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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!

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ No need for $r\ge 0$ since $r^{4/3}=(r^2)^{2/3}$. $\endgroup$ – vadim123 Jun 15 '13 at 14:58
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Simpler proof: $f(r,\theta)=1-r^{4/3}$, which has unique maximum at $r=0$, using ordinary derivatives (no partials necessary).

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