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The second derivative test for functions of two variables.
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Lesson 25 (Chapter 17)Unconstrained Optimization I
Math 20
November 19, 2007
Announcements
I Problem Set 9 on the website. Due November 21.
I There will be class November 21 and homework dueNovember 28.
I next OH: Monday 1-2pm, Tuesday 3-4pm
I Midterm II: Thursday, 12/6, 7-8:30pm in Hall A.
Outline
Single-variable recollections
From one to two dimensionsCritical pointsThe Hessian
The second derivative test
More examplesThe discriminating monopolist
Maximum and Minimum Value in single-variable calculus
Theorem (Fermat’s Theorem)
Let f be a function of one variable. If f has a local maximum orminimum at a, then f ′(a) = 0.
Theorem (Theorem 9.2, a/k/a The Second Derivative Test)
Let f be a function of one variable, and suppose f ′(a) = 0.
I If f ′′(a) > 0, then f has a local minimum at a.
I If f ′′(a) < 0, then f has a local maximum at a.
I (If f ′′(a) = 0, this theorem has nothing to say).
Maximum and Minimum Value in single-variable calculus
Theorem (Fermat’s Theorem)
Let f be a function of one variable. If f has a local maximum orminimum at a, then f ′(a) = 0.
Theorem (Theorem 9.2, a/k/a The Second Derivative Test)
Let f be a function of one variable, and suppose f ′(a) = 0.
I If f ′′(a) > 0, then f has a local minimum at a.
I If f ′′(a) < 0, then f has a local maximum at a.
I (If f ′′(a) = 0, this theorem has nothing to say).
Justification of 2DT
Using Taylor’s Theorem
f (x) = f (a) + f ′(a)(x − a) +1
2f ′′(a)(x − a)2 + R(x),
where R(x)(x−a)2
→ 0 as x → a. (See Sections 5.5 and 7.4) So near a,
f (x) “looks like” a parabola with vertex at (a, f (a)). f ′′(a) is whatdetermines whether this parabola opens up or down.
Outline
Single-variable recollections
From one to two dimensionsCritical pointsThe Hessian
The second derivative test
More examplesThe discriminating monopolist
How do we generalize this to functions of two variables?
The first derivative f ′(x) is replaced by the gradient
Df =(
∂f∂x
∂f∂y
)∇f =
( ∂f∂x∂f∂y
)
Theorem (Fermat’s Theorem)
Let f (x , y) be a function of two variables. If f has a localmaximum or minimum at (a, b), and is differentiable at (a,b), then
∂f
∂x(a, b) = 0
∂f
∂y(a, b) = 0
As in one variable, we’ll call these points critical points.
How do we generalize this to functions of two variables?
The first derivative f ′(x) is replaced by the gradient
Df =(
∂f∂x
∂f∂y
)∇f =
( ∂f∂x∂f∂y
)
Theorem (Fermat’s Theorem)
Let f (x , y) be a function of two variables. If f has a localmaximum or minimum at (a, b), and is differentiable at (a,b), then
∂f
∂x(a, b) = 0
∂f
∂y(a, b) = 0
As in one variable, we’ll call these points critical points.
Example
Example
Let f (x , y) = 8x3 − 24xy + y3. Find the critical points of f .
SolutionWe have
∂f
∂x= 24x2 − 24y = 24(x2 − y)
∂f
∂y= 24x − 3y2 = 3(8x − y2)
Both of these are zero if x2 − y = 0 and 8x − y2 = 0. Substitutingthe first into the second gives
0 = (x2)2 − 8x = x4 − 8x = x(x3 − 8) = x(x − 2)(x2 + 2x + 4)
and the solutions are x = 0 and x = 2 (the third factor has no realroots). If x = 0 then y = 0, and If x = 2 then y = 4. So thecritical points are (0, 0) and (2, 4).
Example
Example
Let f (x , y) = 8x3 − 24xy + y3. Find the critical points of f .
