Prove nonempty, unbounded feasible region has no optimal solution

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In summary: The constraints are x ≥ 0, y ≥ 0 and -x-2y ≥ -8, which is the *same as* x + 2y ≤ 8. The feasible region is nonempty and bounded, and so the problems max 2x + 3y and max -3x - 6y (which is the same as min 3x + 6y) both have feasible, optimal solutions.
  • #1
csc2iffy
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Homework Statement



Let C be the set of all points (x,y) in the plane satisfying x≥0, y≥0, -x-2y≥-8.

a. Show that C is nonempty and unbounded.
b. Prove that the LP problem: Max M=2x+3y subject to the constraint that (x,y) lie in C has no feasible, optimal solution.
c. Show that the LP problem: Max M=-3x-6y subject to the constraint that (x,y) lie in C does have a feasible, optimal solution.


Homework Equations





The Attempt at a Solution



a. I graphed the constraints and showed that the feasible region is the entire first quadrant, and therefore C is nonempty and unbounded (provided attachment of my work - is this enough?)

b. I could "show" this but I have no idea how to "prove" it. Does it involve the simplex method?

c. Same question as above
 

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  • #2
csc2iffy said:

Homework Statement



Let C be the set of all points (x,y) in the plane satisfying x≥0, y≥0, -x-2y≥-8.

a. Show that C is nonempty and unbounded.
b. Prove that the LP problem: Max M=2x+3y subject to the constraint that (x,y) lie in C has no feasible, optimal solution.
c. Show that the LP problem: Max M=-3x-6y subject to the constraint that (x,y) lie in C does have a feasible, optimal solution.


Homework Equations





The Attempt at a Solution



a. I graphed the constraints and showed that the feasible region is the entire first quadrant, and therefore C is nonempty and unbounded (provided attachment of my work - is this enough?)

b. I could "show" this but I have no idea how to "prove" it. Does it involve the simplex method?

c. Same question as above

There is something wrong with the question: what it is asking you is false if you have copied down the constraints correctly. The constraints are x ≥ 0, y ≥ 0 and -x-2y ≥ -8, which is the *same as* x + 2y ≤ 8. Here the feasible region is nonempty and bounded, and so the problems max 2x + 3y and max -3x - 6y (which is the same as min 3x + 6y) both have feasible, optimal solutions.

RGV
 
  • #3
Sorry here is the problem, corrected:
Let C be the set of all points (x,y) in the plane satisfying x≥0, y≥0, -x-2y≤-8.

a. Show that C is nonempty and unbounded.
b. Prove that the LP problem: Max M=2x+3y subject to the constraint that (x,y) lie in C has no feasible, optimal solution.
c. Show that the LP problem: Max M=-3x-6y subject to the constraint that (x,y) lie in C does have a feasible, optimal solution.
 
  • #4
someone please help :(
 
  • #5
csc2iffy said:
someone please help :(

I have already helped, but you have ignored my assistance.

RGV
 
  • #6
No, I had the problem written down wrong. You helped me with the wrong problem
 

Related to Prove nonempty, unbounded feasible region has no optimal solution

What does it mean for a feasible region to be nonempty and unbounded?

A feasible region is a set of all possible solutions that satisfy a given set of constraints. Nonempty means that the feasible region contains at least one point or solution. Unbounded means that the feasible region extends infinitely in one or more directions.

Why does a nonempty, unbounded feasible region have no optimal solution?

In an unbounded feasible region, there is no finite point or solution that can be considered as the "best" or optimal solution. This is because the feasible region extends infinitely, making it impossible to determine a solution that is better than all others.

Can't we just pick a point far enough in the unbounded region to be considered optimal?

No, because the feasible region is unbounded, there will always be a point that is further away and therefore, considered a better solution. This infinite extension of the feasible region makes it impossible to determine a finite optimal solution.

How do we know if a feasible region is nonempty and unbounded?

To determine if a feasible region is nonempty and unbounded, we can graph the constraints and observe if the feasible region extends infinitely in one or more directions. Alternatively, we can also check the constraints to see if they allow for infinitely large or small values, which would indicate an unbounded feasible region.

Does every optimization problem with a nonempty, unbounded feasible region have no optimal solution?

Yes, any optimization problem with a nonempty, unbounded feasible region will have no optimal solution. This is because an optimal solution can only exist in a bounded feasible region, where there is a finite point that is better than all other solutions.

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