Multivariable calculus

Critical Points of x^3 + y^3 - 3xy

Q: Is (1,1) a global minimum?

No. Along the line y = x the function is 2x^{3} - 3x^{2}, which runs to -\infty as x \to -\infty, so f is unbounded below. The point (1,1) is only a local minimum, which is all the second derivative test ever claims.

Q: Why is the mixed partial constant here?

The only term containing both variables is -3xy. Differentiating it once in x leaves -3y, and once more in y leaves -3. The cubes contribute nothing to f_{xy} because each involves a single variable.

f(x,y) = x^3 + y^3 - 3xy has exactly two critical points. At (0,0) the discriminant is D = -9, so that point is a saddle. At (1,1) the discriminant is D = 27 and f_xx = 6 is positive, so that point is a local minimum with value f(1,1) = -1. The general discriminant is D = 36xy - 9.

f(x,y) = x^{3} + y^{3} - 3xy

$\left(0, 0\right)$ saddle pointdiscriminant D = -9
$\left(1, 1\right)$ local minimumdiscriminant D = 27

Solving the system without losing a root

\frac{\partial f}{\partial x} = 3x^{2} - 3y, \qquad \frac{\partial f}{\partial y} = 3y^{2} - 3x

Setting both to zero gives $y = x^{2}$ and $x = y^{2}$ . Substitute the first into the second: $x = (x^{2})^{2} = x^{4}$ , so $x^{4} - x = 0$ .

x^{4} - x = x\left(x^{3} - 1\right) = 0

The real roots are $x = 0$ and $x = 1$ . Since $y = x^{2}$ , the critical points are $(0,0)$ and $(1,1)$ , and there are no others.

One discriminant formula, two verdicts

Compute the second partials once, in general, then evaluate at each point. That is faster and less error prone than starting over twice.

f_{xx} = 6x, \qquad f_{yy} = 6y, \qquad f_{xy} = -3

D(x,y) = (6x)(6y) - (-3)^{2} = 36xy - 9

At the origin $D = -9 < 0$ , a saddle. At $(1,1)$ , $D = 36 - 9 = 27 > 0$ and $f_{xx} = 6 > 0$ , so that is a local minimum with $f(1,1) = -1$ .

The saddle at the origin is easy to feel numerically. Along the line $y = x$ the function is $2x^{3} - 3x^{2}$ , which is negative for small positive $x$ . Along $y = -x$ the two cubes cancel exactly and what is left is $3x^{2}$ , positive for every $x \ne 0$ . So $f$ falls one way and rises the other.

The mistake students make

The usual loss happens at $x^{4} = x$ . Dividing both sides by $x$ gives $x^{3} = 1$ and $x = 1$ , and the critical point at the origin disappears without warning. Never divide an equation by a variable that is allowed to be zero: factor instead and read off every root.

The second slip is treating $x^{4} = x$ as having four solutions because it is a quartic. Over the real numbers $x(x^{3}-1)$ has only the two roots $0$ and $1$ ; the other two roots of $x^{3} = 1$ are complex and are not points of the plane.

Frequently asked questions

Is (1,1) a global minimum?

No. Along the line $y = x$ the function is $2x^{3} - 3x^{2}$ , which runs to $-\infty$ as $x \to -\infty$ , so $f$ is unbounded below. The point $(1,1)$ is only a local minimum, which is all the second derivative test ever claims.

Why is the mixed partial constant here?

The only term containing both variables is $-3xy$ . Differentiating it once in $x$ leaves $-3y$ , and once more in $y$ leaves $-3$ . The cubes contribute nothing to $f_{xy}$ because each involves a single variable.