Find relative maxima/minima/saddle points

Question

I need to find the relative maxima/minima/saddle points of $f(x,y)=x^3-12x+y^3-27y+5$

I found $$ f_x=3x^2-12 \\ f_y=3y^2-27\\ f_{xx}=6x \\ f_{yy}=6y\\ f_{xy}=0 $$ Considering the Hessian matrix, $$ detHf = 36xy = D\\ trHf = 6(x+y) $$ The 4 critical points are $(\pm 2,\pm 3)$

For(2,3):- $D>0 ,f_{xx}>0 \implies$ minimum

For(2,-3):- $D<0 \implies$ saddle point

For(-2,3):- $D<0 \implies$ saddle point

For(-2,-3):- $D>0,f_{xx}<0 \implies$ maximum

Is this correct?

I have a question saying to prove, $detHf<0,trHf>0 \implies minimum $ (I already posted this earlier - still no answer) So in that case, (-2,3) would be a minimum where earlier it was a saddle point. Why is it? What is the difference?

Answer 1

Your calculations are correct. I've answered your last question in your last post, $det Hf <0$ cannot give you a minimum or a maximum, asking you to prove that is wrong.

Answer 2

Your result seems correct. About the added question. We have a minimum if the Hessian is positive definite at the stationary point. In your case this means that the two eigenvalues of the Hessian are positive and, since the eigenvalues a re the solutions of the equation $$ \lambda^2- \lambda\mbox{tr} Hf +\det Hf=0 $$ They can be positive only if $\det Hf$ is positive, and also $\mbox{tr} Hf>0$.