Finding the critical points of $f(x,y) = \sin(x)\sin(y)$

Question

I am attempting to find the critical points of a function for local max min purposes and have gotten stuck.

The function is $$f(x,y) = \sin(x)\sin(y)$$

Bounded by $-\pi < x < \pi$ and $-\pi < y < \pi$.

I have the partial derivative wrt to $x$ as:

\begin{equation} \frac{\partial f}{\partial x} = \cos(x)\sin(y) \end{equation}

and the partial derivate wrt to $y$ as:

\begin{equation} \frac{\partial f}{\partial y} = \cos(y)\sin(x) \end{equation}

I now want $$\cos(y)\sin(x) = 0 = \sin(x)\cos(y)$$

For \begin{equation} \frac{\partial f}{\partial x} = 0 \end{equation}

Then $\sin(x) = 0$ when $x = 0$, OR $\cos(y) = 0$ when $y = -\pi/2$ and $\pi/2$.

For \begin{equation} \frac{\partial f}{\partial y} = 0 \end{equation}

then $\sin(y) = 0$ when $y = 0$, OR $\cos(x) = 0$ when $x = -\pi/2$ and $\pi/2$.

My question is how do I actually arrange these values into critical points properly?

Answer 1

Just arrange the critical points you already got from domain.. first to fourth quadrants. [ Red Max, Blue Min ] of range.

$$ (x_{crit}, y_{crit})= \dfrac{\pi}{2}[ (+1,+1),(-1,+1),(-1,-1),(+1,-1)] $$

Answer 2

Solve,

$$ \frac{\partial(\sin(x)\sin(y))}{\partial \{x,y \}^1} = 0 $$

for $x, y$ without considering the bounds gives:

$$x = \pi(n - \frac{1}{2}) \ \ and \ \ \cos(y) = 0 \ \ and \ \ n \in \mathbb{Z}$$

and

$$ x = n \pi \ \ and \ \ \sin(y) = 0 \ \ and \ \ n \in \mathbb{z} $$

Therefore, critical points within the bounds are $(\frac{-\pi}{2}, \frac{\pi}{2}) $, $(\frac{\pi}{2}, \frac{\pi}{2}) $, $(\frac{-\pi}{2}, \frac{-\pi}{2}) $, $(\frac{\pi}{2}, \frac{-\pi}{2})$ and $(0, 0)$