Critical points in a function of two variables

 Objective: Use partial derivatives to locate critical points for a function of two variables

Critical points. Definition

Examples

Solution

Next, we set each of these expressions equal to zero:

Therefore x = 2 and y = -3, (2, -3) is a critical point of f.

We must also check for the possibility that the denominator of each partial derivative can be equal to zero. In this case, the partial derivative doesn't exist. Since the denominator is the same in both partial derivatives, we need to do this once.

This equation represents also an hyperbola. We should also note that the domain of consists of points satisfying the inequality:

Therefore, any points on the hyperbola are not only critical points, they are also at the boundary of the domain. Let's put the equation of the hyperbola in standard form by completing the square:

Notice that (2, -3) is the center of the hyperbola.

We make each partial derivative equal to zero which give a system of equations with x and y. 

Subtracting the second equation from the first gives: 10 y + 10 = 0  y= -1. Substiuting y in the first

equation gives 2x -2 + 4 = 0, 2x + 2 = 0, x = -1. Therefore (-1, -1) is a critical point of the given function. There are no points in R² that make either partial derivative not to exist since both of them are defined for any point (x, y).

Practice