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I have to do the following problem, and I need help.

Examine the function $f(x,y) = \dfrac{-3x}{x^2+y^2+1}$ with respect to maximum and minimum.

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What have you tried to far? Have you made any progress? –  mixedmath Jun 8 '11 at 19:41
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also, in the future, please do not use the imperative (Examine, Do, Prove, etc...) with the members of this site. It can be thought of as rude. –  mixedmath Jun 8 '11 at 19:42
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2 Answers 2

Maybe you are expected to find the place(s) where the partials are $0$, find out at which of these two places the function is bigger, and where it is less, write down that you have a maximum at the first, a minimum at the second, and go on to the next question.

But in principle, this is not enough. And it is not hard to come up with examples where a mechanical approach of this type leads to the wrong conclusion.

Let's think about $f(x,y)$. It is not hard to see that if you take a big circle around the origin, then outside that circle $f(x,y)$ is close to $0$. The equation $z=f(x,y)$ determines a smooth surface.

Examination of the critical points shows that inside the circle, $f(x,y)$ takes on the values $3/2$ and $-3/2$. Thus there are a global max and a global min, and these must occur inside the circle. So the global max/min must be local max/min, the top of a hill or the bottom of a valley.

It follows that at the global max and min, the surface must have flattened out, that is, the partials must be $0$. But by setting the partials equal to $0$, we found the only two places where such a flattening out occurred, and we are finished.

In this particular case, you could also notice that $f(-x,y)=-f(x,y)$, and that the function is positive when $x<0$ and negative when $x>0$. So when you have located the maximum, the minimum value is automatically the negative of the maximum value. Symmetry is your friend!

There is a "test" you may be taught, involving the second partials, that (usually) enables you to determine whether a critical point gives a local max or a local min. While this test is of great theoretical importance, it is often difficult to use. The second partials for your $f(x,y)$ are not much fun to calculate, so in this case the test that uses second partials would be unpleasant to carry out.

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look for points where the partials vanish simultaneously (called critical points by some), in this case $$f_x=[-3(x^2+y^2+1)+6x^2]/(x^2+y^2+1)^2, f_y=6xy/(x^2+y^2+1)^2$$ which are both zero only at $(x,y)=(1,0), (-1,0)$. now the simplest thing to do is evaluate the function at those points (if a function such as $f$ has a local max/min at some point where it is differentiable, then it is a critical point; note that $f$ is differentiable everywhere). we have $f(1,0)=-3/2, f(-1,0)=3/2$ so those are the min and max respectively.


added for nitpickers

given the behavior at infinity $$ \lim_{|(x,y)|\to\infty}f(x,y)=0 $$ $f$ clearly has a global max/min (continuous on a large closed disk where the boundary values are small wrt the claimed max/min)

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thank you very much :) –  Aline Jun 8 '11 at 23:24
    
@Aline, if you like yoyo's answer, you could vote it up by clicking the up-arrow. If you like it a lot, you could accept it by clicking the checkmark. –  Gerry Myerson Jun 9 '11 at 0:43
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the crucial phrase is "$\it if$ a function has a local max/min...." You have found the critical points, but you haven't established that either one is a maximum or a minimum, not even locally. –  Gerry Myerson Jun 9 '11 at 0:47
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