Multivariable calculus

Critical Points of 2x^3 + xy^2 + 5x^2 + y^2

Q: Why does f_yy vanish at the saddle points?

Because f_{yy} = 2x + 2, which is zero exactly when x = -1, and both saddles sit on that line. It does not obstruct anything: D = f_{xx}f_{yy} - f_{xy}^{2} = -16 is still clearly negative, and a negative D classifies the point on its own.

Q: Is the local maximum at (-5/3, 0) a global maximum?

No. The cubic term dominates for large x, so for example f(3,0) = 54 + 45 = 99, which dwarfs the local maximum value \tfrac{125}{27} \approx 4.63. The function is unbounded above and below.

f(x,y) = 2x^3 + xy^2 + 5x^2 + y^2 has four critical points, and all three verdicts appear. The origin is a local minimum with D = 20. The point (-5/3, 0) is a local maximum with D = 40/3. The points (-1, 2) and (-1, -2) are saddles with D = -16.

f(x,y) = 2x^{3} + xy^{2} + 5x^{2} + y^{2}

$\left(0, 0\right)$ local minimumdiscriminant D = 20
$\left(-1.6666666666666667, 0\right)$ local maximumdiscriminant D = 13.333333333333334
$\left(-1, 2\right)$ saddle pointdiscriminant D = -16
$\left(-1, -2\right)$ saddle pointdiscriminant D = -16

Start with the equation that factors

\frac{\partial f}{\partial x} = 6x^{2} + y^{2} + 10x, \qquad \frac{\partial f}{\partial y} = 2xy + 2y = 2y(x + 1)

The $x$ equation mixes both variables and factors into nothing useful. The $y$ equation factors immediately, so begin there: either $y = 0$ or $x = -1$ .

If $y = 0$ : the $x$ equation becomes $6x^{2} + 10x = 2x(3x + 5) = 0$ , giving $x = 0$ and $x = -\tfrac{5}{3}$ .
If $x = -1$ : the $x$ equation becomes $6 - 10 + y^{2} = 0$ , so $y^{2} = 4$ and $y = \pm 2$ .

That is four critical points: $(0,0)$ , $\left(-\tfrac{5}{3},0\right)$ , $(-1,2)$ and $(-1,-2)$ .

One function, all three verdicts

f_{xx} = 12x + 10, \qquad f_{yy} = 2x + 2, \qquad f_{xy} = 2y

D(x,y) = (12x + 10)(2x + 2) - 4y^{2}

At $(0,0)$ : $D = (10)(2) = 20 > 0$ with $f_{xx} = 10 > 0$ , a local minimum at height $0$ . At $\left(-\tfrac{5}{3},0\right)$ : $D = (-10)\left(-\tfrac{4}{3}\right) = \tfrac{40}{3} > 0$ with $f_{xx} = -10 < 0$ , a local maximum at height $\tfrac{125}{27}$ .

At $(-1,\pm 2)$ the factor $2x + 2$ is zero, so $f_{yy} = 0$ and $D = -4y^{2} = -16 < 0$ . Both are saddles, each at height $f = 3$ .

The mistake students make

Attacking $6x^{2} + y^{2} + 10x = 0$ first is the wasted half hour. It is one equation in two unknowns and describes an ellipse, so on its own it produces nothing. Always start from whichever partial factors, use it to split into cases, and substitute each case into the other equation.

The second slip is stopping at the branch $y = 0$ because it looks like the natural one. The branch $x = -1$ carries both saddles, and a solution set that misses half the critical points is not a solution.

Frequently asked questions

Why does f_yy vanish at the saddle points?

Because $f_{yy} = 2x + 2$ , which is zero exactly when $x = -1$ , and both saddles sit on that line. It does not obstruct anything: $D = f_{xx}f_{yy} - f_{xy}^{2} = -16$ is still clearly negative, and a negative $D$ classifies the point on its own.

Is the local maximum at (-5/3, 0) a global maximum?

No. The cubic term dominates for large $x$ , so for example $f(3,0) = 54 + 45 = 99$ , which dwarfs the local maximum value $\tfrac{125}{27} \approx 4.63$ . The function is unbounded above and below.