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I am having some doubts about this exercise:

Find maximum and minimum of $f(x,y) = y+x-2xy$ restricted to the interior and border of $R = \{(x,y) \in \mathbb{R}^2 : |x| \geq \frac{1}{2}\}$.

I started by looking for critical points, getting:

  • $f_x = 1 - 2y$

  • $f_y = 1 - 2x$

Then $\nabla f(x,y) = 0$ $\iff$ $(x,y) = (\frac{1}{2}, \frac{1}{2})$.

That point, which is the only critical point of $f$, is in $R$, since $|\frac{1}{2}| \geq \frac{1}{2}$.

Also $f(\frac{1}{2}, \frac{1}{2}) = 1/2$.

Now we need to check along the borders (the vertical lines):

  • $f(\frac{1}{2}, y) = \frac{1}{2}$
  • $f(-\frac{1}{2}, y) = 2y - \frac{1}{2}$

Along the first line, $f$ stays constant. Since its value there is the same as the value at the critical point, the whole line is a candidate for extremes.

Along the second line, $f$ behaves like a linear function: monotonically increasing and without any critical points.

Now is the part where I am not entirely sure: I would say that since $(0,0) \in R$, and $f(0,0) = 0 < \frac{1}{2}$, then $f$ doesn't have either maxima nor minima at the points we ruled as candidates for being extremes. But would that be enough to guarantee that $f|_R$ doesn't have any maxima nor minima?

Is there some other approach to this problem?

Thanks a lot.

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  • $\begingroup$ Once you have $2y-0.5$, that assumes every value in $\mathbb{R}$ already, so there are no global maxima nor minima. $\endgroup$ – vadim123 Nov 3 '14 at 18:47
  • $\begingroup$ That and saying that those points are all in R would justify it all right, wouldn't it? $\endgroup$ – Steve Nov 3 '14 at 18:57
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Your function on a given domain is unbounded since $(2y - \frac{1}{2})$ and this means that it does not have minima nor maxima.

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  • $\begingroup$ I see. This all fails (I mean, $f$ doesn't have global extremes), because $R$ is not compact, doesn't it? $\endgroup$ – Steve Nov 3 '14 at 18:58
  • $\begingroup$ You got it right. $\endgroup$ – Jihad Nov 3 '14 at 19:02
  • $\begingroup$ Great! Thank you very much, Jihad. $\endgroup$ – Steve Nov 3 '14 at 19:04
  • $\begingroup$ @Steve, no problems. Peace be upon your math ;) $\endgroup$ – Jihad Nov 3 '14 at 19:07

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