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In some optimization problems with inequality constraints some of the aforementioned constraints can be x>=0 , y>=0 and so on. I think these constraints are called non negativity constraints; they just state that some of the variables of the objective function aren't negative. When I apply the Khun-Tucker method I don't consider as much multipliers as inequality constraints; for example, in the problem max: xy+9x subject to x+4y<=20 and x>=0 and y>=0 , I construct the lagrangian function as this:

L=xy+9x+λ(20-x-4y)

and then I discard all the negative values that I find for x and y. The issue is that somebody told me that this is the correct way to construct the lagrangian function:

L=xy+9x+λ(20-x-4y)+μ(x)+m(y)

and, supposedly, the values for x and y that are found by analyzing each of those lagrangian functions are not the same; the correct ones would be those that are found by analyzing the second function. I'm not sure if that's right.

Also, is it possible for a critical point of a concave function to be a saddle point?

Werther
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1 Answers1

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  1. Both ways are equivalent when the non-negativity constrains are slack but if in the solution either $x=0$ or $y=0$ then as Rahul pointed you need to include the multipliers for them. But you can also use your method, if you don't find a solution just try with $x=0$ and/or $y=0$.
  2. A critical point of a concave function is always a global maximum (assuming a convex domain) as the function must lie below the tangent plane.
Sergio Parreiras
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  • The fact that a function of 2 variables is concave implies that the hessian matrix of the function has the following features: a11<=0 , a22<=0 and the determinant is >=0 for every (x;y). A sufficient condition to determine if a point is a saddle point is for the determinant of the hessian evaluated in that point to be <0 , which can't happen due to the fact that the function is concave. However, it can happen that the determinant be =0, which is a sufficient condition to determine that the point can be everything, either a maximum, minimum or saddle point. What then? – Werther Jul 11 '14 at 04:30
  • No it is not "a sufficient condition to determine that the point can be everything" because the function is concave. If you don't believe me please draw some pictures or check this: http://www.economics.utoronto.ca/osborne/MathTutorial/GOPF.HTM – Sergio Parreiras Jul 13 '14 at 14:23