Step by step
The question
Question: find and classify the critical points of f(x, y) = 2x² + y² − 8x.
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.
2x²: the power 2 comes down: 2 times 2 is 4.
The power goes down by 1.
So 4x.
y² has no x: it is a fixed number, so it gives 0.
8x gives 8.
Drop the zeros: ∂f/∂x = 4x − 8.
Term by term.
2x² has no y: it is a fixed number, so it gives 0.
y²: the power 2 comes down.
The power goes down by 1.
So 2y.
8x has no y: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂y = 2y.
Set ∂f/∂x = 0.
It has only x in it.
Move the number: add 8 on both sides.
Divide by 4: x = 2.
Set ∂f/∂y = 0.
It has only y in it.
Divide by 2: y = 0.
So there is one critical point: (2, 0).
2 The second partials
Term by term.
4x gives 4.
8 has no x: it is a fixed number, so it gives 0.
Drop the zeros: f_xx = 4.
Term by term.
2y gives 2.
So f_yy = 2.
Term by term.
4x has no y: it is a fixed number, so it gives 0.
8 has no y: it is a fixed number, so it gives 0.
Drop the zeros: f_xy = 0.
The test number: D = f_xx × f_yy − (f_xy)².
3 Test each point
At (2, 0): f_xx is 4, f_yy is 2, f_xy is 0.
So D is 8.
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 −8.
4 Check
Let's check.
Here is the surface near (2, 0).
It is a bowl: the point is at the bottom.
Now small steps away from the point.
Near (2, 0), 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.