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I have this function:$$f(x,y)=\frac{10+x^2y^2}{e^{x^2+y^2} }$$ I have found that the only stationary point is $(0,0)$ but I don't know ho to classify it as a max, min or saddle point. Can someone help me? (I could study it using the determinant of the Hessian but I'm looking for another solution)

The Hessian matrix

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    See my edited answer. – 2017-02-19
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    Aside; you could reduce the calculation by using the trick $$ \left( \frac{\partial}{\partial u} g(u,v)\right)_{(u,v)=0} = \left( \frac{d}{du} g(u,0)\right)_{u=0} $$ – 2017-02-19
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    You can tell it's not a saddle point because the two variables are symmetric (ie if you swapped $x$ with $y$ in the function definition it would still be the same thing). – 2017-02-19

3 Answers 3

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Writing in polar form with the substitutions

$$\begin{aligned} x &= r \cos \theta & y &= r \sin \theta \end{aligned}$$

gives

$$ f = \frac{10 + r^4 (\cos \theta \sin \theta)^2}{e^{r^2}} $$

Defining $t = (\cos \theta \sin \theta)^2$ and expanding

$$ e^{-r^2} = 1 - r^2 + \frac{1}{2}r^4 - o(r^6) $$

before multiplying gives

$$\begin{aligned} f &= 10 - 10 r^2 + (5 + t) r^4 - o(r^6) \\ \frac{\partial}{\partial r} f &= -20 r + 4(5+t) r^3 - o(r^5) \\ \frac{\partial^2}{\partial r^2} f &= -20 + 12(5 + t)r^2 - o(r^4) \\ &< 0 &r=0, \forall \theta \end{aligned}$$

That is, $f_\theta(r)$ has a maximum at $r = 0$ for all $\theta$. Therefore $f(x, y)$ must have a maximum at $(x, y) = (0, 0)$.

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    Very good solution. Is it possible to determinate saddle points with this method? – 2017-02-19
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You could taylor expand to second order $$ f(x,y) = 10e^{-(x^2+y^2)}(1+x^2y^2)\approx 10(1-(x^2+y^2))(1) = 10-10(x^2+y^2)$$

and infer that it's a local maximum (since it's an upside down paraboloid).

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Method 1. One may observe that, as $(x,y) \to (0,0)$, $$ \begin{align} 10-f(x,y)&=10-10e^{-(x^2+y^2)}-x^2y^2e^{-(x^2+y^2)} \\\\&=10-10e^{-(x^2+y^2)}-o(x^2+y^2) \\\\&=10-10[1-(x^2+y^2)+o(x^2+y^2)]-o(x^2+y^2) \\\\&=10(x^2+y^2)+o(x^2+y^2) \end{align} $$ which is non-negative as as $(x,y) \to (0,0)$.

Method 2. One may use the second derivative test for functions of two variables, that is evaluating $$ D(x,y)= \frac{\partial^2 f}{\partial x^2}(x,y)\frac{\partial^2 f}{\partial y^2}(x,y) - \left(\frac{\partial^2 f}{\partial xy}(x,y) \right)^2 $$ at $(0,0)$ checking if $D(0,0)>0$ and $\frac{\partial^2 f}{\partial x^2}(0,0)<0$ to know if it is a local maximum.

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    I know. But the I'm trying to avoid to calculate the second derivative. The amount of calculus is huge – 2017-02-18
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    @FedericoZucchi Ok, you can use the fact that $e^{-t}=1-t+t^2/2+o(t^2)$ as $t \to 0$. Here $t=x^2+y^2$ giving an expansion for $\frac{1}{e^{x^2+y^2} }=e^{-(x^2+y^2)} $. – 2017-02-18
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    And what about considering that the function is always positive? Is it important? – 2017-02-18
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    It won't be sufficient. – 2017-02-18
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    Ok. Your method is valid but how am I supposed to manage the $x^2y^2$ with your substition? – 2017-02-18
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    With the expansion I've given in my comment above, one sees that, near $(0,0)$, one has$10-f(x,y)=10(x^2+y^2)+o(x^2+y^2)\ge0$. In fact $x^2y^2$ is much smaller compared to $x^2+y^2$ so it goes into $o(x^2+y^2)$. Do you see that? – 2017-02-18
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    Now I see that. Thanks – 2017-02-18
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    @FedericoZucchi You are welcome. – 2017-02-18
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    I see, and what if the point was a saddle? How can I figure out with your method? – 2017-02-19
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    Here the sign is *constant* near $(0,0)$, if the sign were *non constant* there would be a saddle point... – 2017-02-19