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

1. Jan 16, 2013

1. The problem statement, all variables and given/known data
Find the minimum of f(x,y)= 3x2+y2, subject to the constraints 1<=xy.

2. Relevant 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 16, 2013

### Dick

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.

3. Jan 16, 2013

I'm getting the minimum at a value of 2√3, is this correct?

4. Jan 16, 2013

### Dick

Be best if you showed your work, but yes, that sounds right.

Last edited: Jan 16, 2013
5. Jan 17, 2013

### Ray Vickson

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?