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.
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.
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.
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.
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)