The question
Question: find and classify the critical points of f(x, y) = x² + y² − 4x + 6y.
f(x, y) = x2 + y2 − 4x + 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.
y² has no x: it is a fixed number, so it gives 0.
4x gives 4.
6y has no x: it is a fixed number, so it gives 0.
∂f∂x = 2x + 0 − 4 + 0
Drop the zeros: ∂f/∂x = 2x − 4.
∂f∂x = 2x − 4
Term by term.
x² has no y: it is a fixed number, so it gives 0.
y²: the power 2 comes down.
The power goes down by 1.
So 2y.
4x has no y: it is a fixed number, so it gives 0.
6y gives 6.
∂f∂y = 0 + 2y + 0 + 6
Drop the zeros: ∂f/∂y = 2y + 6.
∂f∂y = 2y + 6
Set ∂f/∂x = 0.
It has only x in it.
2x − 4 = 0
Move the number: add 4 on both sides.
2x = 4
Divide by 2: x = 2.
x = 2
Set ∂f/∂y = 0.
It has only y in it.
2y + 6 = 0
Move the number: take away 6 on both sides.
2y = −6
Divide by 2: y = −3.
y = −3
So there is one critical point: (2, −3).
= (2, −3)
2 The second partials
Term by term.
2x gives 2.
4 has no x: it is a fixed number, so it gives 0.
fxx = 2 + 0
Drop the zeros: f_xx = 2.
fxx = 2
Term by term.
2y gives 2.
6 has no y: it is a fixed number, so it gives 0.
fyy = 2 + 0
Drop the zeros: f_yy = 2.
fyy = 2
Term by term.
2x has no y: it is a fixed number, so it gives 0.
4 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, −3): f_xx is 2, f_yy is 2, f_xy is 0.
So D is 4.
D(2, −3) = 2 × 2 − 0² = 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 positive: a bowl.
A local minimum.
There f is −13.
(2, −3): = local minimum, f = −13
4 Check
Let's check.
Here is the surface near (2, −3).
It is a bowl: the point is at the bottom.
Now small steps away from the point.
Near (2, −3), one step gives a bigger value, the other a bigger one.
That fits a local minimum.
Right!
f(2, −3) = −13: f(2.1, −3) = −12.99, f(2, −2.9) = −12.99
Let's look back.
First: find the critical points.
Then: the second partials.
Then: test each point.
Last: check.
▶ Watch it solved
The question
Question: find and classify the critical points of f(x, y) = x² + 3y² − 6x + 6y.
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.
∂f∂x = 2x + 0 − 6 + 0
Drop the zeros: ∂f/∂x = 2x − 6.
∂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.
∂f∂y = 0 + 6y + 0 + 6
Drop the zeros: ∂f/∂y = 6y + 6.
∂f∂y = 6y + 6
Set ∂f/∂x = 0.
It has only x in it.
2x − 6 = 0
Move the number: add 6 on both sides.
2x = 6
Divide by 2: x = 3.
x = 3
Set ∂f/∂y = 0.
It has only y in it.
6y + 6 = 0
Move the number: take away 6 on both sides.
6y = −6
Divide by 6: y = −1.
y = −1
So there is one critical point: (3, −1).
= (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.
fxx = 2 + 0
Drop the zeros: f_xx = 2.
fxx = 2
Term by term.
6y gives 6.
6 has no y: it is a fixed number, so it gives 0.
fyy = 6 + 0
Drop the zeros: f_yy = 6.
fyy = 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.
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 (3, −1): f_xx is 2, f_yy is 6, f_xy is 0.
So D is 12.
D(3, −1) = 2 × 6 − 0² = 12
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 positive: a bowl.
A local minimum.
There f is −12.
(3, −1): = local minimum, f = −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!
f(3, −1) = −12: f(3.1, −1) = −11.99, f(3, −0.9) = −11.97
Let's look back.
First: find the critical points.
Then: the second partials.
Then: test each point.
Last: check.
▶ Watch it solved
The question
Question: find and classify the critical points of f(x, y) = x³ + y³ − 3x − 12y.
f(x, y) = x3 + y3 − 3x − 12y
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.
x³: the power 3 comes down.
