Multivariable calculus

Critical Points of x^2 + y^2 + x^2 y

Q: Is the origin a global minimum?

No, only a local one. Fix y = -10 and the function becomes x^{2} + 100 - 10x^{2} = 100 - 9x^{2}, which runs to -\infty as x grows. For instance f(10,-10) = -800, well below f(0,0) = 0.

Q: Why are the two saddle points at the same height?

Every x in the formula appears as x^{2}, so f(-x,y) = f(x,y) and the surface is a mirror image across the y axis. That reflection swaps the two saddle points, forcing the same height f = 1 and the same discriminant D = -8.

f(x,y) = x^2 + y^2 + x^2 y has three critical points. The origin is a local minimum with discriminant D = 4 and f_xx = 2. The points (sqrt 2, -1) and (-sqrt 2, -1) are saddles with D = -8. The general discriminant is D = 4 + 4y - 4x^2, which is what separates the three cases.

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

$\left(0, 0\right)$ local minimumdiscriminant D = 4
$\left(1.4142135623730951, -1\right)$ saddle pointdiscriminant D = -8
$\left(-1.4142135623730951, -1\right)$ saddle pointdiscriminant D = -8

Factor the partial that factors

\frac{\partial f}{\partial x} = 2x + 2xy = 2x(1 + y), \qquad \frac{\partial f}{\partial y} = 2y + x^{2}

The $x$ equation splits into two cases, and each has to be followed through separately.

Case $x = 0$ : then $2y = 0$ , so $y = 0$ and the point is $(0,0)$ .
Case $y = -1$ : then $-2 + x^{2} = 0$ , so $x = \pm\sqrt{2}$ and the points are $\left(\sqrt{2},-1\right)$ and $\left(-\sqrt{2},-1\right)$ .

Three critical points, and no others, since the two cases exhaust the solutions of $2x(1+y) = 0$ .

Classifying all three

f_{xx} = 2 + 2y, \qquad f_{yy} = 2, \qquad f_{xy} = 2x

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

At the origin, $D = 4 > 0$ and $f_{xx} = 2 > 0$ , so it is a local minimum with $f(0,0) = 0$ . At $\left(\pm\sqrt{2},-1\right)$ the term $4y$ cancels the leading $4$ and $4x^{2} = 8$ remains, so $D = -8 < 0$ and both are saddles, each at height $f = 1$ .

The mistake students make

Dividing $2x(1+y) = 0$ by $2x$ throws away the origin, and stopping after $x = 0$ throws away both saddles. A factored equation is a signal to branch, not to cancel: every factor gets its own case, and each case is then substituted into the other equation.

The second slip is the sign of $f_{xx}$ at the saddles. There $f_{xx} = 2 + 2(-1) = 0$ , which looks alarming, but $D$ is already negative and that alone settles the classification. The sign of $f_{xx}$ is consulted only when $D > 0$ .

Frequently asked questions

Is the origin a global minimum?

No, only a local one. Fix $y = -10$ and the function becomes $x^{2} + 100 - 10x^{2} = 100 - 9x^{2}$ , which runs to $-\infty$ as $x$ grows. For instance $f(10,-10) = -800$ , well below $f(0,0) = 0$ .

Why are the two saddle points at the same height?

Every $x$ in the formula appears as $x^{2}$ , so $f(-x,y) = f(x,y)$ and the surface is a mirror image across the $y$ axis. That reflection swaps the two saddle points, forcing the same height $f = 1$ and the same discriminant $D = -8$ .