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I'm currently just working through some maxima/minima problems, but came across one that was a bit different from the 'standard' ones.

So they used the usual procedures and ended up finding that the Hessian is zero at the critical point (0,0).

They set $x=y$, which resulted in $f(x,x)=-x^3$, which has an inflection point at the origin, which is the 2D version of the saddle point.

I have a few questions about this.

  1. How did they 'know' to set x=y, or is this a standard technique for these problems? ie: Set $x=f(y)$ and choose some convenient $f(y)$?

  2. In a geometric sense, what does setting $x=y$ mean? I'm having trouble visualising this.

The provided solution

Trogdor
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1 Answers1

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set x=y means evaluating the function on the line x=y. put it another way, evaluating the function along the direction $\mathbf{v}=[1,1]^T$. Since an extremum on the whole must be an extermum along any direction,. If we can find a direction along which this critical point is not an extremum, then we can assert that this point is not an extremum on the whole.

Robert Fan
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  • Does the converse of this also hold true?

    ie: Since we found a direction in which the critical pt is an extremum, then that point therefore is an extremum on the whole?

     – Trogdor Apr 08 '14 at 14:36
  • @Trogdor: No, a single direction is not enough(think about the definition of "saddle point"), but if for any arbitrary direction, the critical point is a directional extremum, then we can justify your assertion. – Robert Fan Apr 08 '14 at 14:42
  • @RobertFan will this then also be true for f(x,y)=e^((x-y)^2 – User69127 Jun 03 '17 at 01:20