[Math] Find the extrema of $\sin(x)\cos(y)$ with the Hessian

hessian-matrixpartial derivativequadratic-formstrigonometry

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:

  1. $\cos(x) = \sin(y) = 0$;

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

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?)

Best Answer

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).

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