Local maxima and minima of $f(x,y,z)=(y+z)^2+(x+z)^2+xyz$

Question

I saw this exam problem but I'm having trouble how to determine the local maxima and the minima of this function.

This is what I did.

I found $f_x=2x+2z+yz \\ f_y=2y+2z+xz \\ f_z=4z+2y+2x+xy$

One stationary point I know it's $A(0,0,0)$ but I can't find the others. I thought of using Sylvester's criterion.

$\Delta _1=f_{xx}=2 \gt 0 \\$

$\Delta_2= \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\\\end{vmatrix} $

$ \Delta _3= \begin{vmatrix} f_{xx} & f_{xy} & f_{xz} \\ f_{yx} & f_{yy} & f_{yz} \\ f_{zx} & f_{yz} & f_{zz} \\ \end{vmatrix}$

For the point, $A$ I get $\Delta _2(A)=4 \gt 0$ and $\Delta _3(A)=0$. This is where I got stuck and I don't know what I should do. Can somebody help me find the other stationary points and help me to underestand what to do when I have the case $\Delta _3=0$?

Answer 1

$f(x,x,-x)=-x^3$ and $f(0,0,0)=0$. This shows: in each neighborhood of $(0,0,0)$ the function $f$ takes values $<0$ and values $>0$.

Consequence: in $(0,0,0)$ the function $f$ does not have a local extremum.

Answer 2

All such candidates for maxima/minima can be found by solving$$f_x=f_y=f_z=0$$From $f_x=f_y=0$ we obtain$$2(x-y)=(x-y)z\\2x+2z+yz=0$$then either$$z=2$$which leads to $$x+y+2=0\\2x+2y+xy+8=0\\\implies xy=-4$$or finally$$(x_1,y_1,z_1)=(\sqrt 5-1,-\sqrt 5-1,2)\\(x_2,y_2,z_2)=(-\sqrt 5-1,\sqrt 5-1,2)$$The other candidates can be found$$x=y\\2x+2z+xz=0\\x^2+4x+4z=0$$$$x^2=4x$$therefore the candidates are$$(x_3,y_3,z_3)=(0,0,0)\\(x_4,y_4,z_4)=\left(-6,-6,-3\right)$$by investigating the eigenvalues of $H$ we obtain $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ and $(x_4,y_4,z_4)$ are saddle points and $H(0,0,0)\succeq 0$ which can not be decided upon since the second order sufficient condition (SOSC) is not fulfilled.

Answer 3

The second order condition $\Delta_3=0$ implies the test is inconclusive and one must use other techniques to classify the stationary point.

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A simple example with $y=x^3$: $$y'=3x^2=0\Rightarrow x=0 \quad \text{(critical point)}\\ y''(0)=6x|_{x=0}=0 \quad \text{(inconclusive)}$$ One method is to look at the small neighborhood of the critical point $x=0$: $$f(0+\epsilon)=(0+\epsilon)^3=\epsilon^3>0\\ f(0-\epsilon)=(0-\epsilon)^3=-\epsilon^3<0$$ It shows that $f(0)=0^3=0$ is not a local maximum nor minimum, but an inflection point.

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Similarly, given $f(x,y,z)=(y+z)^2+(x+z)^2+xyz$, consider the small neighborhood of the critical point $(x,y,z)=(0,0,0)$: $$\begin{align}f(0+\epsilon,0+\epsilon,0+\epsilon)&=(2\epsilon)^2+(2\epsilon)^2+\epsilon^3>0\\ f(0+\epsilon,0+\epsilon,0-\epsilon)&=(0+\epsilon+0-\epsilon)^2+(0+\epsilon+0-\epsilon)^2+(0+\epsilon)(0+\epsilon)(0-\epsilon)=-\epsilon^3<0.\end{align}$$ It shows that $f(0,0,0)=0$ is not a local maximum nor minimum, but a saddle point.