I need to determine if function:

$f(x,y)=x+2y-2\log(xy) $

has global minimum/maximum.

I've found local minimum at $(2,1)$, but that's not any proof of global minimum.

  • $\begingroup$ The function under consideration is unbounded from below: $$ \lim_{x\to -\infty , y\to - \infty} (x+2y-2 \log(xy))= -\infty.$$ $\endgroup$ – user64494 Jun 23 '13 at 13:23
  • $\begingroup$ Local min is correct (validated by wolframalpha) $\endgroup$ – wojteo Jun 23 '13 at 13:29
  • $\begingroup$ I think it is instructive to use the general approach to such problems and be able to determine these things. Regards $\endgroup$ – Amzoti Jun 23 '13 at 13:34

Put $x=1/y$. You get $f(x,1/x)=x+2/x$. Approach $x$ to zero from below and from above. What do you get?

  • $\begingroup$ If $x$ can be negative here, what happens to $f(x,y)$ when $x$ is negative and $y$ is positive? $\endgroup$ – Henry Jun 23 '13 at 13:21
  • $\begingroup$ If $x$ is negative, then $y=1/x$ is negative. $\endgroup$ – OR. Jun 23 '13 at 13:45

Hint: $$f(x,y)=x+2y−2\log(xy) = (x-2 \log(x)) + (2y-2 \log(y))$$ so you have a separable function and can look for minima for $x$ and for $y$ individually.

So long as both are positive, it is obvious that $(2,1)$ is the only point giving a minimum and that there is no maximum.

  • $\begingroup$ One has to be careful. In this case one will get the same answer about unboundedness but notice that the function on the right is not the same as the function on the left. They don't have the same maximal domain of definition. That is precisely the origin of your mistake in the comment you made to my answer. On the right $x$ and $y$ must be positive. On the left you can allow both to be negative. $\endgroup$ – OR. Jun 23 '13 at 13:48
  • $\begingroup$ Well yes, but. If the OP had stated that they could both be negative but that they could not be of different signs, I would completely accept your point. But I suspect that in fact either they both had to be positive, or that the issue had simply been ignored (the possibility of $x$ or $y$ being $0$ is not mentioned in the question). $\endgroup$ – Henry Jun 23 '13 at 16:36
  • $\begingroup$ It is good practice to solve a problem considering all possibilities the statement admits. The only restrictions imposed on the variables are those given by the definition of the given function. By the way, I was wrong, your transformation doesn't give you the correct answer to the problem. The function on the right is bounded from below on its domain, while the function on the left is not. That is the danger of inequivalent transformations. You sometimes get the wrong answer, as in this case. $\endgroup$ – OR. Jun 23 '13 at 17:30

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