Step by step
The question
Question: find and classify the critical points of f(x, y) = x² + 2xy + 3y² − 4y.
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 2xy, 2y is fixed and waits in front.
x gives 1.
So 2y.
3y² has no x: it is a fixed number, so it gives 0.
4y has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 2x + 2y.
Term by term.
x² has no y: it is a fixed number, so it gives 0.
In 2xy, 2x is fixed and waits in front.
y gives 1.
So 2x.
3y²: the power 2 comes down: 3 times 2 is 6.
The power goes down by 1.
So 6y.
4y gives 4.
Drop the zeros: ∂f/∂y = 2x + 6y − 4.
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 x alone: x = −3y + 2.
Put it into ①.
Open the bracket and add the like terms.
So y = 1.
Back into x = −3y + 2: x = −1.
So there is one critical point: (−1, 1).
2 The second partials
Term by term.
2x gives 2.
2y has no x: it is a fixed number, so it gives 0.
Drop the zeros: f_xx = 2.
Term by term.
2x has no y: it is a fixed number, so it gives 0.
6y gives 6.
4 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.
2y gives 2.
Drop the zeros: f_xy = 2.
The test number: D = f_xx × f_yy − (f_xy)².
3 Test each point
At (−1, 1): f_xx is 2, f_yy is 6, f_xy is 2.
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 −2.
4 Check
Let's check.
Here is the surface near (−1, 1).
It is a bowl: the point is at the bottom.
Now small steps away from the point.
Near (−1, 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.