Multivariable calculus

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

Q: Why do all three corner points have the same discriminant?

At each corner two of the three factors of f vanish, which forces xy = 0 and f_{xy} = \pm 1, so D = 0 - 1 = -1 every time. Geometrically each corner is where two of the zero lines of f cross at an angle, and f changes sign as you step across either line, so the four sectors meeting at that corner alternate in sign. That is a saddle. The alternation is what does the work, not the crossing by itself: x^{2}y^{2} also vanishes on two crossing lines, but it never changes sign, and its origin is a minimum.

Q: Is the local maximum a global maximum?

No. Take x = y = -t with t > 0: then f = t^{2}(1 + 2t), which grows without bound. For example f(-5,-5) = 275, far above \tfrac{1}{27}. The value \tfrac{1}{27} is the largest only near that point, and it is the largest on the closed triangle.

f(x,y) = xy - x^2 y - xy^2 factors as xy(1 - x - y) and has four critical points. The three triangle corners (0,0), (1,0) and (0,1) are saddles, each with discriminant D = -1. The centroid (1/3, 1/3) is a local maximum with D = 1/3 and value 1/27.

f(x,y) = xy - x^{2}y - xy^{2} = xy(1 - x - y)

$\left(0, 0\right)$ saddle pointdiscriminant D = -1
$\left(1, 0\right)$ saddle pointdiscriminant D = -1
$\left(0, 1\right)$ saddle pointdiscriminant D = -1
$\left(0.3333333333333333, 0.3333333333333333\right)$ local maximumdiscriminant D = 0.3333333333333333

Two factored equations, four combinations

Differentiate and factor each partial straight away. The factoring is what makes the system tractable.

\frac{\partial f}{\partial x} = y(1 - 2x - y), \qquad \frac{\partial f}{\partial y} = x(1 - x - 2y)

Each equation offers two ways to be zero, so there are four combinations to check.

$y = 0$ and $x = 0$ : the point $(0,0)$ .
$y = 0$ and $1 - x - 2y = 0$ : gives $x = 1$ , the point $(1,0)$ .
$1 - 2x - y = 0$ and $x = 0$ : gives $y = 1$ , the point $(0,1)$ .
$2x + y = 1$ and $x + 2y = 1$ together: gives $x = y = \tfrac{1}{3}$ .

The three corner points are where the lines $x = 0$ , $y = 0$ and $x + y = 1$ meet, and the fourth point is the centroid of that triangle.

Classifying the four

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

D(x,y) = 4xy - (1 - 2x - 2y)^{2}

At each of the three corners the product $xy$ is zero and the mixed partial is $\pm 1$ , so $D = -1 < 0$ and all three are saddles, each at height $f = 0$ . At $\left(\tfrac{1}{3},\tfrac{1}{3}\right)$ you get $4xy = \tfrac{4}{9}$ and $f_{xy} = -\tfrac{1}{3}$ , so $D = \tfrac{4}{9} - \tfrac{1}{9} = \tfrac{1}{3} > 0$ .

There $f_{xx} = -\tfrac{2}{3} < 0$ , so the centroid is a local maximum, with $f\left(\tfrac{1}{3},\tfrac{1}{3}\right) = \tfrac{1}{27}$ . That is the familiar result that the product of three quantities with a fixed sum is largest when they are equal.

The mistake students make

Cancelling the leading factors is fatal here. From $y(1 - 2x - y) = 0$ and $x(1 - x - 2y) = 0$ , dividing by $y$ and by $x$ leaves only the linear system and only the centroid. Three of the four critical points vanish, and they happen to be the three saddles.

The second slip is pairing the cases carelessly. You must combine one factor from the first equation with one factor from the second, all four ways, then solve each pair. Combining $y = 0$ with $x = 0$ is one case; combining $y = 0$ with $1 - x - 2y = 0$ is a different one.

Frequently asked questions

Why do all three corner points have the same discriminant?

At each corner two of the three factors of $f$ vanish, which forces $xy = 0$ and $f_{xy} = \pm 1$ , so $D = 0 - 1 = -1$ every time. Geometrically each corner is where two of the zero lines of $f$ cross at an angle, and $f$ changes sign as you step across either line, so the four sectors meeting at that corner alternate in sign. That is a saddle. The alternation is what does the work, not the crossing by itself: $x^{2}y^{2}$ also vanishes on two crossing lines, but it never changes sign, and its origin is a minimum.

Is the local maximum a global maximum?

No. Take $x = y = -t$ with $t > 0$ : then $f = t^{2}(1 + 2t)$ , which grows without bound. For example $f(-5,-5) = 275$ , far above $\tfrac{1}{27}$ . The value $\tfrac{1}{27}$ is the largest only near that point, and it is the largest on the closed triangle.