SolutionWe have
∂f
∂x= 24x2 − 24y = 24(x2 − y)
∂f
∂y= 24x − 3y2 = 3(8x − y2)
Both of these are zero if x2 − y = 0 and 8x − y2 = 0. Substitutingthe first into the second gives
0 = (x2)2 − 8x = x4 − 8x = x(x3 − 8) = x(x − 2)(x2 + 2x + 4)
and the solutions are x = 0 and x = 2 (the third factor has no realroots). If x = 0 then y = 0, and If x = 2 then y = 4. So thecritical points are (0, 0) and (2, 4).
How do we generalize this to functions of two variables?
The second derivative f ′′(x) is replaced by . . .
a matrix, theHessian of f :
Hf =
∂2f
∂x2
∂2f
∂x∂y∂2f
∂y∂x
∂2f
∂y2
How do we generalize this to functions of two variables?
The second derivative f ′′(x) is replaced by . . . a matrix, theHessian of f :
Hf =
∂2f
∂x2
∂2f
∂x∂y∂2f
∂y∂x
∂2f
∂y2
Compare and contrast the Hessians at (0, 0) for these functions:
(i) f (x , y) = x2 + y2
(ii) f (x , y) = 1− x2 − y2
(iii) f (x , y) = x2 − y2
(iv) f (x , y) = xy
How are they alike and how are they different?
(i) Hf =
(2 00 2
)(ii) Hf =
(−2 00 −2
) (iii) Hf =
(2 00 −2
)(iv) Hf =
(0 11 0
)
Compare and contrast the Hessians at (0, 0) for these functions:
(i) f (x , y) = x2 + y2
(ii) f (x , y) = 1− x2 − y2
(iii) f (x , y) = x2 − y2
(iv) f (x , y) = xy
How are they alike and how are they different?
(i) Hf =
(2 00 2
)(ii) Hf =
(−2 00 −2
) (iii) Hf =
(2 00 −2
)(iv) Hf =
(0 11 0
)
Outline
Single-variable recollections
From one to two dimensionsCritical pointsThe Hessian
The second derivative test
More examplesThe discriminating monopolist
Second order Taylor polynomials in two dimensions
The two-variable analog of
f (x) ≈ f (a) + f ′(a)(x − a) +1
2f ′′(a)(x − a)2
is
f (x , y) ≈ f (a, b) + f ′x(a, b)(x − a) + f ′y (a, b)(y − b)
+ 12 f ′′xx(a, b)(x − a)2 + f ′′xy (a, b)(x − a)(y − b)
+ 12 f ′′yy (a, b)(y − a)2
or
f (x) ≈ f (a) +∇f (a) · (x− a) + 12(x− a) · H(a)(x− a)
Recall
This was the big fact about quadratic forms in two variables:
FactLet f (x , y) = ax2 + 2bxy + cy2 be a quadratic form.
I If a > 0 and ac − b2 > 0, then f is positive definite
I If a < 0 and ac − b2 > 0, then f is negative definite
I If ac − b2 < 0, then f is indefinite
Theorem (The Second Derivative Test)
Let f (x , y) be a function of two variables, and let (a, b) be acritical point of f . Then
I If ∂2f∂x2
∂2f∂y2 −
(∂2f
∂x∂y
)2> 0 and ∂2f
∂x2 > 0, the critical point is a
local minimum.
I If ∂2f∂x2
∂2f∂y2 −
(∂2f
∂x∂y
)2> 0 and ∂2f
∂x2 < 0, the critical point is a
local maximum.
I If ∂2f∂x2
∂2f∂y2 −
(∂2f
∂x∂y
)2< 0, the critical point is a saddle point.
All derivatives are evaluated at the critical point (a, b).
Return to the example
Let f (x , y) = 8x3 − 24xy + y3. Classify the critical points.
∂2f
∂x2= 48x
∂2f
∂x∂y= −24
∂2f
∂y∂x= −24
∂2f
∂y2= 6y
I Hf (0, 0) =
(0 −24−24 0
), which has negative determinant.
Hence (0, 0) is a saddle point.
I Hf (2, 4) = 24
(4 −1−1 1
)which, since the determinant is
positive and the top left entry is positive, indicates a localminimum.
Return to the example
Let f (x , y) = 8x3 − 24xy + y3. Classify the critical points.
∂2f
∂x2= 48x
∂2f
∂x∂y= −24
∂2f
∂y∂x= −24
∂2f
∂y2= 6y
I Hf (0, 0) =
(0 −24−24 0
), which has negative determinant.
