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Exercise. Find and classify the critical points of $f(x,y)=\ln{(2+\sin{xy})}$.

I want to verify my work until now, and I also have a question about classifying the infinitely many critical points I'm getting. Here are the details:

To begin we compute $f_x(x,y)=\frac{y\cos{xy}}{\sin{xy}+2}$ and $f_y(x,y)=\frac{x\cos{xy}}{\sin{xy}+2}$. Then, to find the critical points, we solve the following system of equations:

$$\frac{y\cos{xy}}{\sin{xy}+2}=0$$ $$\frac{x\cos{xy}}{\sin{xy}+2}=0$$

which is equivalent to

$$y\cos{xy}=0\label{1}\tag{1}$$ $$x\cos{xy}=0\label{2}\tag{2}$$

From $\eqref{1}$ follows that $y=0$ or $\cos{xy}=0$.

If $y=0$, then $x=0$. Thus, (0,0) is a critical point.

If $\cos{xy}=0$, then $xy=\frac{\pi+4\pi{}n}{2} \Rightarrow x=\frac{\pi(4n+1)}{2y}$, for all $n\in\mathbb{Z}, y\ne{0}$. So, there are infinitely many critical points in the form $\big(\frac{\pi(4n+1)}{2y},y)$, for all $n\in\mathbb{Z}, y\ne{0}$.

Questions: Is this right? If so, how do I proceed to classify the critical points? (I know of the Hessian method, should I use it for all critical points with an analysis by cases?).

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    $\begingroup$ Don't you want $$2n\pi$ instead of $4n\pi$? There are two zeros in an interval of length $2\pi$ Other wise, it looks good to me. I don't think you need to look at the Hessians. What are the possible values of $\sin xy$ when $\cos xy=0?$ $\endgroup$
    – saulspatz
    Apr 2, 2018 at 5:34
  • $\begingroup$ I think you're correct about using $2n\pi$ instead of $4n\pi$, but I thought that $\cos{xy}=0$ implies $xy=\frac{\pi}{2}+2\pi{}n=\frac{\pi(4n+1)}{2}$. How do you get $\frac{\pi(2n+1)}{2}$? $\endgroup$
    – mathman
    Apr 2, 2018 at 6:14
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    $\begingroup$ It's $xy=\frac{\pi}{2}+n\pi$ How will get the zero at $\frac{3\pi}{2}?$ BTW, I think you may have to look at the Hessian at $(0,0)$ $\endgroup$
    – saulspatz
    Apr 2, 2018 at 6:20
  • $\begingroup$ What is the relevance of looking at the values of $\sin{xy}$ when $\cos{xy}=0$? $\endgroup$
    – mathman
    Apr 2, 2018 at 7:37
  • $\begingroup$ Please see the answer I've just posted. $\endgroup$
    – saulspatz
    Apr 2, 2018 at 15:05

1 Answer 1

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Maxima and minima of $\ln (2+\sin (xy))$ are same as those of $2+\sin (xy)$ because $\ln$ is strictly increasing. When $\cos (xy)=0$ either $\sin (xy)=1$ or $\sin (xy)=-1$. The former give local maxima and the latter give local minima.

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  • $\begingroup$ I can't see a way to use the last fact ($\sin{xy}=\pm{1}$) to find the critical points. What am I missing? $\endgroup$
    – mathman
    Apr 2, 2018 at 7:44
  • $\begingroup$ Maximum and minimum of $2+\sin x$ are 3 and 1 right? $\endgroup$
    – Kavi Rama Murthy
    Apr 2, 2018 at 10:05

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