Multivariable calculus

Critical Points of x^4 + y^4 - 4xy

Q: Why do the two minima have exactly the same value?

Because f(-x,-y) = x^{4} + y^{4} - 4xy = f(x,y): the function is unchanged by a half turn about the origin. That symmetry carries (1,1) onto (-1,-1), so the two points must have the same height, -2, and the same discriminant, 128.

Q: Are these local minima also global minima?

Yes. The quartic terms dominate, so f \to +\infty in every direction and the global minimum has to be attained at a critical point. Among the three candidates, the heights are 0 at the origin and -2 at the other two, so -2 is the global minimum.

f(x,y) = x^4 + y^4 - 4xy has three critical points. The origin is a saddle with discriminant D = -16. The points (1,1) and (-1,-1) are local minima with D = 128 and f_xx = 12, and both sit at height -2, which is the global minimum of the function.

f(x,y) = x^{4} + y^{4} - 4xy

$\left(0, 0\right)$ saddle pointdiscriminant D = -16
$\left(1, 1\right)$ local minimumdiscriminant D = 128
$\left(-1, -1\right)$ local minimumdiscriminant D = 128

Chasing the system down to x^9 = x

\frac{\partial f}{\partial x} = 4x^{3} - 4y, \qquad \frac{\partial f}{\partial y} = 4y^{3} - 4x

Setting both to zero gives $y = x^{3}$ and $x = y^{3}$ . Substituting the first into the second gives $x = \left(x^{3}\right)^{3} = x^{9}$ .

x^{9} - x = x\left(x^{8} - 1\right) = 0

The real solutions are $x = 0$ , $x = 1$ and $x = -1$ . Pairing each with $y = x^{3}$ gives the critical points $(0,0)$ , $(1,1)$ and $(-1,-1)$ .

The second derivative test

f_{xx} = 12x^{2}, \qquad f_{yy} = 12y^{2}, \qquad f_{xy} = -4

D(x,y) = 144x^{2}y^{2} - 16

At the origin both pure second partials vanish, so $D = -16 < 0$ and the origin is a saddle. At $(1,1)$ and at $(-1,-1)$ the product $x^{2}y^{2}$ is $1$ , so $D = 144 - 16 = 128 > 0$ , and $f_{xx} = 12 > 0$ makes both of them local minima at height $f = -2$ .

The mistake students make

Cancelling $x$ from $x^{9} = x$ is the standard way to lose the saddle at the origin. Factor the equation as $x(x^{8}-1) = 0$ and both cases survive.

The other trap is $x^{8} = 1$ . A degree eight equation has eight roots in the complex numbers, but only $x = 1$ and $x = -1$ are real, and only real numbers are coordinates of points in the plane. Three critical points in total, not nine.

Frequently asked questions

Why do the two minima have exactly the same value?

Because $f(-x,-y) = x^{4} + y^{4} - 4xy = f(x,y)$ : the function is unchanged by a half turn about the origin. That symmetry carries $(1,1)$ onto $(-1,-1)$ , so the two points must have the same height, $-2$ , and the same discriminant, $128$ .

Are these local minima also global minima?

Yes. The quartic terms dominate, so $f \to +\infty$ in every direction and the global minimum has to be attained at a critical point. Among the three candidates, the heights are $0$ at the origin and $-2$ at the other two, so $-2$ is the global minimum.