Step by step
The question
Question: find and classify the critical points of f(x, y) = x² + 4xy + y².
1 Find the critical points
At a top, a bottom or a saddle, the surface is flat: both partial derivatives are 0.
Term by term.
x²: the power 2 comes down.
The power goes down by 1.
So 2x.
In 4xy, 4y is fixed and waits in front.
x gives 1.
So 4y.
y² has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 2x + 4y.
Term by term.
x² has no y: it is a fixed number, so it gives 0.
In 4xy, 4x is fixed and waits in front.
y gives 1.
So 4x.
y²: the power 2 comes down.
The power goes down by 1.
So 2y.
Drop the zeros: ∂f/∂y = 4x + 2y.
Both partial derivatives must be 0.
Call them ① and ②.
Divide ① by 2: smaller numbers.
Divide ② by 2: smaller numbers.
From ②, get y alone: y = −2x.
Put it into ①.
Open the bracket and add the like terms.
So x = 0.
Back into y = −2x: y = 0.
So there is one critical point: (0, 0).
2 The second partials
Term by term.
2x gives 2.
4y has no x: it is a fixed number, so it gives 0.
Drop the zeros: f_xx = 2.
Term by term.
4x has no y: it is a fixed number, so it gives 0.
2y gives 2.
Drop the zeros: f_yy = 2.
Term by term.
2x has no y: it is a fixed number, so it gives 0.
4y gives 4.
Drop the zeros: f_xy = 4.
The test number: D = f_xx × f_yy − (f_xy)².
3 Test each point
At (0, 0): f_xx is 2, f_yy is 2, f_xy is 4.
So D is −12.
Careful!
f_xx alone is not enough.
If D is negative, it is a saddle, whatever f_xx is.
D is negative: up one way, down the other.
A saddle point.
4 Check
Let's check.
Here is the surface near (0, 0).
It is a saddle: up one way, down the other.
Now small steps away from the point.
Near (0, 0), one step gives a bigger value, the other a smaller one.
That fits a saddle point.
Right!
Let's look back.
First: find the critical points.
Then: the second partials.
Then: test each point.
Last: check.