Karush Kuhn Tucker problem (minimizing a function subject to a constraint)

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Homework Help Overview

The discussion revolves around minimizing the function f(x,y) = 3x² + y² under the constraint 1 ≤ xy. Participants are exploring the application of the Karush-Kuhn-Tucker (KKT) theorem to identify potential solutions and critical points.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of KKT conditions, examining cases where the constraint is inactive and active. There are questions about the consistency of derived equations and the implications of the constraint on the existence of solutions.

Discussion Status

Some participants have provided guidance on how to approach the problem using Lagrange multipliers, suggesting that the only critical point found does not satisfy the constraint. Others have proposed that the minimum value may be found along the boundary defined by the constraint, while acknowledging the lack of explicit consensus on the existence of a maximum.

Contextual Notes

There is a mention of the constraint defining an open set, and some participants express uncertainty about the compactness of the set, which may affect the existence of minima or maxima.

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Homework Statement


Find the minimum of f(x,y)= 3x2+y2, subject to the constraints 1<=xy.


Homework Equations


I thought I would use Karush Kuhn Tucker's theorem to solve this.
∇f=(6x, 2y) and ∇h=(-y,-x)
The general equation according to KKT is ∇f=λ∇h.

First case: h<0. According to KKT theorem when h is inactive (<0) then λ=0.
So the equation becomes 6x=0, therefore x=0 and 2y=0 therefore y=0 which cannot happen due to the constraint 1<=xy (since 1 is not <=0). Therefore this is not a solution.

Second case: h=0
The equation is 6x=-y and 2y=-x. These seem like inconsistent equations to me?
I also got 1-xy=0.

What is the next step or am I right in thinking there are again no solutions and there is no minimum or maximum? I'm pretty sure the set is not compact so the min/max may not even exist.

Some help would be greatly appreciated. Thank you.
 
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porroadventum said:

Homework Statement


Find the minimum of f(x,y)= 3x2+y2, subject to the constraints 1<=xy.


Homework Equations


I thought I would use Karush Kuhn Tucker's theorem to solve this.
∇f=(6x, 2y) and ∇h=(-y,-x)
The general equation according to KKT is ∇f=λ∇h.

First case: h<0. According to KKT theorem when h is inactive (<0) then λ=0.
So the equation becomes 6x=0, therefore x=0 and 2y=0 therefore y=0 which cannot happen due to the constraint 1<=xy (since 1 is not <=0). Therefore this is not a solution.

Second case: h=0
The equation is 6x=-y and 2y=-x. These seem like inconsistent equations to me?
I also got 1-xy=0.

What is the next step or am I right in thinking there are again no solutions and there is no minimum or maximum? I'm pretty sure the set is not compact so the min/max may not even exist.

Some help would be greatly appreciated. Thank you.

I'd just reason using ordinary Lagrange multipliers. 1<xy is an open set. So setting the derivatives equal to 0 gives the only critical point at (0,0) as you showed, which doesn't satisfy the constraint. So the only possible extrema will be on 1=xy. You should be able to find them using a Lagrange multiplier. No, there obviously won't be a max but you'll certainly be able to find a min.
 
I'm getting the minimum at a value of 2√3, is this correct?
 
porroadventum said:
I'm getting the minimum at a value of 2√3, is this correct?

Be best if you showed your work, but yes, that sounds right.
 
Last edited:
porroadventum said:

Homework Statement


Find the minimum of f(x,y)= 3x2+y2, subject to the constraints 1<=xy.


Homework Equations


I thought I would use Karush Kuhn Tucker's theorem to solve this.
∇f=(6x, 2y) and ∇h=(-y,-x)
The general equation according to KKT is ∇f=λ∇h.

First case: h<0. According to KKT theorem when h is inactive (<0) then λ=0.
So the equation becomes 6x=0, therefore x=0 and 2y=0 therefore y=0 which cannot happen due to the constraint 1<=xy (since 1 is not <=0). Therefore this is not a solution.

Second case: h=0
The equation is 6x=-y and 2y=-x. These seem like inconsistent equations to me?
I also got 1-xy=0.

What is the next step or am I right in thinking there are again no solutions and there is no minimum or maximum? I'm pretty sure the set is not compact so the min/max may not even exist.

Some help would be greatly appreciated. Thank you.

Your "inconsistent" equations are not correct, so their being inconsistent is not important.

You should have 6x = -λy and 2y = -λx, so λ can be found. You should take the positive root (why?) Then, putting the value of λ into one of those equations and using the other equation xy = 1, you get the complete solution. There are two optimal solutions in (x,y) space; do you see why?
 

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