# Critical points of $f(x,y) = \sin x \sin y, -\pi<x<\pi, -\pi<y<\pi$

$\frac{df}{dx} = \cos x\sin y = 0$

$\frac{df}{dy} = \cos y\sin x = 0$

$\cos x\sin y = \cos y\sin x$

$\frac{\cos x}{\sin x} = \frac{\cos y}{\sin y}$

$\cot x = \cot y$

$P_1 = (\pi/2,\pi/2)$

$P_2 = (-\pi/2,\pi/2)$

$P_3 = (\pi/2,-\pi/2)$

$P_4 = (-\pi/2,-\pi/2)$

$P5 = (0,0)$

This gives me that $P_1$ and $P_4$ are maximum, $P_2$ and $P_3$ are minimum and $P_5$ is saddle. But when testing with Wolfram Alpha I get something different:

http://www.wolframalpha.com/input/?i=local+maxima+senxseny+%2C+-pi%3Cx%3Cpi+%2C+-pi%3Cy%3Cpi

http://www.wolframalpha.com/input/?i=local+minima+senxseny+%2C+-pi%3Cx%3Cpi+%2C+-pi%3Cy%3Cpi

Did I do anything wrong?

EDIT: Can people please stop removing the second wolfram link? They aren't the same, one is for maximum point and the other is for minimum.

• I edited your title, changing $F$ to $f$, so its consistent with the body of your post. Cheers! Commented Jul 15, 2015 at 6:26

The only dubious thing I can see is the introduction of the equation $$\cot x = \cot y$$; this appears problematic since (i.) the cotangent functions have singularities in $$(-\pi, \pi)$$; and (ii.) I can see no way to infer the values of $$x$$ and $$y$$ given by $$P_1$$-$$P_5$$ from this equation; indeed, all it directly tells us is that $$x = y$$, $$x = y + \pi$$ or $$x = y -\pi$$ on $$(-\pi, \pi) \setminus \{ 0 \}$$. As far as issues with Wolfram Alpha are concerned, I can’t say much as I don’t know that much about it; but I did notice xsr’s input syntax appears different than Mary Star’s, so perhaps there is a clue to any mysteries.

These things being said, here’s my analysis of this problem:

As our O.P. xsr correctly showed,

$$f_x =\dfrac{\partial f}{\partial x} = (\cos x)(\sin y), \tag{1}$$

and similarly,

$$f_y = (\sin x)(\cos y). \tag{2}$$

Thus we seek those points

$$(x, y) \in (-\pi, \pi) \times (-\pi, \pi) \tag{3}$$

such that

$$(\cos x)(\sin y) = 0, \tag{4}$$

and

$$(\sin x)(\cos y) = 0. \tag{5}$$

Then from (4),

$$\cos x = 0 \;\; \text{or} \;\; \sin y = 0, \tag{6}$$

whilst from (5)

$$\sin x = 0 \;\; \text{or} \;\; \cos y = 0. \tag{7}$$

We recall that $$\sin x$$, $$\cos x$$ cannot simultaneously vanish, and likewise for $$\sin y$$, $$\cos y$$; thus if

$$\cos x = 0, \tag{8}$$

then $$\sin x \ne 0$$, so we must have

$$\cos y = 0; \tag{9}$$

it follows from (8) and (9) that

$$x, y = \pm \dfrac{\pi}{2}. \tag{10}$$

Next, if

$$\cos x \ne 0, \tag{11}$$

(4) yields

$$\sin y = 0, \tag{12}$$

hence

$$\cos y \ne 0, \tag{13}$$

hence, from (5),

$$\sin x = 0. \tag{14}$$

The only solutions to (12), (14) with $$x, y \in (-\pi, \pi)$$ are

$$x = y = 0. \tag{15}$$

We have thus covered all possible cases according to whether $$\cos x = 0$$ or not; we see the critical points are

$$(x, y) = \left(\pm \dfrac{\pi}{2}, \pm \dfrac{\pi}{2}\right), (0, 0); \tag{16}$$

we next proceed to classify them according to type. We do this by evaluating the Hessian of $$f$$, $$H_f$$:

$$H_f(x, y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} -(\sin x)(\sin y) & (\cos x)(\cos y) \\ (\cos x)(\cos y) & -(\sin x)(\sin y) \end{bmatrix}. \tag{17}$$

From (17), we calculate that

$$H_f\left(\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) = H_f\left(-\dfrac{\pi}{2}, -\dfrac{\pi}{2}\right) = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}; \tag{18}$$

$$H_f\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) = H_f\left(\dfrac{\pi}{2}, -\dfrac{\pi}{2}\right) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \tag{19}$$

and

$$H_f(0, 0) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}. \tag{20}$$

We immediately see from (18), (19) that the points $$P_1$$, $$P_4$$ are local maxima, whilst $$P_2$$, $$P_3$$ are local minima; these assertions follow from the manifest fact that the eigenvalues of $$H_f(P_1)$$ and $$H_f(P_4)$$ are $$-1, -1$$, and those of $$H_f(P_2)$$, $$H_f(P_3)$$ are $$1, 1$$; as for $$P_5 = (0, 0)$$͵ the characteristic polynomial of $$H_f(P_0)$$ is

$$\det(H_f(0, 0) - \lambda I) = \det\left(\begin{bmatrix} -\lambda & 1 \\ 1 & -\lambda \end{bmatrix}\right)$$ $$= \lambda^2 - 1; \tag{21}$$

the eigenvalues are thus $$\lambda = \pm 1$$, being the roots of $$\lambda^2 - 1 = 0$$; $$P_5$$ is indeed a saddle point.

It is correct!!

For the minimum

For the maximum

Locating maximum and minimum values are straight forward as also replied by others.

• I think you're talking about saddles of the shape operator in a region where $K < 0$, which are not the same as saddles of a function such as $f$. Cheers! Commented Jul 15, 2015 at 6:46