Find the extrema and saddle points of $f$
$f(x,y)=e^x \sin(y)$
My attempt:
$f_x=e^x \sin(y)$
$f_y=e^x \cos(y)$
Since $f_x$ and $f_y$ exist for all $(x,y)$ then the only critical points are the solutions of the following system:
$e^x \sin(y)=0$
$e^x \cos(y)=0$
But I can’t solve this system, can you help please?
Thanks.
Best Answer
You are quite close to the answer:
Since you've already deduced that the critical points are where the following equations hold: $$ \begin{align} e^x sin(y) &= 0\\ e^x cos(y) &= 0 \end{align} $$ Thus all that's left to do is to solve it. Consider what happens when we set $f_x$ to $0$:
$$ f_x = e^x sin(y) = 0 $$ Thus, $f_x$ takes on the value of zero when either $e^x = 0$ (not possible for any value of x, since $e^x > 0 \space \forall x$), or $sin(y)= 0$. Hence, we require $sin(y) = 0$. For which values of $y$ does this hold for? Well, it holds for: $$ y = k\pi, \quad k\in \mathbb{Z} $$ Applying the same treatment with $f_y$, we have $f_y = 0$ when $cos(y) = 0$, i.e. $$ y = (k +\frac{1}{2})\pi $$ Now, what this means geometrically is that there will not be any point $(x,y)$ in which the two equations $$ f_x = 0\\ f_y = 0 $$ hold, since $f_x \neq 0$ when $f_y = 0$, and $f_y \neq 0$ when $f_x = 0$.
Thus, there are no saddle points nor any critical points.