1. ## critical points multivariable

show that the function $f(x,y)=xy^2-x^2y$ has a unique critical point. what kind is it?

I tried using the second derivative test. So from first derivatives I got the critical points (0,0) (1,0) and (1,2)

I then tried using the determinant where $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2$ but it was always inconclusive. For example D=0 for the point (0,0) and $f_{xx}=0$ for (1,0) and I don't know how to interpret that.

I haven't learned Larange multipliers yet.

2. Originally Posted by superdude
show that the function $f(x,y)=xy^2-x^2y$ has a unique critical point. what kind is it?

I tried using the second derivative test. So from first derivatives I got the critical points (0,0) (1,0) and (1,2)

How? $f_x=y^2-2xy=y(y-2x)=0\,,\,\,f_y=2xy-x^2=x(2y-x)=0\,\Longrightarrow\,x=y=0$ $\Longrightarrow\,(0,0)$ is the only solution to this system, as can be checked...or better: check that $(1,0)\,,\,(1,2)$ are not zeroes of this sytem but $(,0) is.$[tex]

I then tried using the determinant where $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2$ but it was always inconclusive. For example D=0 for the point (0,0) and $f_{xx}=0$ for (1,0) and I don't know how to interpret that.

And you don't need to: critical points are points where the first order partial derivatives vanish, and that's what's been done above

I haven't learned Lagrange multipliers yet.
You don't need them and anyway you can't use them here.

Tonio

3. ok I follow you so far. But that doesn't answer the question what type of critical point is at (0,0)

4. Originally Posted by superdude
ok I follow you so far. But that doesn't answer the question what type of critical point is at (0,0)

No, it doesn't...and according to you that wasn't asked. I suspect it is a saddle point, but you'll have to go into higher partial derivatives (perhaps the Wronskian matrix...I just don't remember) to find out.

Tonio

5. could someone define "unique critical point"?