Looking to find critical points and classify them as max/min or saddles for the following multivariate function


Computed the partial derivatives with respect to $x$ and $y$ and equated them to zero and got the following critical points $(-4\sqrt(3),0), (4\sqrt(3),0),(0,-4),(0,4)$

Using the formula for second derivative test I found that all critical points above were saddle points except for $(0,4)$ which turned out to be a min as $D(0,4)>0$ and $f_{xx}>0$.

After doing so I found that apparently there is still something wrong with this answer. Can anyone give a hint as to what that is as I have spent a long time at this and still can't find it?


$$\frac{\partial^2f}{\partial x^2}\cdot\frac{\partial^2f}{\partial y^2}-\left(\frac{\partial^2f}{\partial x\partial y}\right)^2=2y\cdot6y-(2x)^2=4(3y^2-x^2),$$ which gives that $(\pm4\sqrt3,0)$ are not extreme points.

Now, $$\frac{\partial^2f}{\partial x^2}(0,4)>0,$$ which gives that $(0,4)$ is a minimum point and $$\frac{\partial^2f}{\partial x^2}(0,-4)<0,$$ which gives that $(0,-4)$ is a maximum point.

  • 1
    $\begingroup$ Thank you very much! I see what I was doing wrong was just basic algebra. The propensity of Math to make me feel stupid is seemingly infinite :P $\endgroup$ – esc123 Jul 23 at 18:52

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