Multivariable calculus

Critical Points of x^2 + xy + y^2 - 3y

Q: Why does the cross term not stop this from being a minimum?

For ax^{2} + bxy + cy^{2} you get f_{xx} = 2a, f_{yy} = 2c and f_{xy} = b, so the discriminant is exactly 4ac - b^{2}. The quadratic part therefore stays bowl shaped as long as b^{2} < 4ac. Here b^{2} = 1 and 4ac = 4, so there is room to spare and (-1,2) is a genuine local minimum.

Q: Does this function have a maximum anywhere?

No. Along the line y = x the function is 3x^{2} - 3x, which grows without bound as x increases, so f has no global maximum. It also has no other critical point, so there is no local maximum either.

The only critical point of f(x,y) = x^2 + xy + y^2 - 3y is (-1, 2). The cross term gives f_xy = 1, so the discriminant is D = (2)(2) - 1^2 = 3, which is positive, and f_xx = 2 is positive, so (-1, 2) is a local minimum. The value there is f(-1, 2) = -3.

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

$\left(-1, 2\right)$ local minimumdiscriminant D = 3

Solving a linear system, not two separate equations

The $xy$ term means each partial involves both variables, so the two equations have to be solved together rather than one at a time.

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

From $2x + y = 0$ you get $y = -2x$ . Substituting into $x + 2y - 3 = 0$ gives $x - 4x - 3 = 0$ , so $-3x = 3$ and $x = -1$ , hence $y = 2$ . The system is linear with a nonzero coefficient determinant, so this is the only solution: $(-1, 2)$ , where $f = -3$ .

Running the test with a nonzero mixed partial

f_{xx} = 2, \qquad f_{yy} = 2, \qquad f_{xy} = 1

D = (2)(2) - 1^{2} = 3 > 0

Since $D > 0$ and $f_{xx} = 2 > 0$ , the point $(-1, 2)$ is a local minimum. The cross term costs you $f_{xy}^{2} = 1$ of discriminant, which is not enough here to change the verdict, but it is exactly the quantity that decides these cases.

The mistake students make

The error is dropping $f_{xy}$ , either by forgetting the cross term when differentiating or by using $D = f_{xx}f_{yy}$ from habit. That shortcut gives $4$ here instead of $3$ , which happens to reach the same conclusion, so the mistake goes unnoticed and then ruins the next problem.

See how little it takes to flip the answer. Change the function to $x^{2} + 3xy + y^{2}$ and the origin is its critical point, with $f_{xx} = 2$ , $f_{yy} = 2$ and $f_{xy} = 3$ .

D = (2)(2) - 3^{2} = -5 < 0

Same pure second partials, opposite verdict: that one is a saddle. The cross term is doing all the work.

Frequently asked questions

Why does the cross term not stop this from being a minimum?

For $ax^{2} + bxy + cy^{2}$ you get $f_{xx} = 2a$ , $f_{yy} = 2c$ and $f_{xy} = b$ , so the discriminant is exactly $4ac - b^{2}$ . The quadratic part therefore stays bowl shaped as long as $b^{2} < 4ac$ . Here $b^{2} = 1$ and $4ac = 4$ , so there is room to spare and $(-1,2)$ is a genuine local minimum.

Does this function have a maximum anywhere?

No. Along the line $y = x$ the function is $3x^{2} - 3x$ , which grows without bound as $x$ increases, so $f$ has no global maximum. It also has no other critical point, so there is no local maximum either.