Step by step
The question
Question: find and classify the critical points of f(x, y) = x² + 3y² − 6x + 6y.
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.
3y² has no x: it is a fixed number, so it gives 0.
6x gives 6.
6y has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 2x − 6.
Term by term.
x² has no y: it is a fixed number, so it gives 0.
3y²: the power 2 comes down: 3 times 2 is 6.
The power goes down by 1.
So 6y.
6x has no y: it is a fixed number, so it gives 0.
6y gives 6.
Drop the zeros: ∂f/∂y = 6y + 6.
Set ∂f/∂x = 0.
It has only x in it.
Move the number: add 6 on both sides.
Divide by 2: x = 3.
Set ∂f/∂y = 0.
It has only y in it.
Move the number: take away 6 on both sides.
Divide by 6: y = −1.
So there is one critical point: (3, −1).
2 The second partials
Term by term.
2x gives 2.
6 has no x: it is a fixed number, so it gives 0.
Drop the zeros: f_xx = 2.
Term by term.
6y gives 6.
6 has no y: it is a fixed number, so it gives 0.
Drop the zeros: f_yy = 6.
Term by term.
2x has no y: it is a fixed number, so it gives 0.
6 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 (3, −1): f_xx is 2, f_yy is 6, f_xy is 0.
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 positive and f_xx is positive: a bowl.
A local minimum.
There f is −12.
4 Check
Let's check.
Here is the surface near (3, −1).
It is a bowl: the point is at the bottom.
Now small steps away from the point.
Near (3, −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.