Math Problem Statement

(*) Given the following feasible region: X={x:x_1+x_2≥2,x_1-x_2≤0,-3x_1+x_2≥-6,x_1≥0,x_2≥0}⊂R^2
Calculate (i) extreme points and (ii) extreme directions. 

Solution

Let’s break the problem into two parts:

Feasible region XX:

Given:

  1. x1+x22x_1 + x_2 \geq 2
  2. x1x20x_1 - x_2 \leq 0 or x1x2x_1 \leq x_2
  3. 3x1+x26-3x_1 + x_2 \geq -6
  4. x10x_1 \geq 0
  5. x20x_2 \geq 0

This is a polyhedral region in R2\mathbb{R}^2. The extreme points of the feasible region correspond to vertices of this polyhedron, while the extreme directions arise from unbounded edges of the region.


(i) Finding Extreme Points:

Extreme points occur at the intersection of constraints. We find all intersections of constraint boundaries and verify which satisfy all conditions.

  1. Boundary equations:

    • Line 1: x1+x2=2x_1 + x_2 = 2
    • Line 2: x1x2=0x_1 - x_2 = 0 or x1=x2x_1 = x_2
    • Line 3: 3x1+x2=6-3x_1 + x_2 = -6
    • Lines 4 & 5: x1=0x_1 = 0 and x2=0x_2 = 0
  2. Finding intersections:

    • Intersection of x1+x2=2x_1 + x_2 = 2 and x1=x2x_1 = x_2: Substitute x1=x2x_1 = x_2 into x1+x2=2x_1 + x_2 = 2: 2x1=2    x1=1,  x2=12x_1 = 2 \implies x_1 = 1, \; x_2 = 1 Point: (1,1)(1, 1)

    • Intersection of x1+x2=2x_1 + x_2 = 2 and 3x1+x2=6-3x_1 + x_2 = -6: Solve the system: x1+x2=2(1)x_1 + x_2 = 2 \quad \text{(1)} 3x1+x2=6(2)-3x_1 + x_2 = -6 \quad \text{(2)} From (1): x2=2x1x_2 = 2 - x_1. Substitute into (2): 3x1+(2x1)=6-3x_1 + (2 - x_1) = -6 4x1+2=6    4x1=8    x1=2,  x2=0-4x_1 + 2 = -6 \implies -4x_1 = -8 \implies x_1 = 2, \; x_2 = 0 Point: (2,0)(2, 0)

    • Intersection of x1=x2x_1 = x_2 and 3x1+x2=6-3x_1 + x_2 = -6: Substitute x1=x2x_1 = x_2 into 3x1+x2=6-3x_1 + x_2 = -6: 3x1+x1=6    2x1=6    x1=3,  x2=3-3x_1 + x_1 = -6 \implies -2x_1 = -6 \implies x_1 = 3, \; x_2 = 3 Point: (3,3)(3, 3)

  3. Verify against all constraints:

    • (1,1)(1, 1), (2,0)(2, 0), and (3,3)(3, 3) satisfy all constraints. These are the extreme points.

(ii) Finding Extreme Directions:

Extreme directions exist if the polyhedron is unbounded. To find extreme directions, examine unbounded constraints.

  1. Since x1,x20x_1, x_2 \geq 0, the polyhedron is in the first quadrant.
  2. Directions that satisfy the equality part of the constraints:
    • From x1+x22x_1 + x_2 \geq 2: Direction (1,1)(1, 1)
    • From 3x1+x26-3x_1 + x_2 \geq -6: Direction (3,1)(3, 1)

These directions lead to unbounded edges of the region and form the extreme directions.


Final Answer:

(i) Extreme points: (1,1),(2,0),(3,3)(1, 1), (2, 0), (3, 3)
(ii) Extreme directions: (1,1),(3,1)(1, 1), (3, 1)


Would you like detailed verification of the constraints, a visualization of the feasible region, or additional examples?


5 Related Questions:

  1. What are the geometric interpretations of extreme points and extreme directions in R2\mathbb{R}^2?
  2. How can we generalize this problem to higher dimensions?
  3. What is the relationship between extreme points and convex combinations in polyhedra?
  4. How does the presence of equality constraints affect the feasible region?
  5. Can you identify the boundedness of a feasible region without explicitly finding extreme directions?

Tip: Always plot the constraints on a graph for better visualization when solving such problems!

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Geometry of Polyhedra

Formulas

x_1 + x_2 ≥ 2
x_1 - x_2 ≤ 0
-3x_1 + x_2 ≥ -6
x_1 ≥ 0
x_2 ≥ 0

Theorems

Convex Hull Theorem
Fundamental Theorem of Linear Programming

Suitable Grade Level

Grades 11-12