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I'm struggling a bit with my multivariable optimization. Assuming the determinant of the hessian matrix ≠ 0 I have no issue, though when the det = 0 I get stumped. Example- $$f(x,y)=x^4+y^4-(x+y)^2$$

The Hessian comes to: $$ \begin{matrix} 12x^2-2 & -2 \\ -2 & 12y^2-2 \\ \end{matrix} $$ The critical points come to (1,1,-2) -Local Minima, (-1,-1,-2) - Local Minima, (0,0,0) - ?$$-$$ The det for (0,0,0) comes to 0 in the Hessian and as a result is a negative semi definite so could be a saddle point, max or neither. Beyond that I don't know what to do. Any help is appreciated. Thanks

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1) On the line $y=-x$, $\;\;f(x,-x)=2x^4>0=f(0,0)$ for $x\ne0$,

$\hspace{.3 in}$so $f(0,0)$ is not a local maximum.

2) On the line $y=0$, $\;\;f(x,0)=x^4-x^2=x^2(x^2-1)<0=f(0,0)$ for $0<x<1$,

$\hspace{.3 in}$so $f(0,0)$ is not a local minimum.

Therefore $f$ has a saddle point at $(0,0,0)$.

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  • $\begingroup$ Thanks for taking the time to answer, could you explain the intuition behind how you answer this? I'm still not fully clear as to why we've set y=-x or y=0 Is the same method/approach used everytime when faced with this sort of situation. $\endgroup$ – Nik-D Jun 26 '15 at 8:26
  • $\begingroup$ I was trying to find 2 lines (or other curves) through the critical point which ruled out having a maximum and a minimum; usually I start with the x- and y-axes (assuming the point is the origin), and then look at other lines if necessary. In this case, I used $y=-x$ since it made the second term vanish. $\endgroup$ – user84413 Jun 26 '15 at 16:08

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