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Let $\mathbb{D}=\{(x,y) \mid xy<1\}$ and $f:\mathbb{D}\rightarrow \mathbb{R}$, $f(x,y)=\sqrt{1-xy}$.

I want to find the stationary points of $f$.

I have done the following:

The partial derivatives are: \begin{align*}f_x=\frac{-y}{2\sqrt{1-xy}}\\ f_y=\frac{-x}{2\sqrt{1-xy}}\end{align*}

The roots are: \begin{align*}f_x=0 \Rightarrow \frac{-y}{2\sqrt{1-xy}}=0 \Rightarrow y=0 \in \mathbb{D}\\ f_y=0 \Rightarrow \frac{-x}{2\sqrt{1-xy}}=0 \Rightarrow x=0 \in \mathbb{D}\end{align*}

Therefore, the stationary point is $(0,0)$.

According to the second order condition we have that \begin{equation*}f_{xx}(0,0)=0 , \ f_{yy}(0,0)=0 , \ f_{xy}(0,0)=f_{yx}(0,0)=-\frac{1}{2}\end{equation*}

So, we get \begin{equation*}f_{xx}(0,0)f_{yy}(0,0)-\left (f_{xy}(0,0)\right )^2=-\left (-\frac{1}{2}\right )^2=-\frac{1}{4}<0\end{equation*}

This means that $(0,0)$ is a saddle point, right?

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We consider a bigger domain: $\mathbb{D}=\{(x,y) \mid xy\leq 1 \}$.

From above we have shown that the function has no extrema in the set $\{(x,y) \mid xy< 1 \}$. So we have to check if there are extrema at $xy=1$, right?

We have that $x=\frac{1}{y}$. Replacing this in the function we get \begin{equation*}h(y)=f\left (\frac{1}{y}, y\right )=\sqrt{1-\frac{1}{y}\cdot y}=\sqrt{1-1}=0\end{equation*}

Every point of a constant function is a global maximm and a global minimum.

Therefore, the function $f(x,y)$ has at the points $\left (\frac{1}{y}, y\right )$ global extrema.

Is this correct?

What are strict global extrema?

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No. The region $$\mathbb{D}=\{(x,y) \mid xy\leq 1 \}$$ is unbounded, and thus $f$ may be unbounded on $\mathbb D$.

Indeed, let $x_n=n$ and $y_n=-1$. Then $(x_n,y_n)\in\mathbb D$. But $$f(x_n,y_n)=\sqrt{1+n}\to\infty.$$

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  • $\begingroup$ So, it has also no extrema at this set? $\endgroup$ – Mary Star Apr 19 '17 at 12:18
  • $\begingroup$ It has an obvious lower bound $f\ge 0$, and no upper bound as I have shown. $\endgroup$ – Eclipse Sun Apr 19 '17 at 12:20
  • $\begingroup$ Ah ok. So, the function has a minimum that is equal to $0$. This minimum is at all the points $(\frac{1}{y}, y)$, or not? Is this minimum global? $\endgroup$ – Mary Star Apr 19 '17 at 12:25
  • $\begingroup$ That's correct. $\endgroup$ – Eclipse Sun Apr 19 '17 at 12:27
  • $\begingroup$ Ah ok!! What are strict global extrema? $\endgroup$ – Mary Star Apr 19 '17 at 12:31

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