Stationary Points of a Surface

Level in both directions

At the top of a hill the ground is level: whichever way you step, it neither rises nor falls at first. On a surface z = f(x, y) that needs the slope in the x direction and the slope in the y direction to be 0 at the same point. One is not enough. For $f=x^2+y$f = x² + y, $\frac{\partial f}{\partial x}=2x$∂f/∂x = 2x is 0 all along the y-axis, but $\frac{\partial f}{\partial y}=1$∂f/∂y = 1 everywhere, so the surface keeps climbing in y and is level nowhere.

A stationary point of f is a point where $\frac{\partial f}{\partial x}=0$∂f/∂x = 0 and $\frac{\partial f}{\partial y}=0$∂f/∂y = 0 together, so that $\nabla f=\left(0,0\right)$∇f = (0, 0). Then the slope $\nabla f\cdot u$∇f · u is 0 in every direction u, and the tangent plane there is horizontal.

The dome $z=16\left(1-x^2-y^2\right)$z = 16(1 − x² − y²). At its top, above the origin, $\frac{\partial f}{\partial x}=-32x$∂f/∂x = −32x and $\frac{\partial f}{\partial y}=-32y$∂f/∂y = −32y are both 0, and the gold patch of tangent plane there is horizontal, at height 16.

Solve both equations together

To find the stationary points, set both partial derivatives to 0 and solve the two equations as a pair. For $f=x^2+2y^2-4x+4y$f = x² + 2y² − 4x + 4y, $\frac{\partial f}{\partial x}=2x-4$∂f/∂x = 2x − 4 and $\frac{\partial f}{\partial y}=4y+4$∂f/∂y = 4y + 4. The first is 0 only when x = 2 and the second only when y = −1, so the one stationary point is (2, −1), where f = 4 + 2 − 8 − 4 = −6.

For $f=x^2+y^2-2x-4y$f = x² + y² − 2x − 4y, 2x − 2 = 0 and 2y − 4 = 0 give the point (1, 2), not (2, 4): differentiating halves each coefficient. And x comes from the x equation, so the point is not (2, 1) either.

Bowl, dome, or neither

A stationary point can be the bottom of a bowl, a local minimum, where the surface rises in every direction. It can be the top of a dome, a local maximum, where it falls in every direction.

Or it can be neither. The saddle $z=x^2-y^2$z = x² − y² has $\frac{\partial f}{\partial x}=2x$∂f/∂x = 2x and $\frac{\partial f}{\partial y}=-2y$∂f/∂y = −2y, both 0 at the origin, so the origin is stationary. Along the x-axis the surface is $z=x^2$z = x², rising on both sides; along the y-axis it is $z=-y^2$z = −y², falling on both sides. A stationary point that is a minimum along one line and a maximum along another is a saddle point.

The saddle $z=x^2-y^2$z = x² − y² over the square with $|x|\le 2$|x| ≤ 2 and $|y|\le 2$|y| ≤ 2. Its tangent plane at the origin, the gold patch, is horizontal, at height 0, yet the surface rises above it along x and falls below it along y.

The second-derivative test

The test in two variables uses all three second partial derivatives at the stationary point. Work out $D=\left(\frac{{\partial }^{2}f}{\partial x^2}\right)\left(\frac{{\partial }^{2}f}{\partial y^2}\right)-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^2$D = (∂²f/∂x²)(∂²f/∂y²) − (∂²f/∂x∂y)². If D > 0 and $\frac{{\partial }^{2}f}{\partial x^2}>0$∂²f/∂x² > 0, the point is a local minimum. If D > 0 and $\frac{{\partial }^{2}f}{\partial x^2}<0$∂²f/∂x² < 0, it is a local maximum. If D < 0, it is a saddle point. If D = 0, the test gives no conclusion.

For $x^2+2y^2-4x+4y$x² + 2y² − 4x + 4y at (2, −1): $\frac{{\partial }^{2}f}{\partial x^2}=2$∂²f/∂x² = 2, $\frac{{\partial }^{2}f}{\partial y^2}=4$∂²f/∂y² = 4 and $\frac{{\partial }^{2}f}{\partial x\partial y}=0$∂²f/∂x∂y = 0, so $D=2\times 4-0^2=8$D = 2 × 4 − 0² = 8. D is positive and $\frac{{\partial }^{2}f}{\partial x^2}$∂²f/∂x² is positive, so (2, −1) is a minimum. Completing the square agrees: $f={\left(x-2\right)}^2+2{\left(y+1\right)}^2-6$f = (x − 2)² + 2(y + 1)² − 6, which is never below −6.

For the saddle $x^2-y^2$x² − y² at the origin: $D=2\times \left(-2\right)-0^2=-4$D = 2 × (−2) − 0² = −4, negative, so the origin is a saddle point.

