# What are the values and types of the critical points, if any, of f(x,y) = x(2-x-y)?

Dec 22, 2017

Saddle point at $\left(0 , 2\right)$

#### Explanation:

I will assume by "critical points" you are referring to the values of $x$ and $y$ where the function is stationary.

First we need to find the partial derivatives of the function:

The $x$ partial derivative:

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

by use of the chain rule.

The $y$ partial derivative:

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

We now take both of these equations, and set them equal to 0 to obtain a pair of simultaneous equations which can then be solved to obtain one or more pairs of co-ordinates where our stationary point lies, so:

$2 - 2 x - y = 0$
$- x = 0$

Immediately we can obviously see $x = 0$ from the bottom equation. That leaves us with:

$2 - y = 0$

for the top equation. This has the solution:

$y = 2$

So we have only one stationary point at $\left(0 , 2\right)$. Here $f \left(0 , 2\right) = 0$

In order to classify the point we now need to find the higher order derivatives, those will be:

$\frac{{\partial}^{2} f}{\partial {x}^{2}} = - 2$
$\frac{{\partial}^{2} f}{\partial {y}^{2}} = 0$
$\frac{{\partial}^{2} f}{\partial x \partial y} = - 1$

We can classify the points by:

$\frac{{\partial}^{2} f}{\partial {x}^{2}} \frac{{\partial}^{2} f}{\partial {y}^{2}} - {\left(\frac{{\partial}^{2} f}{\partial x \partial y}\right)}^{2} < 0$ at $\left(x , y\right)$

then it will be a saddle point, or else if:

$\frac{{\partial}^{2} f}{\partial {x}^{2}} \frac{{\partial}^{2} f}{\partial {y}^{2}} - {\left(\frac{{\partial}^{2} f}{\partial x \partial y}\right)}^{2} > 0$ as $\left(x , y\right)$

then it will be a maximum or minimum. Substituting the values of our derivative in we get:

$\left(2\right) \left(0\right) - {\left(- 1\right)}^{2} = - 1 < 0$ So it is a saddle point. Hence our critical point is a saddle point at $\left(0 , 2\right)$.

I have included a contour plot of the function below (I did try a 3d plot at first but the shallowness of the gradient did not make the turning point clear enough). On the image the darker regions represent values of low $f \left(x , y\right)$ and the brighter regions represent high $f \left(x , y\right)$ in accordance with the plot legend.

We see that going from the bottom left to the top right you go up "over a hill" reaching a maximum at the calculated stationary point. Similarly going from the top left to the bottom right you go through the "bottom of a valley" with the minimum being at the calculated stationary point.

In other words, we notice that the contours from all directions converge at $\left(0 , 2\right)$, hence the stationary point of the function. Hopefully this helps to see the idea more clearly.