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

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The discussion revolves around solving the optimization problem of minimizing the function f(x,y) = 3x² + y² under the constraint 1 ≤ xy using the Karush Kuhn Tucker (KKT) theorem. Initially, the user explores two cases: when the constraint is inactive (h < 0) and when it is active (h = 0), concluding that the first case yields no valid solutions due to the constraint. In the second case, confusion arises regarding the equations derived from KKT conditions, leading to questions about the existence of a minimum or maximum. Another participant clarifies that the equations can be solved correctly to find optimal solutions, emphasizing the need to consider the constraint and the positive root for λ. The conversation highlights the complexity of applying KKT in constrained optimization problems.
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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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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