Multivariable calculus

Partial Derivatives of x^4 + y^4 - 4xy

Q: Why are there two separate minimum points?

The cross term -4xy is most negative when x and y have the same sign and similar size. That happens on both the first-quadrant diagonal and the third-quadrant diagonal, so the function dips to the same depth of -2 in each.

For f(x, y) = x^4 + y^4 - 4xy the partial with respect to x is 4x^3 - 4y, and the partial with respect to y is 4y^3 - 4x. Setting both to zero gives y = x^3 and x = y^3, so the critical points are (0, 0), (1, 1), and (-1, -1). The value at both non-zero critical points is -2.

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

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

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

Differentiate, one variable at a time

With $y$ frozen, $y^4$ is a constant and contributes nothing, while $-4xy$ contributes $-4y$ .

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

The function is symmetric under swapping the variables, so the $y$ partial is the mirror image.

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

At $(2,1)$ these are $32 - 4 = 28$ and $4 - 8 = -4$ . The steep positive value in $x$ reflects how quickly a fourth power grows once you leave the origin.

The mistake: solving the critical point system carelessly

Setting $4x^3 - 4y = 0$ and $4y^3 - 4x = 0$ gives $y = x^3$ and $x = y^3$ . Substituting the first into the second gives $x = x^9$ , and the temptation is to divide by $x$ and report only $x = \pm 1$ .

Dividing by $x$ throws away the solution $x = 0$ . Factor instead.

x^9 - x = x\left(x^8 - 1\right) = x\left(x^4 - 1\right)\left(x^4 + 1\right) = 0

$x = 0$ gives $y = 0$ , the critical point $(0,0)$ .
$x = 1$ gives $y = 1$ , and $x = -1$ gives $y = -1$ .
$x^4 + 1$ has no real roots, so there are exactly three critical points.

Never divide an equation by a variable that could be zero. Factoring keeps every root in view.

Classifying the three points

The second partials are $f_{xx} = 12x^2$ , $f_{yy} = 12y^2$ , and $f_{xy} = -4$ , so the discriminant is $144x^2y^2 - 16$ .

At $(0,0)$ the discriminant is $-16 < 0$ , so the origin is a saddle point. At $(1,1)$ and $(-1,-1)$ the discriminant is $144 - 16 = 128 > 0$ with $f_{xx} = 12 > 0$ , so both are local minima.

f(1,1) = 1 + 1 - 4 = -2, \qquad f(-1,-1) = 1 + 1 - 4 = -2

Since $x^4 + y^4$ eventually overwhelms $4xy$ far from the origin, these two local minima are in fact the global minimum value of the function.

Frequently asked questions

Why are there two separate minimum points?

The cross term $-4xy$ is most negative when $x$ and $y$ have the same sign and similar size. That happens on both the first-quadrant diagonal and the third-quadrant diagonal, so the function dips to the same depth of $-2$ in each.

What makes this a standard textbook example?

It puts a saddle and two separate minima in one problem, so the second derivative test has to be run three times with two different verdicts, and the critical point system has a hidden root at the origin that punishes dividing by $x$ . That combination is why it appears in so many Calc 3 courses.