2
$\begingroup$

Let $f(x,y) = 4x^3y - 4xy^3$. This has first derivatives

$$f_x = 12x^2y - 4y^3$$

and

$$f_y = 4x^3 - 12xy^2$$

from which we can conclude that there is an extremum at $(0,0)$. Upon attempting to classify this extremum we obtain a discriminant of $D = 0$ and have to proceed by using Taylor's Theorem. My problem is that even to the fourth order and above I keep getting zeroes everywhere and am not learning anything about the function. Is there a point at which I should just stop and conclude that I can't say anything about the extreme point or am I doing something wrong? I'm expanding

$$f(\varepsilon \cos\theta, \varepsilon\sin\theta) = \sum_{k=0}^{\infty}\frac{1}{k!}\left( \varepsilon\cos\theta\frac{\partial}{\partial x} + \varepsilon\sin\theta\frac{\partial}{\partial y}\right)^kf\bigg\vert_{(0,0)}$$

and getting zeroes at all orders.

  • 0
    How do you conclude it is an extremum? – 2017-01-10
  • 0
    Solving the equations $f_x = 0$ and $f_y = 0$ simultaneously. I was under the impression that this is how to do such problems! – 2017-01-10
  • 2
    It means it is a critical point,. An extremum requires more, just like in the one-variable case: the function $x^3$ has its first derivative v$0$ at $0$, but the function is increasing, and it has no maximum and no minimum on $\mathbf R$. – 2017-01-10
  • 0
    @Bernard I see, so I am to conclude that the point at $(0,0)$ is a saddle? Looking at the graph this makes sense.. – 2017-01-10
  • 0
    You can study the sign of the function and conclude that $(0,0)$ is a saddle point. – 2017-01-10
  • 0
    I would check that, for the intersections with planes $y=tx$, one obtains curves that have either a maximum or a minimum, depending on the value of $t$. – 2017-01-10

1 Answers 1

1

As suggested in the comments above, consider $y = tx$. Then, $f(x, tx) = 4tx^4 - 4t^3x^4 = 4tx^4(1-t^2)$. Taking first-order conditions, we get $f_x = 16tx^3(1-t^2)$. This is equal to zero obviously when $x = 0$. However, notice that $f_{xx} = 48 t x^2(1-t^2)$, which is non-positive everywhere when $t > 1$ and non-negative everywhere when $0 < t < 1$. Thus we conclude that $(x, y) = (x, tx) = (0, 0)$ is a saddle point.