Steply

Classify the critical points of f(x, y) = 2xy - x^2 - 2y^2 + 4y

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) = 2xy − x² − 2y² + 4y.

f(x, y) = 2xy − x2 − 2y2 + 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.

In 2xy, 2y is fixed and waits in front.

x gives 1.

So 2y.

x²: the power 2 comes down: −1 times 2 is −2.

The power goes down by 1.

So 2x.

2y² 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.

∂f∂x = 2y − 2x + 0 + 0

Drop the zeros: ∂f/∂x = 2y − 2x.

∂f∂x = 2y − 2x

Term by term.

In 2xy, 2x is fixed and waits in front.

y gives 1.

So 2x.

x² has no y: it is a fixed number, so it gives 0.

2y²: the power 2 comes down: −2 times 2 is −4.

The power goes down by 1.

So 4y.

4y gives 4.

∂f∂y = 2x + 0 − 4y + 4

Drop the zeros: ∂f/∂y = 2x − 4y + 4.

∂f∂y = 2x − 4y + 4

Both partial derivatives must be 0.

Call them ① and ②.

① 2y − 2x = 0
② 2x − 4y + 4 = 0

Divide ① by 2: smaller numbers.

① −x + y = 0

Divide ② by 2: smaller numbers.

② x − 2y + 2 = 0

From ②, get x alone: x = 2y − 2.

x = 2y − 2

Put it into ①.

−(2y − 2) + y = 0

Open the bracket and add the like terms.

−y + 2 = 0

So y = 2.

y = 2

Back into x = 2y − 2: x = 2.

x = 2

So there is one critical point: (2, 2).

= (2, 2)

2 The second partials

Term by term.

2y has no x: it is a fixed number, so it gives 0.

2x gives 2.

fxx = 0 − 2

Drop the zeros: f_xx = −2.

fxx = −2

Term by term.

2x has no y: it is a fixed number, so it gives 0.

4y gives 4.

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

fyy = 0 − 4 + 0

Drop the zeros: f_yy = −4.

fyy = −4

Term by term.

2y gives 2.

2x has no y: it is a fixed number, so it gives 0.

fxy = 2 + 0

Drop the zeros: f_xy = 2.

fxy = 2

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

D = fxx × fyy − (fxy)²

3 Test each point

At (2, 2): f_xx is −2, f_yy is −4, f_xy is 2.

So D is 4.

D(2, 2) = (−2) × (−4) − 2² = 4

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 positive and f_xx is negative: a dome.

A local maximum.

There f is 4.

(2, 2): = local maximum, f = 4

4 Check

Let's check.

Here is the surface near (2, 2).

It is a dome: the point is at the top.

Now small steps away from the point.

Near (2, 2), one step gives a smaller value, the other a smaller one.

That fits a local maximum.

Right!

f(2, 2) = 4: f(2.1, 2) = 3.99, f(2, 2.1) = 3.98

Let's look back.

First: find the critical points.

Then: the second partials.

Then: test each point.

Last: check.

Answer
= (2, 2)

Your turn

Try this one yourself, then check:

Find and classify the critical points of f(x, y) = 3xy − 3x2 − 2y2 + 5y
Show the answer
local maximum at (1, 2)

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