Steply

Critical points of f(x, y) = 12x - x^3 - 3y^2

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) = 12x − x³ − 3y².

f(x, y) = 12x − x3 − 3y2

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.

12x gives 12.

x³: the power 3 comes down: −1 times 3 is −3.

The power goes down by 1.

So 3x².

3y² has no x: it is a fixed number, so it gives 0.

∂f∂x = 12 − 3x2 + 0

Drop the zeros: ∂f/∂x = 12 − 3x².

∂f∂x = 12 − 3x2

Term by term.

12x has no y: it is a fixed number, so it gives 0.

x³ 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.

∂f∂y = 0 + 0 − 6y

Drop the zeros: ∂f/∂y = −6y.

∂f∂y = −6y

Set ∂f/∂x = 0.

It has only x in it.

12 − 3x2 = 0

Move the number and divide by −3: x² = 4.

x2 = 4

Two numbers square to 4: x = 2 or −2.

x = 2 or x = −2

Set ∂f/∂y = 0.

It has only y in it.

−6y = 0

Divide by −6: y = 0.

y = 0

Every x goes with every y: 2 critical points.

= (2, 0), (−2, 0)

2 The second partials

Term by term.

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

3x²: the power 2 comes down: −3 times 2 is −6.

The power goes down by 1.

So 6x.

fxx = 0 − 6x

Drop the zeros: f_xx = −6x.

fxx = −6x

Term by term.

6y gives 6.

So f_yy = −6.

fyy = −6

Term by term.

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

3x² has no y: it is a fixed number, so it gives 0.

fxy = 0 + 0

Drop the zeros: f_xy = 0.

fxy = 0

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

D = fxx × fyy − (fxy)²

3 Test each point

At (2, 0): f_xx is −12, f_yy is −6, f_xy is 0.

So D is 72.

D(2, 0) = (−12) × (−6) − 0² = 72

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 16.

(2, 0): = local maximum, f = 16

At (−2, 0): f_xx is 12, f_yy is −6, f_xy is 0.

So D is −72.

D(−2, 0) = 12 × (−6) − 0² = −72

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

A saddle point.

(−2, 0): = saddle point

4 Check

Let's check.

Here is the surface near (2, 0).

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

Now small steps away from the point.

Near (2, 0), one step gives a smaller value, the other a smaller one.

That fits a local maximum.

Right!

f(2, 0) = 16: f(2.1, 0) = 15.939, f(2, 0.1) = 15.97

Near (−2, 0), one step gives a bigger value, the other a smaller one.

That fits a saddle point.

Right!

f(−2, 0) = −16: f(−1.9, 0) = −15.941, f(−2, 0.1) = −16.03

Let's look back.

First: find the critical points.

Then: the second partials.

Then: test each point.

Last: check.

Answer
= (2, 0), (−2, 0)

Your turn

Try this one yourself, then check:

Find and classify the critical points of f(x, y) = 6x − 2x3 − 3y2
Show the answer
local maximum at (1, 0); saddle point at (−1, 0)

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