What are the extrema and saddle points of f(x, y) = xy(1-x-y)?

Aug 19, 2017

The points $\left(0 , 0\right) , \left(1 , 0\right)$, and $\left(0 , 1\right)$ are saddle points. The point $\left(\frac{1}{3} , \frac{1}{3}\right)$ is a local maximum point.

Explanation:

We can expand $f$ to $f \left(x , y\right) = x y - {x}^{2} y - x {y}^{2}$. Next, find the partial derivatives and set them equal to zero.

$\setminus \frac{\setminus \partial f}{\setminus \partial x} = y - 2 x y - {y}^{2} = y \left(1 - 2 x - y\right) = 0$

$\setminus \frac{\setminus \partial f}{\setminus \partial y} = x - {x}^{2} - 2 x y = x \left(1 - x - 2 y\right) = 0$

Clearly, $\left(x , y\right) = \left(0 , 0\right) , \left(1 , 0\right) ,$ and $\left(0 , 1\right)$ are solutions to this system, and so are critical points of $f$. The other solution can be found from the system $1 - 2 x - y = 0$, $1 - x - 2 y = 0$. Solving the first equation for $y$ in terms of $x$ gives $y = 1 - 2 x$, which can be plugged into the second equation to get $1 - x - 2 \left(1 - 2 x\right) = 0 \implies - 1 + 3 x = 0 \implies x = \frac{1}{3}$. From this, $y = 1 - 2 \left(\frac{1}{3}\right) = 1 - \frac{2}{3} = \frac{1}{3}$ as well.

To test the nature of these critical points, we find second derivatives:

$\setminus \frac{\setminus {\partial}^{2} f}{\setminus \partial {x}^{2}} = - 2 y$, $\setminus \frac{\setminus {\partial}^{2} f}{\setminus \partial {y}^{2}} = - 2 x$, and $\setminus \frac{\setminus {\partial}^{2} f}{\setminus \partial x \setminus \partial y} = \setminus \frac{\setminus {\partial}^{2} f}{\setminus \partial y \setminus \partial x} = 1 - 2 x - 2 y$.

The discriminant is therefore:

$D = 4 x y - {\left(1 - 2 x - 2 y\right)}^{2}$

$= 4 x y - \left(1 - 2 x - 2 y - 2 x + 4 {x}^{2} + 4 x y - 2 y + 4 x y + 4 {y}^{2}\right)$

$= 4 x + 4 y - 4 {x}^{2} - 4 {y}^{2} - 4 x y - 1$

Plugging the first three critical points in gives:

$D \left(0 , 0\right) = - 1 < 0$, $D \left(1 , 0\right) = 4 - 4 - 1 = - 1 < 0$, and $D \left(0 , 1\right) = 4 - 4 - 1 = - 1 < 0$, making these points saddle points.

Plugging in the last critical point gives $D \left(\frac{1}{3} , \frac{1}{3}\right) = \frac{4}{3} + \frac{4}{3} - \frac{4}{9} - \frac{4}{9} - \frac{4}{9} - 1 = \frac{1}{3} > 0$. Also note that $\setminus \frac{\setminus {\partial}^{2} f}{\setminus \partial {x}^{2}} \left(\frac{1}{3} , \frac{1}{3}\right) = - \frac{2}{3} < 0$. Therefore, $\left(\frac{1}{3} , \frac{1}{3}\right)$ is a location of a local maximum value of $f$. You can check that the local maximum value itself is $f \left(\frac{1}{3} , \frac{1}{3}\right) = \frac{1}{27}$.

Below is a picture of the contour map (of level curves) of $f$ (the curves where the output of $f$ is constant), along with the 4 critical points of $f$. 