Hence (0, 0) is a saddle point.
I Hf (2, 4) = 24
(4 −1−1 1
)which, since the determinant is
positive and the top left entry is positive, indicates a localminimum.
Plotting the function
-2
0
2
4
0
5
-10
-5
0
5
10
-1 0 1 2 3-1
0
1
2
3
4
5
Plotting the function
-2
0
2
4
0
5
-10
-5
0
5
10
-1 0 1 2 3-1
0
1
2
3
4
5
Online Demo
Try this site (thanks to Tony Pino):
http://www.slu.edu/classes/maymk/banchoff/LevelCurve.html
Launch the applet and enter:
I f (x , y) = x^3 - 3 * x * y + y^3/8 (1/3 of f from theexample)
I x from −1 to 10 in 50 steps
I y from −1 to 10 in 50 steps
I z from −10 to 10 in 50 steps
Remarks
I The Hessian matrix will always be symmetric in our cases.I If the Hessian has determinant zero, nothing can be said from
this theorem:I f (x , y) = x4 + y4 has a local min at (0, 0)I f (x , y) = −x4 − y4 has a local max at (0, 0)I f (x , y) = x4 − y4 has a saddle point at (0, 0)
In each case Hf (x , y) =
(±12x2 0
0 ±12y2
), so Hf (0, 0) is the
zero matrix.
Outline
Single-variable recollections
From one to two dimensionsCritical pointsThe Hessian
The second derivative test
More examplesThe discriminating monopolist
Example
A firm sells a product in two separate areas with distinct lineardemand curves, and has monopoly power to decide how much tosell in each area. How does its maximal profit depend on thedemand in each area?
Let the demand curves be given by
P1 = a1 − b1Q1 P2 = a2 − b2Q2
And the cost function by C = α(Q1 + Q2). The profit is therefore
π = P1Q1 + P2Q2 − α(Q1 + Q2)
= (a1 − b1Q1)Q1 + (a2 − b2Q2)Q2 − α(Q1 + Q2)
= (a1 − α)Q1 − b1Q21 + (a2 − α)Q2 − b2Q2
2
Example
A firm sells a product in two separate areas with distinct lineardemand curves, and has monopoly power to decide how much tosell in each area. How does its maximal profit depend on thedemand in each area?
Let the demand curves be given by
P1 = a1 − b1Q1 P2 = a2 − b2Q2
And the cost function by C = α(Q1 + Q2). The profit is therefore
π = P1Q1 + P2Q2 − α(Q1 + Q2)
= (a1 − b1Q1)Q1 + (a2 − b2Q2)Q2 − α(Q1 + Q2)
= (a1 − α)Q1 − b1Q21 + (a2 − α)Q2 − b2Q2
2
π(Q1,Q2) = (a1 − α)Q1 − b1Q21 + (a2 − α)Q2 − b2Q2
2
Solution
We have
∂π
∂Q1= a1 − α− 2b1Q1
∂π
∂Q2= a2 − α− 2b2Q2
So
Q∗1 =a1 − α
2b1Q∗2 =
a2 − α2b2
is the critical point. Also,
Hπ =
(−2b1 0
0 −2b2
)So the critical point (Q∗1 ,Q
∗2 ) is a local maximum.
π(Q1,Q2) = (a1 − α)Q1 − b1Q21 + (a2 − α)Q2 − b2Q2
2
SolutionWe have
∂π
∂Q1= a1 − α− 2b1Q1
∂π
∂Q2= a2 − α− 2b2Q2
So
Q∗1 =a1 − α
2b1Q∗2 =
a2 − α2b2
is the critical point. Also,
Hπ =
(−2b1 0
0 −2b2
)So the critical point (Q∗1 ,Q
∗2 ) is a local maximum.
π(Q1,Q2) = (a1 − α)Q1 − b1Q21 + (a2 − α)Q2 − b2Q2
2
SolutionWe have
∂π
∂Q1= a1 − α− 2b1Q1
∂π
∂Q2= a2 − α− 2b2Q2
So
Q∗1 =a1 − α
2b1Q∗2 =
a2 − α2b2
is the critical point.
