Multivariable calculus

Critical Points of x^2 + y^2 - 4x + 6y

Q: Is (2, -3) a global minimum as well as a local one?

Yes. Completing the square gives f = (x-2)^{2} + (y+3)^{2} - 13, so f \ge -13 everywhere, with equality only at (2,-3). The second derivative test on its own only ever certifies local behaviour, so the algebra is what upgrades the claim.

Q: Why is the discriminant the same at every point here?

Because f is a quadratic polynomial. Differentiating twice kills every x and y, leaving f_{xx} = 2, f_{yy} = 2 and f_{xy} = 0 as constants, so D = 4 no matter where you evaluate it. For a cubic or higher, D varies from point to point and has to be recomputed at each one.

The only critical point of f(x,y) = x^2 + y^2 - 4x + 6y is (2, -3). There the discriminant is D = f_xx f_yy - f_xy^2 = (2)(2) - 0 = 4, which is positive, and f_xx = 2 is also positive, so the point is a local minimum. The value there is f(2, -3) = -13, and it is the global minimum too.

f(x,y) = x^{2} + y^{2} - 4x + 6y

$\left(2, -3\right)$ local minimumdiscriminant D = 4

Setting both partials to zero

A critical point is a place where every first partial is zero at once, so start by writing them both down. Each one here is linear, and neither contains the other variable, so the two equations do not interact.

\frac{\partial f}{\partial x} = 2x - 4, \qquad \frac{\partial f}{\partial y} = 2y + 6

Setting $2x - 4 = 0$ gives $x = 2$ , and setting $2y + 6 = 0$ gives $y = -3$ . Those are the only solutions each equation has, so $(2, -3)$ is the only critical point, and the value there is $f(2,-3) = -13$ .

The second derivative test at (2, -3)

Differentiate a second time. Because $f$ is a quadratic, all three second partials are constants, so the discriminant is the same number at every point in the plane.

f_{xx} = 2, \qquad f_{yy} = 2, \qquad f_{xy} = 0

D = f_{xx}f_{yy} - f_{xy}^{2} = (2)(2) - 0^{2} = 4 > 0

A positive $D$ says the point is an extremum rather than a saddle, and then the sign of $f_{xx}$ decides which. Here $f_{xx} = 2 > 0$ , so the surface curves upward in the $x$ direction and $(2,-3)$ is a local minimum.

The mistake students make

The common slip is to compute $D > 0$ and then guess between maximum and minimum, or to look at the sign of $f$ instead of the sign of $f_{xx}$ . The value $f(2,-3) = -13$ is negative and the point is still a minimum: the height of the surface has nothing to do with the classification.

You can confirm the answer without the test at all by completing the square, which is worth doing once so you trust the machinery.

f(x,y) = (x-2)^{2} + (y+3)^{2} - 13

Both squares are zero only at $(2,-3)$ and positive everywhere else, so $f \ge -13$ across the whole plane.

Frequently asked questions

Is (2, -3) a global minimum as well as a local one?

Yes. Completing the square gives $f = (x-2)^{2} + (y+3)^{2} - 13$ , so $f \ge -13$ everywhere, with equality only at $(2,-3)$ . The second derivative test on its own only ever certifies local behaviour, so the algebra is what upgrades the claim.

Why is the discriminant the same at every point here?

Because $f$ is a quadratic polynomial. Differentiating twice kills every $x$ and $y$ , leaving $f_{xx} = 2$ , $f_{yy} = 2$ and $f_{xy} = 0$ as constants, so $D = 4$ no matter where you evaluate it. For a cubic or higher, $D$ varies from point to point and has to be recomputed at each one.