Multivariable calculus

Critical Points of sin(x) sin(y)

Q: Why is the second derivative test never inconclusive on this surface?

Because D only ever takes the values -1 and 1 at a critical point. On the \sin x = \sin y = 0 grid, f_{xx} = f_{yy} = 0 and f_{xy} = \pm 1, giving D = -1. On the \cos x = \cos y = 0 grid, f_{xy} = 0 and f_{xx} = f_{yy} = \pm 1, giving D = 1. Since D is never zero there, the test always decides.

Q: Are the local maxima also global maxima?

Yes. Both factors are bounded by 1 in absolute value, so |f| \le 1 everywhere, and f\left(\frac{\pi}{2}, \frac{\pi}{2}\right) = 1 reaches that ceiling. The same argument makes every local minimum a global minimum with value -1. A bounded periodic surface like this one has no distinction between local and global at its extreme values.

The critical points of f(x,y) = sin(x) sin(y) fall into two families. Where x and y are both integer multiples of pi, D = -1 and the point is a saddle, as at the origin. Where both are odd multiples of pi/2, D = 1: (pi/2, pi/2) is a local max of value 1 and (pi/2, -pi/2) is a local min of value -1.

f(x,y) = \sin x \, \sin y

$\left(0, 0\right)$ saddle pointdiscriminant D = -1
$\left(1.5707963267948966, 1.5707963267948966\right)$ local maximumdiscriminant D = 1
$\left(1.5707963267948966, -1.5707963267948966\right)$ local minimumdiscriminant D = 1

These are not necessarily all of them; the page says which it covers.

Setting both partials to zero at once

A critical point needs $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ to vanish at the same point. Differentiate one variable at a time, holding the other fixed, and each derivative comes out as a single product.

\frac{\partial f}{\partial x} = \cos x \, \sin y, \qquad \frac{\partial f}{\partial y} = \sin x \, \cos y

A product is zero when a factor is zero, so each equation splits in two. The first forces $\cos x = 0$ or $\sin y = 0$ . The second forces $\sin x = 0$ or $\cos y = 0$ . That is four combinations to test, and two of them die immediately: $\cos x$ and $\sin x$ are never zero at the same $x$ , and the same goes for $\cos y$ and $\sin y$ .

The two surviving combinations are the entire critical set.

$\cos x = 0$ and $\cos y = 0$ , so $x = \frac{\pi}{2} + m\pi$ and $y = \frac{\pi}{2} + n\pi$ for integers $m$ and $n$ .
$\sin x = 0$ and $\sin y = 0$ , so $x = m\pi$ and $y = n\pi$ .

Both families are infinite grids, which is why the points listed on this page are one representative of each behaviour rather than a finished catalogue.

One discriminant for each grid

Because $f$ is a function of $x$ alone times a function of $y$ alone, the second partials are short.

f_{xx} = -\sin x \, \sin y, \qquad f_{yy} = -\sin x \, \sin y, \qquad f_{xy} = \cos x \, \cos y

On the grid where $\sin x = \sin y = 0$ , both diagonal entries vanish and $f_{xy} = \pm 1$ , so $D = -f_{xy}^{2} = -1$ at every one of those points. A negative discriminant is a saddle and no further work is needed.

D(0,0) = (0)(0) - (1)^{2} = -1 < 0

On the grid where $\cos x = \cos y = 0$ the mixed partial is zero and $f_{xx} = f_{yy} = -\sin x \sin y = -f$ . At $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$ that value is $-1$ , giving $D = 1 > 0$ with $f_{xx} < 0$ : a local maximum, and $f = 1$ there. At $\left(\frac{\pi}{2}, -\frac{\pi}{2}\right)$ the product $\sin x \sin y$ flips sign, so $f_{xx} = 1 > 0$ and the point is a local minimum with $f = -1$ .

Those two values are also the global extremes. Since $|\sin x| \le 1$ and $|\sin y| \le 1$ , the product never leaves $[-1, 1]$ , and both bounds are attained.

The mistake students make

The usual wrong turn is treating the two equations one at a time instead of together. A student reads $\cos x \sin y = 0$ , writes $x = \frac{\pi}{2}$ , then reads $\sin x \cos y = 0$ , writes $x = 0$ , sees a contradiction and concludes there are no critical points. The contradiction is manufactured: the two equations can be satisfied by different factors, and enumerating the four cases is what keeps that straight.

Factor both equations before substituting anything.
List all four pairings of one factor from each equation.
Discard the pairings that ask a sine and a cosine of the same variable to vanish together.
Solve what is left, and remember that each solution repeats with period $\pi$ .

The second slip is calling the origin a minimum because $f(0,0) = 0$ and $f$ looks positive nearby. Check the sign in two directions: $f(0.1, 0.1) > 0$ but $f(0.1, -0.1) < 0$ . Both signs occur arbitrarily close to the origin, which is exactly what $D < 0$ is telling you. Near $(0,0)$ the surface behaves like $xy$ .

Frequently asked questions

Why is the second derivative test never inconclusive on this surface?

Because $D$ only ever takes the values $-1$ and $1$ at a critical point. On the $\sin x = \sin y = 0$ grid, $f_{xx} = f_{yy} = 0$ and $f_{xy} = \pm 1$ , giving $D = -1$ . On the $\cos x = \cos y = 0$ grid, $f_{xy} = 0$ and $f_{xx} = f_{yy} = \pm 1$ , giving $D = 1$ . Since $D$ is never zero there, the test always decides.

Are the local maxima also global maxima?

Yes. Both factors are bounded by 1 in absolute value, so $|f| \le 1$ everywhere, and $f\left(\frac{\pi}{2}, \frac{\pi}{2}\right) = 1$ reaches that ceiling. The same argument makes every local minimum a global minimum with value $-1$ . A bounded periodic surface like this one has no distinction between local and global at its extreme values.