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

Question

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?)

Answer 1

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).

Answer 2

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 $2)$?

Answer 3

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.