Find critical points, minima and maxima of several variable function

Question

I'm having a problem while deciding which points I should take into consideration as critical points, the function I'm working with is:

$\ f(x,y)=(ax^2 +by^2)e^{(-x^2 -y^2)}$

I found that

$\ \frac{\partial f}{\partial x}=f_x=-2xe^{(-x^2 -y^2)}(ax^2 -a+by^2)$

$\ f_y=-2ye^{(-x^2 -y^2)}(ax^2 -b+by^2)$

After that, I tried to find the points where $\ f_y=f_x=0$ and found:

$\ f_x=0$ if and only if $\ y=±\frac{\sqrt(a-ax^2)}{\sqrt(b)}$ or $\ a=0$ and $\ b=0$ or $\ x=0$

$\ f_y=0$ if and only if $\ y=±\frac{\sqrt(b-ax^2)}{\sqrt(b)}$ or $\ a=0$ and $\ b=0$ or $\ y=0$

So, the criticals points are (I think):

$\ (0,\frac{\sqrt(b-ax^2)}{\sqrt(b)})$, $\ (0,-\frac{\sqrt(b-ax^2)}{\sqrt(b)})$, $\ (0,0)$

Question is, am I right? Are these all the critical points of $\ f$?

Answer 1

You are giving as critical points lines in the plane. It is so only when some $a$ or $b$ vanish.

It has five critical points assuming that $a\neq0\neq b$

$(0,0)$

With $f_x=0$, $y=\pm\sqrt{\dfrac{a-ax^2}{b}}$ or $x=0$. With the first, taking $x=\pm1$ implies directly that $f_y=0$. With $x=0$, the condition $f_y=0$ leads to $-b+by^2=0$ or $y=\pm1$.

With $f_y=0$, better isolate $x$, so we have $x=\pm\sqrt{\dfrac{b-by^2}{b}}$ or $y=0$. With the first, we get the previous result of $x=0$ chosing $y=\pm1$. With $y=0$, $f_x=0$ leads to the previous result too of $x=\pm1$

The critical points are $(0,0)\;,(-1,0)\;,(1,0)\;,(0,-1)\;,(0,1)$

With $a=0$ both partials are zero for the line $y=0$ and for $(-1,0)\;,(1,0)$

With $b=0$ both partials are zero for the line $x=0$ and for $(0,-1)\;,(0,1)$