Find the local max/min or saddle points on the function $f(x, y) = xy$

My attempt I've got the first partial derivatives to be

$$f_x = y$$ $$f_y = x$$

Giving a critical point at (0,0)

For the second derivative test, I've got $f_{xx} = 0$, $f_{yy} = 0$, $f_{xy} = 1$

I know that if $$f_{xx} \times f_{yy} < (f_{xy})^2$$ implies a saddle point. But is this true if the second derivatives $f_{xx}$ or $f_{yy}$ is zero.

Thanks in advance.

  • 1
    $\begingroup$ Yes its true. Take a look at a plot of $f(x,y) = xy$ near 0 and you'll see the characteristic (horse) saddle shape $\endgroup$
    – Gregory
    Oct 28 '17 at 15:47

Yes, $(0,0)$ is a saddle point. The condition that you write in the question come from the general test of the Hessian determinant, and in this case we have a saddle because the Hessian determinant at this point is $H=-1$.


On the line $y=x$, the function becomes $f(x)=x^2$ and therefore $(0,0)$ is a minimum.

On the other hand, when you take $y=-x$, the function becomes $f(x)=-x^2$ and hence $(0,0)$ is a maximum.

These imply that $(0,0)$ is neither a maximum nor a minimum.

However, you have already found that it is a critical point since $f_x(0,0)=f_y(0,0)=0$.

Thus it must be a saddle point.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.