# "Find all 'stationary points'.

• Sep 3rd 2009, 09:09 AM
courteous
"Find all 'stationary points'.
Quote:

And for each tell whether it is a local extreme. Given function: $f(x,y)=ye^{x^2-y^2}$ "
(a,b) is stationary $\Leftrightarrow gradient(a,b)=0$

So, both $f_x=\frac{\partial f}{\partial x}=2xye^{x^2-y^2}$ and $f_y=\frac{\partial f}{\partial y}=(1-2y^2)e^{x^2-y^2}$ must equal $0$.
Thus, two stationary points are $T_1=(0,\frac{1}{\sqrt2})$ and $T_2=(0,-\frac{1}{\sqrt2})$.

Now, how do I find if there are any "local extremes" among these two points, and what are they (MAX, min)?

I know you need Hesse's matrix but don't know how.
• Sep 3rd 2009, 10:00 AM
ynj
Quote:

Originally Posted by courteous
(a,b) is stationary $\Leftrightarrow gradient(a,b)=0$

So, both $f_x=\frac{\partial f}{\partial x}=2xye^{x^2-y^2}$ and $f_y=\frac{\partial f}{\partial y}=(1-2y^2)e^{x^2-y^2}$ must equal $0$.
Thus, two stationary points are $T_1=(0,\frac{1}{\sqrt2})$ and $T_2=(0,-\frac{1}{\sqrt2})$.

Now, how do I find if there are any "local extremes" among these two points, and what are they (MAX, min)?

I know you need Hesse's matrix but don't know how.

One theorem says that if the Hessian is positive definite, then it will be a local minimal extremes. If it is negative definite, then it will be a local maximal extremes.
• Sep 3rd 2009, 10:06 AM
HallsofIvy
Quote:

Originally Posted by courteous
(a,b) is stationary $\Leftrightarrow gradient(a,b)=0$

So, both $f_x=\frac{\partial f}{\partial x}=2xye^{x^2-y^2}$ and $f_y=\frac{\partial f}{\partial y}=(1-2y^2)e^{x^2-y^2}$ must equal $0$.
Thus, two stationary points are $T_1=(0,\frac{1}{\sqrt2})$ and $T_2=(0,-\frac{1}{\sqrt2})$.

Now, how do I find if there are any "local extremes" among these two points, and what are they (MAX, min)?

I know you need Hesse's matrix but don't know how.

Well, you don't really need the Hessian. There are two ways to determine whether a stationary point is an extremum and, if so, whether it is a max or min, equivalent to the "first derivative test" and "second derivative test" for functions of one variable.

1) If there is some neighborhood of the point such that, for each (x,y) in the neighborhood, the dot product of $\nabla f(x,y)$ with the vector from p to (x,y) to p is positive, (so that $\nabla f(x,y)$ is pointing "away from" p) p is a minimum.

If there is some neighborhood of the point such that, for each (x,y) in the neighborhood, the dot product of $\nabla f(x,y)$ with the vector from p to (x,y) to p is negative, (so that $\nabla f(x,y)$ is pointing "toward" p) p is a minimum.

Of course, it may well happen that there is no such neighborhood, in which case p is not an extremum.

2) If $\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}- \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2$ (this is the Hessian), evaluated at p, is greater than 0 then either
a) If either $\frac{\partial^2 f}{\partial x^2}> 0$ or $\frac{\partial^2 f}{\partial y^2}> 0$ then the point is a minimum.
b) If either $\frac{\partial^2 f}{\partial x^2}< 0$ or $\frac{\partial^2 f}{\partial y^2}< 0$ then the point is a minimum.
(Here, the two second derivatives must have the same sign so you can check either.)

These are equivalent to ynj's "positive definite" and "negative definite".

If $\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}- \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2$, evaluated at p, is less than 0, then p is not an extremum.

If $\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}- \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2$, evaluated at p, is equal to 0, this test is inconclusive. For example, $f(x,y)= x^4+ y^4$, $g(x,y)= -x^4- y^4$, and $h(x,y)= x^4- y^4$ all have a singular point at (0, 0), all second derivatives are equal to 0, but f has a minimum, g a maximum, and h a "saddle point".

I would be very surprised if these were not in your textbook.