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I am trying to find the critical points of the function:

$f(x,y)=2x^4-3x^2y+y^2$

and find the Max, Min and saddle points. What I've done so far is:

$f_x=8x^3-6xy=0 , f_y=-3x^2+2y=0 , f_{xx}=24x^2-6y , f_{yy}=2 , f_{xy}=-6x$

So (0,0) is the only critical point. But using the second partial derivative test:

$\Delta(0,0)=f_{xx} . f_{yy} - f^2_{xy}=0$

Which is inconclusive. Without using MATLAB or similar software and based on calculation, how can we determine whether (0,0) is Max, Min or saddle point?

Or the general question is what to do when it's inconclusive (without using software)?

beep-boop
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Sam
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4 Answers4

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Plug in $x = \epsilon$, $y = 3\epsilon^2$ and you will get that the function is greater than $0$ for all $\epsilon > 0$. Plug in $x = (3/4) \epsilon$, $y = \epsilon ^2$ and you will get that the function is less than $0$ for all $\epsilon > 0$. This implies that $(0,0)$ must be a saddle point because you take take $\epsilon > 0$ arbitrarily small.

user2566092
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  • If $x=\dfrac{2\epsilon}{3}$ and $y=\epsilon^2$, then $f(x,y)=\dfrac{5\epsilon^4}{81}$, which is positive, not negative. – Steve Kass Apr 13 '14 at 21:43
  • Your suggestion works if you change $\frac{2}{3}$ to $\frac{3}{4}$. For any $\epsilon>0$, $,f(\dfrac{3\epsilon}{4},\epsilon^2)=-\dfrac{7\epsilon^4}{128}$. – Steve Kass Apr 13 '14 at 21:59
  • How did you calculate y if $x=\dfrac{3\epsilon}{4}$? Shouldn't we use the equation $y=\dfrac{3x^2}{2}$? @SteveKass – Sam Apr 13 '14 at 22:06
  • I didn’t calculate $y$. I calculated $f(x,y)=2x^4-3x^2y+y^2$ at the point $(x,y)=(\dfrac{3\epsilon}{4},\epsilon^2)$. – Steve Kass Apr 13 '14 at 22:08
  • So you chose an arbitrary value ($\varepsilon^2$) for y? (I can't figure out how you got $y=\varepsilon^2$) @SteveKass – Sam Apr 13 '14 at 22:11
  • I was following up on @user2566092's answer. They didn’t “get” $y=\epsilon^2$, they chose it. It turns out that close to zero, you can use numerical tools to see that $f(x,y)$ is less than zero for points like $(0.03162, 0.001491)$, where it looks like $y$ is around $\frac{3}{2}x^2$. So it makes sense to see if for points like $(a,\tfrac{3}{2}a^2)$, you always get negative values of $f$, which you do. It turns out that $f(a,\tfrac{3}{2}a^2)=-\tfrac{a^4}{4}$, so there are points arbitrarily close to $(0,0)$ (take $a$ very small) at which the value of $f$ is negative. – Steve Kass Apr 13 '14 at 22:21
  • Thanks for pointing out the problem and the fix @SteveKass . I've updated my answer to reflect it. – user2566092 Apr 14 '14 at 14:30
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Hint: $$2x^4−3x^2y+y^2=(x^2-y)(2x^2-y).$$

Martín-Blas Pérez Pinilla
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Possibly easier hint (complete the square): $$ 2x^4 - 3x^2 + y^2 = (y-\frac32 x^2)^2 - \frac14x^2 $$

mrf
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Added: Here's a way not to solve the problem (because it gives the wrong answer!)

Look at the concavity of the function at $(0,0)$ in a non-axis direction $y=ax$. $$\left.\frac{d^2}{dx^2}\right\rvert_{x=0}\left(f(x,ax)\right) = \left.\frac{d^2}{dx^2}\right\rvert_{x=0}\left(2x^4-3x^2 (a x)+(a x)^2\right)=\left. 2a^2-18ax+24x^2\right\rvert_{x=0}=2a^2.$$ This tells you that the function is concave up at $(0,0)$ in every non-axis direction, so $(0,0)$ is not a saddle point. It's not a local maximum, because $f_{yy}(0,0)>0$, so it’s a local minimum.

(The right answer is that this is a saddle point. See the answer @user2566092 gave, and the correction in the comments, if the answer hasn't been fixed.)

Steve Kass
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  • Wait sorry why doesn't this method work? Is it because you've only considered the concavity of the function in a non-axis linear direction, (so it looks concave up in every linear direction) however its like concave down along some non-linear curves? – QCD_IS_GOOD May 08 '16 at 11:07
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    Sort of. I just wanted to point out that up-concavity in every linear direction through a critical point is not sufficient evidence for a local minimum. You cannot conclude that you have a local minimum $f(0,0)=0$ just because along every linear direction through $(0,0)$, the function is either constant or concave up — that’s not a theorem. (This function is a counterexample to show that that conclusion is not valid in general.) You have here a function that is never concave down in a linear direction through $(0,0)$ but nevertheless has points arbitrarily close to zero where $f(x,y) < 0$. – Steve Kass May 09 '16 at 01:37