Question

How do you find the critical point(s) of `f(x,y) = (x-y)^2`?

Answer 1

The critical points of a two-variables functions are to be found using the gradient.

The gradient is a vector which has dimension equal to the number of variables: in this case, 2.

The coordinates of the gradient are the derivatives with respect to each variable the function depends on. In this case, the vector will be a 2-dimensional vector, where the first coordinate is the derivative with respect to `x`, and the second is the derivative with respect to `y`.

Note that deriving with respect to a variable means to consider the other as a constant.

Now, expand the square in the definition of `f(x,y)` to get

`(x-y)^2=x^2-2xy+y^2`

Deriving with respect to `x`, we get
`d/dx f(x,y)= 2x-2y=2(x-y)`

Deriving with respect to `y`, we get
`d/dy f(x,y)= -2x+2y=-2(x-y)`

Now, critical points of a functions are the points in which the gradient equals the zero vector. This happens if the following system is solved:
`2(x-y)=0`
`-2(x-y)=0`

Both equations yield the line `x=y` for solutions, which means that this line is the set of the critical points for `f(x,y)`