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I wanted to check with you if my reasoning to this problem was right.

Find the global maxima and minima of $f(x,y) = xy$ inside the set $A = \{ (x,y) \in \mathbb{R}^2: \frac{|xy|}{|xy|+1} \leq 1\}$.

Testing the boundary $\frac{|xy|}{|xy|+1} = 1$, we get that $0=1$. That would imply that the set $A$ doesn't have a border.

Deriving $f$:

  • $f_x = y$

  • $f_y = x$

So the only critical point is $(0,0)$, where $f(0,0) = 0$.

Now, the function that defines $A$ seems complicated to work with. So we check the second derivatives:

$Hf(x,y) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \Rightarrow \det(Hf(x,y)) = -1 < 0$.

The second derivative test says that every point in $f$ is a saddle point. By that, is it correct to say that $f$ will not have global maxima or minima at $A$?

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To see why there is no global maximum or minimum, look at the line $\{(x,1)\}_{x\in\mathbb{R}}$. $\frac{\lvert x\rvert }{\lvert x\rvert +1 }\leq 1$, so the line is contained in $A$; but $f(x,1)=x\xrightarrow[x\to\pm\infty]{}\pm\infty$.

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  • $\begingroup$ Great! That's a nice and simple justification. Thanks a lot, Clement! $\endgroup$ – Steve Nov 3 '14 at 21:47

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