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my task is to figure out the critical points of $f(x,y)=e^y(x^4-x^2+y)$, $\ $$\mathbb{R}^2 \rightarrow \mathbb{R}$, and show which of them is a maximum or minimum. As far as I got, I've shown that the critical points are:

1.: $(0,-1)$ which is neither max. nor min. (char. pol. of Hessian is indefinite)

2.: $\left(\frac{1}{ \sqrt2},-\frac{3}{4}\right)$, which is a local minimum and

3.: $\left(-\frac{1}{ \sqrt2},-\frac{3}{4}\right)$ which is the second local minimum.

Moving towards my question, is there any way to easily show if any global maximum or minimum exists (in general and/or concerning this example)? I’ve used the char. poly. of the Hessian, is there any faster possibility to finding local min./max.?

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Is this over the entire $\mathbb{R}^2$? –  Ataraxia May 25 '13 at 14:26
    
Yes, I will add this information! –  Josh May 25 '13 at 14:29
    
You might want to review my answer here: math.stackexchange.com/questions/369945/… –  Amzoti May 25 '13 at 16:35
    
Will definitly read through it, thanks. –  Josh May 25 '13 at 16:44

1 Answer 1

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First find the limit as x and y approach infinity. In single variable functions, you only have to check to "ends", $-\infty$ and $\infty$. In functions of two variables, there are four, $x\to-\infty$, $x\to\infty$, $y\to-\infty$, and $y\to\infty$.

$$\lim_{x\to-\infty}f(x,y)=\infty$$

$$\lim_{x\to\infty}f(x,y)=\infty$$

$$\lim_{y\to-\infty}f(x,y)=-\lim_{y\to\infty}\frac{e^{-y}}{\frac{1}{y}}=0$$

$$\lim_{y\to\infty}f(x,y)=\infty$$

Evaluate $f(x,y)$ at the critical points you calculated and compare them to these values.

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$f(0,-1)= -e^{-1}$,$f(\left(\frac{1}{ \sqrt2},-\frac{3}{4}\right))= -e^{- \frac{3}{4} }$. This means the values of f(x,y) at the critical points aren't equal to these limits, so I can conclude that there is non global max./min.? –  Josh May 25 '13 at 14:59
    
@Josh It would appear that $-e^{-\frac{3}{4}}$ is the global minimum and $\infty$ is the global maximum. Are you sure the Hessian at $(0,−1)$ is indefinite? I just checked, and I'm getting that the Hessian at $(0,-1)$ is negative semi-definite. –  Ataraxia May 25 '13 at 15:23
    
Well, I calculated the eigenvalues $\lambda_1 = \frac{1}{e}$ and $\lambda_2 = - \frac{2}{e}$ for the char. pol. of the Hessian at (0,-1). WolframAlpha also tells me that there are just two local minima( $\left(-\frac{1}{ \sqrt2},-\frac{3}{4}\right)$ and $\left(\frac{1}{ \sqrt2},-\frac{3}{4}\right)$). –  Josh May 25 '13 at 15:30
    
@Josh Ok, I was just wondering because I tried it and got that it was negative semi-definite, but it doesn't really matter since $-e^{-1}$ isn't a minimum. So since the only other three critical values are $-e^{-1}$, $0$, and $\infty$, the global minimum is...? –  Ataraxia May 25 '13 at 15:35
    
I'd say $-e^{-1}$? I'm still a little bit confused: the step I have to make is evaluating f(x,y) at the critical points and also the limits? What tells me then if global min./max. exists? –  Josh May 25 '13 at 15:41

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