given system of inequalities graphically and identify the corner points, we can follow these steps:

  1. Rewrite the inequalities:

    • 3x+y153x + y \geq 15
    • x+2y10x + 2y \geq 10
    • x0x \geq 0
    • y0y \geq 0
  2. Convert inequalities to equations:

    • From 3x+y=153x + y = 15
    • From x+2y=10x + 2y = 10
  3. Find the intersection points:

    • Solve 3x+y=153x + y = 15 and x+2y=10x + 2y = 10 simultaneously.
  4. Graph the lines:

    • Plot the equations 3x+y=153x + y = 15 and x+2y=10x + 2y = 10 in the first quadrant (since x0x \geq 0 and y0y \geq 0).
  5. Identify corner points:

    • The points where the lines intersect, as well as where they intersect the axes, will provide the corner points.

Step-by-Step Calculations:

  1. Finding intersection points:

    • From 3x+y=153x + y = 15: y=153xy = 15 - 3x
    • Substitute into x+2y=10x + 2y = 10: x+2(153x)=10    x+306x=10    5x=20    x=4x + 2(15 - 3x) = 10 \implies x + 30 - 6x = 10 \implies -5x = -20 \implies x = 4 Then substituting x=4x = 4 back to find yy: y=153(4)=1512=3y = 15 - 3(4) = 15 - 12 = 3
    • So one intersection point is (4,3)(4, 3).
  2. Finding intercepts:

    • For 3x+y=153x + y = 15:

      • xx-intercept: y=03x=15x=5y = 0 \Rightarrow 3x = 15 \Rightarrow x = 5 (5,0)\Rightarrow (5, 0)
      • yy-intercept: x=0y=15x = 0 \Rightarrow y = 15 (0,15)\Rightarrow (0, 15)
    • For x+2y=10x + 2y = 10:

      • xx-intercept: y=0x=10y = 0 \Rightarrow x = 10 (10,0)\Rightarrow (10, 0)
      • yy-intercept: x=02y=10y=5x = 0 \Rightarrow 2y = 10 \Rightarrow y = 5 (0,5)\Rightarrow (0, 5)
  3. Identifying the feasible region:

    • We plot the points: (4,3)(4, 3), (5,0)(5, 0), (0,15)(0, 15), (10,0)(10, 0), and (0,5)(0, 5) on a graph.
    • The region satisfying all inequalities will be bounded by these lines and the axes.

Summary of Corner Points:

The corner points of the feasible region are:

  1. (5,0)(5, 0)
  2. (0,5)(0, 5)
  3. (4,3)(4, 3)
  4. (0,15)(0, 15)
  5. (10,0)(10, 0) (but does not satisfy the 3x+y153x + y \geq 15 constraint)

Conclusion:

  • The feasible region is bounded.
  • The valid corner points are:
    • (5,0)(5, 0)
    • (0,5)(0, 5)
    • (4,3)(4, 3)

If you have any further questions or need more details, let me know! Here are some follow-up questions you might consider:

  1. How do I graph these inequalities on a coordinate plane?
  2. What does it mean for a region to be bounded or unbounded?
  3. Can you provide examples of other inequalities and their graphical solutions?
  4. How do the corner points relate to optimization problems?
  5. What other methods can be used to solve systems of inequalities besides graphical methods?

Tip: Always check that the corner points satisfy all original inequalities to confirm they are part of the feasible region.

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Math Problem Analysis

Mathematical Concepts

Linear Inequalities
Graphing
Coordinate Geometry

Formulas

3x + y = 15
x + 2y = 10

Theorems

Feasible Region Theorem
Corner Point Theorem

Suitable Grade Level

Grades 9-12