Finding the maximum and minimum points of $f(x,y)=\frac{2\sin(y)}{1+(x-\pi)^2}$ over the region $R=\{(x,y) \in \mathbb{R}^2 : 0 \le x \le 2\pi, \ 0 \le y \le \pi\}$.

I have found $f_x=\frac{-4(x-\pi)\sin(y)}{(1+(x-\pi)^2)^2}$ and $f_y=\frac{2\cos(y)}{1+(x-\pi)^2}$

For $f_x=0$ it should be that $-4(x-\pi)\sin(y)=0$ so $x=\pi$ or $y=0, \pi$

For $f_y=0$, $\ 2\cos(y)=0$ so $y=\frac{\pi}{2}$

But how do I find the points themselves? And how do I evaluate the function in the "borders" delimited by $R$?

Thanks for your time!


The critical point is where both derivatives are zero, so it is $(\pi,\pi/2)$

For the rectangular borders, observe that $x$ is constant along the left and right borders, and $y$ is constant along the other 2, so you have $$ f(0,y) = f(2\pi,y) = \frac{2\sin y}{1+\pi^2} $$ $$ f(x,0) = f(x,\pi) = 0 $$

For two of these boundaries, the function is identically $0$, so you don't need to do anything else. For the other two, the critical points corresponds the critical points of $\sin y$, which is maximum when $y=\pi/2$. This gives you two other critical points on $(0,\pi/2)$ and $(2\pi,\pi/2)$

We can conclude that the maximum is on $f(\pi,\pi/2) = 2$, and the minima are all points on the boundaries $y=0$, $y=\pi$

  • $\begingroup$ Thank you! Just a detail, $x$ goes from $0$ to $2\pi$ so $f(0,y)=f(2\pi, y)=\frac{2\sin(y)}{1+\pi^2}$ in any case the result is the same :) $\endgroup$ Dec 17 '17 at 19:33
  • $\begingroup$ Thanks! I didn't read good. Updated answer. $\endgroup$
    – Dylan
    Dec 17 '17 at 22:32

The maximum is $2$ and this occurs when $x=\pi, y=\pi/2$ (note that $\sin y\leq 1$ for all $y$, and $(x-\pi)^2\geq 0$ for all $x$).

For $0\leq y\leq \pi$, $\sin y\geq 0$, so $\frac{\sin y}{1+(x-\pi)^2}\geq 0$. The minimum is $0$. This occurs whenever $y=0$ or $y=\pi$, regardless of the value of $x$.


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