For the function

$ f(x,y) = 1 - (x^2 + y^2)^{2/3} $

one has to find extrema and saddle points. Without applying much imagination, it is obvious that the global maximum is at $ (0,0)$.

To prove that, I set up the Jacobian as

$$ Df(x,y) = \left( -\frac{4}{3} x (x^2 y^2)^{-1/3}, -\frac{4}{3} y (x^2 y^2)^{-1/3}\right)$$

The only solution for $ Df(x,y) = 0$ yields indeed $(0,0)$. In this point the Hessian is $$ D^2f(0,0) = \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right)$$

which is both positive and negative semi-definite, suggesting that $(0,0)$ is both a minimum and a maximum - can that be? Or is it actually a saddle point? Thanks for your hints!

  • 1
    $\begingroup$ I believe you omitted the addition signs in typing your derivatives, but probably not in your work. While it is true that the limits of the first derivatives are zero at $ \ (0,0) \ $ , those derivatives are undefined there. The "cusp" is likely a source of trouble for this technique of characterizing critical points; the point is unquestionably a global maximum, however. $\endgroup$ – colormegone Jun 14 '13 at 20:03
  • 3
    $\begingroup$ The positive or negative definedness of the Hessian is only a sufficient condition for the point being a minimum or a maximum respectively. And it holds when the Hessian is defined, of course. $\endgroup$ – egreg Jun 14 '13 at 20:16

When analyzing this example forget about derivatives and Hessians. It is obvious that the function $$g(r):=1-r^{4/3}\qquad(r\geq0)$$ takes its maximum at $r=0$ and then strictly decreases with increasing $r$.

Introducing polar coordinates $(r,\phi)$ in the $(x,y)$-plane your function $$f:\ (x,y)\mapsto 1-(x^2+y^2)^{2/3}$$ appears as $$\tilde f(r,\phi)=g(r)\ .$$ Therefore the graph of $f$ resembles a "paraboloid looking downwards", and we just have the global maximum you spotted right away and no saddle points whatever.

Since $f(x,0)=1-x^{4/3}$ $\>(x\geq0)$ the given function is not twice differentiable at $(0,0)$; therefore the Hessian is not even defined there.

  • $\begingroup$ No need for $r\ge 0$ since $r^{4/3}=(r^2)^{2/3}$. $\endgroup$ – vadim123 Jun 15 '13 at 14:58

Simpler proof: $f(r,\theta)=1-r^{4/3}$, which has unique maximum at $r=0$, using ordinary derivatives (no partials necessary).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.