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