The power goes down by 1.
So 3x².
y³ has no x: it is a fixed number, so it gives 0.
3x gives 3.
12y has no x: it is a fixed number, so it gives 0.
∂f∂x = 3x2 + 0 − 3 + 0
Drop the zeros: ∂f/∂x = 3x² − 3.
∂f∂x = 3x2 − 3
Term by term.
x³ has no y: it is a fixed number, so it gives 0.
y³: the power 3 comes down.
The power goes down by 1.
So 3y².
3x has no y: it is a fixed number, so it gives 0.
12y gives 12.
∂f∂y = 0 + 3y2 + 0 − 12
Drop the zeros: ∂f/∂y = 3y² − 12.
∂f∂y = 3y2 − 12
Set ∂f/∂x = 0.
It has only x in it.
3x2 − 3 = 0
Move the number and divide by 3: x² = 1.
x2 = 1
Two numbers square to 1: x = 1 or −1.
x = 1 or x = −1
Set ∂f/∂y = 0.
It has only y in it.
3y2 − 12 = 0
Move the number and divide by 3: y² = 4.
y2 = 4
Two numbers square to 4: y = 2 or −2.
y = 2 or y = −2
Every x goes with every y: 4 critical points.
= (1, 2), (1, −2), (−1, 2), (−1, −2)
2 The second partials
Term by term.
3x²: the power 2 comes down: 3 times 2 is 6.
The power goes down by 1.
So 6x.
3 has no x: it is a fixed number, so it gives 0.
fxx = 6x + 0
Drop the zeros: f_xx = 6x.
fxx = 6x
Term by term.
3y²: the power 2 comes down: 3 times 2 is 6.
The power goes down by 1.
So 6y.
12 has no y: it is a fixed number, so it gives 0.
fyy = 6y + 0
Drop the zeros: f_yy = 6y.
fyy = 6y
Term by term.
3x² has no y: it is a fixed number, so it gives 0.
3 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 (1, 2): f_xx is 6, f_yy is 12, f_xy is 0.
So D is 72.
D(1, 2) = 6 × 12 − 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 positive: a bowl.
A local minimum.
There f is −18.
(1, 2): = local minimum, f = −18
At (1, −2): f_xx is 6, f_yy is −12, f_xy is 0.
So D is −72.
D(1, −2) = 6 × (−12) − 0² = −72
D is negative: up one way, down the other.
A saddle point.
(1, −2): = saddle point
At (−1, 2): f_xx is −6, f_yy is 12, f_xy is 0.
So D is −72.
D(−1, 2) = (−6) × 12 − 0² = −72
D is negative: up one way, down the other.
A saddle point.
(−1, 2): = saddle point
At (−1, −2): f_xx is −6, f_yy is −12, f_xy is 0.
So D is 72.
D(−1, −2) = (−6) × (−12) − 0² = 72
D is positive and f_xx is negative: a dome.
A local maximum.
There f is 18.
(−1, −2): = local maximum, f = 18
4 Check
Let's check.
Here is the surface near (1, 2).
It is a bowl: the point is at the bottom.
Now small steps away from the point.
Near (1, 2), one step gives a bigger value, the other a bigger one.
That fits a local minimum.
Right!
f(1, 2) = −18: f(1.1, 2) = −17.969, f(1, 2.1) = −17.939
Near (1, −2), one step gives a bigger value, the other a smaller one.
That fits a saddle point.
Right!
f(1, −2) = 14: f(1.1, −2) = 14.031, f(1, −1.9) = 13.941
Near (−1, 2), one step gives a bigger value, the other a smaller one.
That fits a saddle point.
Right!
f(−1, 2) = −14: f(−1, 2.1) = −13.939, f(−0.9, 2) = −14.029
Near (−1, −2), one step gives a smaller value, the other a smaller one.
That fits a local maximum.
Right!
f(−1, −2) = 18: f(−0.9, −2) = 17.971, f(−1, −1.9) = 17.941
Let's look back.
First: find the critical points.
Then: the second partials.
Then: test each point.
Last: check.
Answer = (1, 2), (1, −2), (−1, 2), (−1, −2)
▶ Watch it solved