Also,
Hπ =
(−2b1 0
0 −2b2
)So the critical point (Q∗1 ,Q
∗2 ) is a local maximum.
π(Q1,Q2) = (a1 − α)Q1 − b1Q21 + (a2 − α)Q2 − b2Q2
2
SolutionWe have
∂π
∂Q1= a1 − α− 2b1Q1
∂π
∂Q2= a2 − α− 2b2Q2
So
Q∗1 =a1 − α
2b1Q∗2 =
a2 − α2b2
is the critical point. Also,
Hπ =
(−2b1 0
0 −2b2
)
So the critical point (Q∗1 ,Q∗2 ) is a local maximum.
π(Q1,Q2) = (a1 − α)Q1 − b1Q21 + (a2 − α)Q2 − b2Q2
2
SolutionWe have
∂π
∂Q1= a1 − α− 2b1Q1
∂π
∂Q2= a2 − α− 2b2Q2
So
Q∗1 =a1 − α
2b1Q∗2 =
a2 − α2b2
is the critical point. Also,
Hπ =
(−2b1 0
0 −2b2
)So the critical point (Q∗1 ,Q
∗2 ) is a local maximum.
Example
Find the critical points of f (x , y) = xx2+y2+1
and classify them.
SolutionThe derivatives are
f ′x =1− x2 + y2
(1 + x2 + y2)2f ′y = − 2xy
(1 + x2 + y2)2
The only way these can both be zero is if y = 0 and x = ±1. Thesecond derivatives are
f ′′xx =2x(x2 − 3
(y2 + 1
))(x2 + y2 + 1)3
f ′′xy =2y(−3x2 + y2 + 1
)(x2 + y2 + 1)3
f ′′yy = −2(x3 − 3y2x + x
)(x2 + y2 + 1)3
Example
Find the critical points of f (x , y) = xx2+y2+1
and classify them.
SolutionThe derivatives are
f ′x =1− x2 + y2
(1 + x2 + y2)2f ′y = − 2xy
(1 + x2 + y2)2
The only way these can both be zero is if y = 0 and x = ±1. Thesecond derivatives are
f ′′xx =2x(x2 − 3
(y2 + 1
))(x2 + y2 + 1)3
f ′′xy =2y(−3x2 + y2 + 1
)(x2 + y2 + 1)3
f ′′yy = −2(x3 − 3y2x + x
)(x2 + y2 + 1)3
Example
Find the critical points of f (x , y) = xx2+y2+1
and classify them.
SolutionThe derivatives are
f ′x =1− x2 + y2
(1 + x2 + y2)2f ′y = − 2xy
(1 + x2 + y2)2
The only way these can both be zero is if y = 0 and x = ±1.
Thesecond derivatives are
f ′′xx =2x(x2 − 3
(y2 + 1
))(x2 + y2 + 1)3
f ′′xy =2y(−3x2 + y2 + 1
)(x2 + y2 + 1)3
f ′′yy = −2(x3 − 3y2x + x
)(x2 + y2 + 1)3
Example
Find the critical points of f (x , y) = xx2+y2+1
and classify them.
SolutionThe derivatives are
f ′x =1− x2 + y2
(1 + x2 + y2)2f ′y = − 2xy
(1 + x2 + y2)2
The only way these can both be zero is if y = 0 and x = ±1. Thesecond derivatives are
f ′′xx =2x(x2 − 3
(y2 + 1
))(x2 + y2 + 1)3
f ′′xy =2y(−3x2 + y2 + 1
)(x2 + y2 + 1)3
f ′′yy = −2(x3 − 3y2x + x
)(x2 + y2 + 1)3
So
Hf (1, 0) =
(−1
2 00 −1
2
)Hf (−1, 0) =
(12 00 1
2
)
So we have a local max and a local min.
So
Hf (1, 0) =
(−1
2 00 −1
2
)Hf (−1, 0) =
(12 00 1
2
)So we have a local max and a local min.
Plotting the function
-2
-1
0
1
2-2
-1
0
1
2
-0.5
0.0
0.5
-2 -1 0 1 2-2
-1
0
1
2
Plotting the function
-2
-1
0
1
2-2
-1
0
1
2
-0.5
0.0
0.5
-2 -1 0 1 2-2
-1
0
1
2