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I have the following exercise:

Define $f:\mathbb{R}^2\to\mathbb{R}$ by $f(x,y)=(x^2y)^{1/3}$. Is $f$ differentiable at $(0,0)?$. Has $f$ absolute extrema in $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}$?.

The function $f$ is continous and by the compactness of $D$ I know because of Weierstrass theorem that $f$ has absolute maxima and minima in $D$, but how to find them?. I'm considering the possibility of study the existence of critical points inside $D$ (when $x^2+y^2<1$) and for the boundary of $D$ separately, but for the last I'd use Lagrange multipliers and the calculations I'm doing doesn't seem right.

I have $\displaystyle\frac{\partial f}{\partial x}= \displaystyle\frac{2y^{1/3}}{3(x^2y)^{2/3}}=\displaystyle\frac{2y^{1/3}}{3x^{1/3}}$ and $\displaystyle\frac{\partial f}{\partial y}= \displaystyle\frac{x^2}{3(x^2y)^{2/3}}=\displaystyle\frac{x^{2/3}}{3y^{2/3}}$. Now define $g(x) = x^2+y^2-1=0$ in order to find extrema in $\partial D$ and for the last one $\displaystyle\frac{\partial g}{\partial x} = 2x$ and $\displaystyle\frac{\partial g}{\partial y}=2y$.

Using Lagrange multipliers I have to solve

$$\begin{cases} \frac{\partial f}{\partial x}(x,y) = \lambda\frac{\partial g}{\partial x}(x,y) \\ \frac{\partial f}{\partial y}(x,y) = \lambda\frac{\partial g}{\partial y}(x,y)\\ g(x,y)=0\end{cases}$$

which is

$$\begin{cases} \frac{2y^{1/3}}{3x^{1/3}} = 2\lambda x & (1)\\ \frac{x^{2/3}}{3y^{2/3}} = 2\lambda y & (2)\\ x^2+y^2=1 & (3) \end{cases}$$

Using $(1)$ I get $y$ in terms of $x,\lambda$ having $y^{1/3}=3\lambda x^{4/3}\implies y= 9\lambda ^3 x^4$; and by $(2)$ I have $x^{2/3}=6\lambda y^{5/3}\implies x= (6\lambda y ^{5/3})^{3/2}=6\sqrt{6}\lambda^{3/2}y^{5/2}$.

Then $y=9\lambda ^3 x^4 = 9\lambda^3 (6\sqrt{6}\lambda ^{3/2} y^{5/2})^4=9\cdot6^6\lambda ^9 y ^{10}\implies y^2=\displaystyle\frac{1}{9\lambda^3 (6\sqrt{6}\lambda ^{3/2} y^{5/2})^{4/9}}$.

Isn't this too weird as a solution?. I still would have to use it to find possibles $\lambda$ but it seems the former system has -if my calculations are correct- strange solutions; did I had another mistake?.

Cure
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  • What do you get if you do not use LMs? – Amzoti Jan 30 '14 at 04:23
  • Are you also asking about differentiability? From the looks of it, it isn't going to be. – IAmNoOne Jan 30 '14 at 04:24
  • Observe that for any real valued $f$, the extrema of $f$ and $f^{1/3}$ occur at the same points. – Jeff Snider Jan 30 '14 at 04:27
  • @user117818 I know, I proved that part taking a directional derivative and showing that it depends of the directional vector. One thing I did not understand was about the evaluation of partial derivatives, the book says that $f(0,y)=0$ then $\frac{\partial f}{\partial x}=0$ if $(x,y)=(0,0)$, but according to the partial derivative I calculated above, doesn't that mean that $\frac{\partial f}{\partial x}|_{(x,y)=(0,0)}$ doesn't exist?. – Cure Jan 30 '14 at 04:30
  • @Amzoti If I don't use LMs, how could I study the extrema of $f$ at the boundary of $D$? – Cure Jan 30 '14 at 04:34
  • @JeffSnider Thanks!, then I could just find the extrema of $x^2y$ which would make the work easier? – Cure Jan 30 '14 at 04:34
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    @Dante: See my answer here for a very similar example: http://math.stackexchange.com/questions/415486/find-the-absolute-minimum-and-absolute-maximum – Amzoti Jan 30 '14 at 04:37
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    @Dante, I can't think of a witty answer at the moment, but rigourously computing the limit of $f(\mathbf{h} ) / |\mathbf{h}|$ does show the limit does not exist, and hence no differentiable at $(0,0)$ – IAmNoOne Jan 30 '14 at 04:39

1 Answers1

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The minimizers and maximizers of $f(x,y)=(x^2y)^{1/3}$ are the same as for $g(x,y)=x^2y$.

We have $$\frac{\partial g}{\partial x}=2xy,$$ and $$\frac{\partial g}{\partial y}=x^2.$$ Observe that $g$ takes positive and negative values, so any stationary point where $g(x,y)=0$ is not a minimizer or maximizer, ruling out points where $x=0$ and $y=0$. Thus $g$ is unbounded above and below as we move away from the origin: the extrema will lie on the unit circle.

We now know $x^2=1-y^2$, giving us $g(y)=(1-y^2)y$, for which $g'(y)=-3y^2+1$. The stationary points are at $$y=\pm\frac{1}{\sqrt{3}},$$ and I think you can handle it from there.

Jeff Snider
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  • Thanks, but why $g(y)=(1-y^2)y$?. From $x^2=1-y^2$ I thought that would follow two functions given by $\sqrt{1-y^2}$ and $-\sqrt{1-y^2}$. – Cure Jan 30 '14 at 15:26
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    It's direct substitution of $x^2=1-y^2$ into $x^2y$, rather than writing $x=\pm\sqrt{1-y^2}$ then substituting into $x^2y$, for which the positive and negative possibilities of $x$ both give the same value of $g$. – Jeff Snider Jan 31 '14 at 20:38