Why the mixed derivative counts

Take $f=x^2+3xy+y^2$f = x² + 3xy + y². Its partial derivatives 2x + 3y and 3x + 2y are both 0 only at the origin. There $\frac{{\partial }^{2}f}{\partial x^2}=2$∂²f/∂x² = 2 and $\frac{{\partial }^{2}f}{\partial y^2}=2$∂²f/∂y² = 2, both positive, so the surface curves upward along the x-axis and along the y-axis. But $\frac{{\partial }^{2}f}{\partial x\partial y}=3$∂²f/∂x∂y = 3, and $D=2\times 2-3^2=-5$D = 2 × 2 − 3² = −5. The origin is a saddle point.

The slices show it. Along the line y = x, $f=x^2+3x^2+x^2=5x^2$f = x² + 3x² + x² = 5x², rising. Along the line y = −x, $f=x^2-3x^2+x^2=-x^2$f = x² − 3x² + x² = −x², falling. A test that looked only at the two pure second derivatives would have called this a minimum.

Three slices of $z=x^2+3xy+y^2$z = x² + 3xy + y² through the origin, each plotted against x. Along the x-axis it is $z=x^2$z = x²; along y = x it is the gold curve $z=5x^2$z = 5x²; along y = −x it is $z=-x^2$z = −x², which falls on both sides. One slice falling is enough to rule out a minimum.

Two stationary points

For $f=3x-x^3-y^2$f = 3x − x³ − y², $\frac{\partial f}{\partial x}=3-3x^2$∂f/∂x = 3 − 3x² and $\frac{\partial f}{\partial y}=-2y$∂f/∂y = −2y. The first is 0 at x = 1 and x = −1, and the second at y = 0, so there are two stationary points, (1, 0) and (−1, 0). The second derivatives are $\frac{{\partial }^{2}f}{\partial x^2}=-6x$∂²f/∂x² = −6x, $\frac{{\partial }^{2}f}{\partial y^2}=-2$∂²f/∂y² = −2 and $\frac{{\partial }^{2}f}{\partial x\partial y}=0$∂²f/∂x∂y = 0.

At (1, 0): $D=\left(-6\right)\left(-2\right)-0^2=12$D = (−6)(−2) − 0² = 12, positive, with $\frac{{\partial }^{2}f}{\partial x^2}=-6$∂²f/∂x² = −6 negative, so it is a local maximum, with f = 3 − 1 = 2. At (−1, 0): $D=\left(6\right)\left(-2\right)-0^2=-12$D = (6)(−2) − 0² = −12, negative, so it is a saddle point, with f = −3 + 1 = −2. Each stationary point is classified on its own.

When D = 0

At the origin, $f=x^4+y^4$f = x⁴ + y⁴ has every second derivative 0, so D = 0, and the origin is a minimum. $f=-x^4-y^4$f = −x⁴ − y⁴ also has D = 0, and the origin is a maximum. $f=x^3+y^2$f = x³ + y² has $\frac{{\partial }^{2}f}{\partial x^2}=6x=0$∂²f/∂x² = 6x = 0 there, so D = 0 again, and the origin is neither: along the x-axis $f=x^3$f = x³ is negative on one side and positive on the other.

So D = 0 decides nothing. The point must be examined another way, by completing the square or by looking at slices.

The contours of $z=x^2-y^2$z = x² − y² and a vertical cut through the origin at $\theta =30^{\circ }$θ = 30°. The slice is $z=\left(\cos 60^{\circ }\right){\rho }^2=0.5{\rho }^2$z = (cos 60°)ρ² = 0.5ρ², an upward parabola. Turn the cut to 45° and the slice is flat; past 45° it opens downward, to $z=-{\rho }^2$z = −ρ² at 90°. The origin is the bottom of some slices and the top of others: a saddle, with D = −4.

The usual mistakes

Setting only one partial derivative to 0. The point must make both 0 at once.

Reading the point off the coefficients. For $x^2+y^2-6x-8y$x² + y² − 6x − 8y the stationary point is (3, 4), from 2x − 6 = 0 and 2y − 8 = 0, not (6, 8).

Classifying from $\frac{{\partial }^{2}f}{\partial x^2}$∂²f/∂x² and $\frac{{\partial }^{2}f}{\partial y^2}$∂²f/∂y² alone. $x^2+3xy+y^2$x² + 3xy + y² has both positive and is still a saddle, because D = −5.

Reading D = 0 as a saddle. The test is silent at D = 0: $x^4+y^4$x⁴ + y⁴ has a minimum there.

A coffee roaster

In the application below, a roaster’s weekly profit depends on the prices of two blends, because customers switch between them. Setting both partial derivatives of the profit to 0 gives two linear equations for the two prices, and D, with the sign of the second derivative, shows that the point is a maximum.