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Critical points of f(x, y) = xy

Step by step, with the reason for every step · Local maximum, minimum and saddle points · Calculus 3

Step by step

The question

Question: find and classify the critical points of f(x, y) = xy.

f(x, y) = xy

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.

In xy, y is fixed and waits in front.

x gives 1.

So y.

So ∂f/∂x = y.

∂f∂x = y

Term by term.

In xy, x is fixed and waits in front.

y gives 1.

So x.

So ∂f/∂y = x.

∂f∂y = x

Both partial derivatives must be 0.

Call them ① and ②.

① y = 0
② x = 0

② gives x = 0 straight away.

x = 0

① has only y in it.

y = 0

So y = 0.

y = 0

So there is one critical point: (0, 0).

= (0, 0)

2 The second partials

Term by term.

y has no x: it is a fixed number, so it gives 0.

fxx = 0

Drop the zeros: f_xx = 0.

fxx = 0

Term by term.

x has no y: it is a fixed number, so it gives 0.

fyy = 0

Drop the zeros: f_yy = 0.

fyy = 0

Term by term.

y gives 1.

So f_xy = 1.

fxy = 1

The test number: D = f_xx × f_yy − (f_xy)².

D = fxx × fyy − (fxy)²

3 Test each point

At (0, 0): f_xx is 0, f_yy is 0, f_xy is 1.

So D is −1.

D(0, 0) = 0 × 0 − 1² = −1

Careful!

f_xx alone is not enough.

If D is negative, it is a saddle, whatever f_xx is.

Careful: D first, then the sign of fxx

D is negative: up one way, down the other.

A saddle point.

(0, 0): = 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!

f(0, 0) = 0: f(0.1, 0.1) = 0.01, f(0.1, −0.1) = −0.01

Let's look back.

First: find the critical points.

Then: the second partials.

Then: test each point.

Last: check.

Answer
= (0, 0)

Your turn

Try this one yourself, then check:

Find and classify the critical points of f(x, y) = 3xy
Show the answer
saddle point at (0, 0)

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