Find the extrema of $\sin(x)\cos(y)$ with the Hessian I've got the following function: $f(x,y) = \sin(x)\cos(y)$.
I performed the Hessian matrix correctly, with the second derivatives:
\begin{bmatrix}
-\sin(x)\cos(y) & -\cos(x)\sin(y)\\
-\cos(x)\sin(y) & -\sin(x)\cos(y)
\end{bmatrix}
But the trouble comes when I want to find the extrema (critical points and saddle points). By solving $\cos(x)\cos(y) = 0$ and $-\sin(x)\sin(y) = 0$ (the first partial derivatives) I obtain:

$\cos(x) = 0$ or $\cos(y) = 0$
$\sin(x) = 0$ or $\sin(y) = 0$.

If $\cos(x) = 0$, then $\sin(x) = \pm 1$ and if $\sin(x) = 0$ then $\cos(x) = \pm1$. Well, here I think that we have got just two possibilities:


*

*$\cos(x) = \sin(y) = 0$;

*$\sin(x) = \cos(y) = 0$.
Now, how do I find the extrema?
I mean, with $\cos(x) = \sin(y) = 0$, which is the point $(x, y)$ and how do I substitute it in the Hessian matrix? (The solution says it must a saddle point but I don't get it, why?)
 A: The critical points are


*

*$f_x=\cos x \cos y=0$

*$f_y=-\sin x \sin y=0$


and thus


*

*$x=k\pi \quad y=\frac{\pi}2+j\pi$

*$y=k\pi \quad x=\frac{\pi}2+j\pi$


the Hessian matrix is
$$\begin{bmatrix}
-\sin x \cos y & -\cos x \sin y \\
-\cos x \sin y & -\sin x \cos y 
\end{bmatrix}$$
and for $x=k\pi \quad y=\frac{\pi}2+j\pi$ we obtain
$$H_1=\begin{bmatrix}
0 & 1 \\
1 & 0 
\end{bmatrix} \quad H_2=\begin{bmatrix}
0 & -1 \\
-1 & 0 
\end{bmatrix}$$
and for $y=k\pi \quad x=\frac{\pi}2+j\pi$ we obtain
$$H_3=\begin{bmatrix}
1 & 0 \\
0 & 1 
\end{bmatrix} \quad H_4=\begin{bmatrix}
-1 & 0 \\
0 & -1 
\end{bmatrix}$$
and $H_1$,$H_2$ have signature (1,1), $H_3$ have signature (2,0) and $H_4$ have signature (0,2).
A: Find the extrema of $f(x,y) = \sin(x)\cos(y)$.
$$\begin{cases} f_x=\cos x\cos y=0 \\ f_y=-\sin x\sin y=0\end{cases} \Rightarrow 1)\begin{cases}\cos x=0 \\ \sin y=0\end{cases} \ \ \text{or} \ \ 2)\begin{cases}\cos y=0 \\ \sin x=0\end{cases}.$$
The Hessian is:
$$H=\begin{bmatrix}
-\sin x \cos y & -\cos x \sin y \\
-\cos x \sin y & -\sin x \cos y 
\end{bmatrix}\\
H_1=-\sin x\cos y; \ \ \ H_2=\sin^2 x\cos^2y-\cos^2x\sin^2y.$$
For $1)$: 
$$a) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=1 \\ \cos y=1\end{cases} \Rightarrow H_1=-1<0, H_2=1>0 \Rightarrow max; \\ 
b) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=-1 \\ \cos y=1\end{cases} \Rightarrow H_1=1>0, H_2=1>0 \Rightarrow min; \\
c) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=1 \\ \cos y=-1\end{cases} \Rightarrow H_1=1>0, H_2=1>0 \Rightarrow min; \\
d) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=-1 \\ \cos y=-1\end{cases} \Rightarrow H_1=-1<0, H_2=1>0 \Rightarrow max. \\$$
Can you handle $2)$?
A: You compute the Hessian wrongly; it is actually
$$
H(x,y)=
\begin{bmatrix}
-\sin(x)\cos(y) & -\cos(x)\sin(y)\\
-\cos(x)\sin(y) & -\sin(x)\cos(y)
\end{bmatrix}
$$
At the points where $\cos x=0$ and $\sin y=0$, the determinant of the Hessian is $1$, so these are either maxima or minima (the eigenvalues are either both positive or both negative).
At the points where $\sin x=0$ and $\cos y=0$, the determinant of the Hessian is $-1$, so these are saddle points (one eigenvalue is positive, the other one is negative).
If you limit yourself to $x$ and $y$ in $[0,2\pi)$, which is simpler, the points are
\begin{align}
&(\pi/2,0),\quad
(\pi/2,\pi),\quad
(3\pi/2,0),\quad
(3\pi/2,\pi)\\
&(0,\pi/2),\quad
(\pi,\pi/2),\quad
(0,3\pi/2),\quad
(\pi,3\pi/2)
\end{align}
Can you assign each of these points the property of being maximum, minimum or saddle?
Since the function is periodic of period $2\pi$ in each variable, adding integer multiples of $2\pi$ to the points above will give the whole set.
