Multivariable calculus

Gradient of x^3 + y^3 - 3xy: Steps and Critical Points

Q: Why does x^3 + y^3 - 3xy have no global maximum or minimum?

The cubes dominate far from the origin. Along y = 0 the function is x^3, which goes to +\infty as x grows and to -\infty as x falls. A function unbounded in both directions can only have local extremes.

Q: What is the direction of steepest descent at (2, 1)?

The negative gradient, -\nabla f(2,1) = \langle -9, 3\rangle, or as a unit vector \frac{1}{\sqrt{10}}\langle -3, 1\rangle. The function falls at rate \left|\nabla f\right| = \sqrt{81 + 9} = 3\sqrt{10} in that direction.

For f(x, y) = x^3 + y^3 - 3xy the gradient is (3x^2 - 3y, 3y^2 - 3x). At (2, 1) it equals (9, -3). Setting both components to zero gives exactly two critical points, the origin and (1, 1). The origin is a saddle and (1, 1) is a local minimum with value -1.

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

\nabla f(x,y) = \left\langle 3x^2 - 3y,\ 3y^2 - 3x \right\rangle

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

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

Three terms, and each partial keeps two of them

Hold $y$ fixed. The term $y^3$ is a pure constant and drops out, while $-3xy$ has $-3y$ sitting in front of a single $x$ .

\frac{\partial f}{\partial x} = 3x^2 + 0 - 3y = 3x^2 - 3y

Hold $x$ fixed instead. Now $x^3$ is the constant that vanishes, and $-3xy$ leaves $-3x$ behind.

\frac{\partial f}{\partial y} = 0 + 3y^2 - 3x = 3y^2 - 3x

\nabla f(x,y) = \left\langle 3x^2 - 3y,\ 3y^2 - 3x \right\rangle

At $(2,1)$ this gives $\langle 12 - 3,\ 3 - 6\rangle = \langle 9, -3\rangle$ . Notice the pleasing symmetry: swapping $x$ and $y$ in $f$ leaves it unchanged, and it swaps the two gradient components.

The mistake: letting the cube die in the wrong partial

The two frequent wrong answers are $\langle 3x^2 - 3y,\ 3y^2\rangle$ and $\langle 3x^2,\ 3y^2 - 3x\rangle$ . In each case the $-3xy$ term has been used once and then forgotten, as though a term can only be spent on one component.

Each partial derivative is a fresh calculation on the entire function. The term $-3xy$ contains both variables, so it appears in both components, contributing $-3y$ to one and $-3x$ to the other.

$y^3$ vanishes under $\frac{\partial}{\partial x}$ because it has no $x$ ; $-3xy$ does not vanish because it does.
The symmetry of $f$ is a free check. If your two components are not mirror images under swapping $x$ and $y$ , one of them is wrong.
Slice check at $y = 1$ : $f(x,1) = x^3 + 1 - 3x$ has derivative $3x^2 - 3$ , which matches $3x^2 - 3y$ at $y = 1$ .

Solving for the critical points

Set both components to zero. Dividing each by $3$ gives $x^2 = y$ and $y^2 = x$ , a pair of parabolas that cross where the gradient dies.

Substitute the first into the second: $\left(x^2\right)^2 = x$ , so $x^4 - x = 0$ and $x\left(x^3 - 1\right) = 0$ . Over the reals that leaves $x = 0$ and $x = 1$ , giving the points $(0,0)$ and $(1,1)$ .

Classify them with the discriminant $D = f_{xx}f_{yy} - f_{xy}^2$ , where $f_{xx} = 6x$ , $f_{yy} = 6y$ and $f_{xy} = -3$ .

D(0,0) = 0 - 9 = -9, \qquad D(1,1) = 36 - 9 = 27

So the origin is a saddle, and $(1,1)$ has $D > 0$ with $f_{xx} = 6 > 0$ , making it a local minimum with $f(1,1) = 1 + 1 - 3 = -1$ . It is not a global minimum: along the line $y = 0$ , $f = x^3$ runs off to negative infinity.

Frequently asked questions

Why does x^3 + y^3 - 3xy have no global maximum or minimum?

The cubes dominate far from the origin. Along $y = 0$ the function is $x^3$ , which goes to $+\infty$ as $x$ grows and to $-\infty$ as $x$ falls. A function unbounded in both directions can only have local extremes.

What is the direction of steepest descent at (2, 1)?

The negative gradient, $-\nabla f(2,1) = \langle -9, 3\rangle$ , or as a unit vector $\frac{1}{\sqrt{10}}\langle -3, 1\rangle$ . The function falls at rate $\left|\nabla f\right| = \sqrt{81 + 9} = 3\sqrt{10}$ in that direction.