# Find saddle point/local min max of $f(x,y)= (x^2-64)^2-(y^2-16)^2$

Question: Find local minima/maxima or saddle points of $$f(x,y)= (x^2-64)^2-(y^2-16)^2$$ $f_x=4x^3-256=0 \longrightarrow x=0,-8,8;$ $f_y=-4y^3+64=0 \longrightarrow y=0,-4,4$

I am confused about the critical points. Are there 9 critical points? That is, (0,0),(0,-4),(0,4),(-8,0),(-8,-4),(-8,4),(8,0),(8,-4),(8,4)?

$D=f_{xx}*f_{yy}-[f_{xy}]^2<0$ for all 9 critical values. So these are all saddle points. This just seems unusual that a function has 9 critical points. So, any feedback is much appreciated.

• hint: look at the graph and level curves at wolframalpha.com/input/…. – EDZ Feb 19 '18 at 0:34
• I would do those derivatives again. I'd probably start by expanding the brackets. – Theo Bendit Feb 19 '18 at 0:40
• $$f_x = 4 x (-64 + x^2), f_y = -4 y (-16 + y^2)$$ There are nine critical points. Am I missing something? – Moo Feb 19 '18 at 1:15
• @user477343 Partial derivatives of $f$ with respect to $x$ and $y$ respectively. – Theo Bendit Feb 19 '18 at 2:22
• @TheoBendit so essentially,$$f_x := \frac{\partial}{\partial x}\, f(x)$$ with the same applying for $y$? – Mr Pie Feb 19 '18 at 2:28

Yes, there are nine stationary points. You are looking for points at which $f_x$ and $f_y$ take the value $0$ simultaneously. Computing the partial derivatives yields \begin{align*} f_x(x, y) &= 4x(x^2 - 64) \\ f_y(x, y) &= 4y(y^2 - 16). \end{align*} Solving $f_x(x, y) = 0$ yields $x = 0$ or $x = 8$ or $x = -8$. This doesn't say anything about $y$, so from this equation alone, $y$ can be any number.
On the other hand, solving $f_y(x, y) = 0$ yields $y = 0$ or $y = 4$ or $y = -4$, and $x$ can be any number. The only points at which both equations are simultaneously solved are:
If you're still not sure, try plugging in all nine of these points into $f_x(x, y)$ and $f_y(x, y)$ and verify that both are simultaneously $0$ (the definition of a stationary point).