Lesson 28: Lagrange Multipliers II

Lesson 28 (Sections 18.2–5)
Lagrange Multipliers II

Math 20

November 28, 2007

Announcements
Problem Set 11 assigned today. Due December 5.
next OH: Today 1–3 (SC 323)
Midterm II review: Tuesday 12/4, 7:30-9:00pm in Hall E
Midterm II: Thursday, 12/6, 7-8:30pm in Hall A

Outline

A homework problem

Restating the Method of Lagrange Multipliers
Statement
Justiﬁcations

Second order conditions
Compact feasibility sets
Ad hoc arguments
Analytic conditions

Example: More than two variables

More than one constraint

Problem 17.1.10

Problem
Maximize the quantity f (x, y , z) = Ax a y b z c subject to the
constraint that px + qy + rz = m. (Here A, a, b, c, p, q, r , m are
positive constants.)

Problem 17.1.10

Problem

Solution (By elimination)
Solving the constraint for z in terms of x and y , we get
m − px − qy
z=
r
So we optimize the unconstrained function
Aab
x y (m − px − qy )c
f (x, y ) =
rc

We have
∂f (x, y ) A
= c ax a−1 y b (m − px − qy )c + x a y b c(m − px − qy )c−1 (−p)
x r
A
= c x a−1 y b (m − px − qy )c−1 [a(m − px − qy ) − cpx]
r
Likewise
∂f (x, y ) A
= c x a y b−1 (m − px − qy )c−1 [b(m − px − qy ) − cqy ]
y r
So throwing out the critical points where x = 0, y = 0, or z = 0
(these give minimal values of f , not maximal), we get

(a + c)px + aqy = am
bpx + (b + c)qy = bm

This is a fun exercise in Cramer’s Rule:
am aq 1 1
amq
bm (b + c)q b b+c
x= =
(a + c)p aq a+c a
pq
bp (b + c)q b b+c
amqc m a
= =
pq(ac + bc + c 2 ) p a+b+c

It follows that
m b m c
y= z=
q a+b+c r a+b+c
If this is a utility-maximization problem subject to ﬁxed budget,
the portion spent on each good ( px , for instance) is the relative
m
a
degree to which that good multiplies utility ( a+b+c ).

Theorem (The Method of Lagrange Multipliers)
Let f (x1 , x2 , . . . , xn ) and g (x1 , x2 , . . . , xn ) be functions of several
variables. The critical points of the function f restricted to the set
g = 0 are solutions to the equations:
∂f ∂g
(x1 , x2 , . . . , xn ) = λ (x1 , x2 , . . . , xn ) for each i = 1, . . . , n
∂xi ∂xi
g (x1 , x2 , . . . , xn ) = 0.

Note that this is n + 1 equations in n + 1 variables x1 , . . . , xn , λ.

Graphical Justiﬁcation
In two variables, the critical points of f restricted to the level curve
g = 0 are found when the tangent to the the level curve of f is
parallel to the tangent to the level curve g = 0.

These tangents have slopes

dy fx dy gx
=− =−
and
dx fy dx gy
f g

These tangents have slopes

dy fx dy gx
=− =−
and
dx fy dx gy
f g

So they are equal when

fy
fx g f
= x =⇒ x =
fy gy gx gy
or

fx = λgx
fy = λgy

Symbolic Justiﬁcation

Suppose that we can use the relation g (x1 , . . . , xn ) = 0 to solve for
xn in terms of the the other variables x1 , . . . , xn−1 , after making
some choices. Then the critical points of f (x1 , . . . , xn ) are
unconstrained critical points of f (x1 , . . . , xn (x1 , . . . , xn−1 )).

f

x1 x2 xn
···

xn−1
x1 x2 ···

Now for any i = 1, . . . , n − 1,

∂f ∂f ∂f ∂xn
= +
∂xi ∂xi ∂xn ∂xi
g g
∂f ∂f ∂g /∂xi
−
=
∂xi ∂xn ∂g /∂xn

∂f
If = 0, then
∂xi g

∂f /∂xi ∂g /∂xi ∂f /∂xi ∂f /∂xn
⇐⇒
= =
∂f /∂xn ∂g /∂xn ∂g /∂xi ∂g /∂xn

∂f ∂g
So as before, =λ for all i.
∂xi ∂xi

Another perspective

To ﬁnd the critical points of f subject to the constraint that
g = 0, create the lagrangian function

L = f (x1 , x2 , . . . , xn ) − λg (x1 , x2 , . . . , xn )

If L is restricted to the set g = 0, L = f and so the constrained
critical points are unconstrained critical points of L . So for each i,
∂L ∂f ∂g
= 0 =⇒ =λ .
∂xi ∂xi ∂xi
But also,
∂L
= 0 =⇒ g (x1 , x2 , . . . , xn ) = 0.
∂λ

Second order conditions

The Method of Lagrange Multipliers ﬁnds the constrained critical
points, but doesn’t determine their “type” (max, min, neither).
So what then?

A dash of topology
Cf. Sections 17.2–3

Deﬁnition
A subset of Rn is called closed if it includes its boundary.

A dash of topology
Cf. Sections 17.2–3

Deﬁnition
A subset of Rn is called closed if it includes its boundary.

x2 + y2 ≤ 1
x2 + y2 ≤ 1 y ≥0
not closed closed
closed

Basically, if a subset is described by ≤ or ≥ inequalities, it is closed.

Deﬁnition
A subset of Rn is called bounded if it is contained within some
ball centered at the origin.

x2 + y2 ≤ 1
x2 + y2 ≤ 1 y ≥0
bounded not bounded
bounded

Deﬁnition
A subset of Rn is called compact if it is closed and bounded.

x2 + y2 ≤ 1
x2 + y2 ≤ 1 y ≥0
not compact not compact
compact

Optimizing over compact sets

Theorem (Compact Set Method)
To find the extreme values of function f on a compact set D of
Rn , it suffices to find
the (unconstrained) critical points of f “inside” D
the (constrained) critical points of f on the “boundary” of D.

Ad hoc arguments

If D is not compact, sometimes it’s still easy to argue that as x
gets farther away, f becomes larger, or smaller, so the critical
points are “obviously” maxes, or mins.

Ad hoc arguments

If D is not compact, sometimes it’s still easy to argue that as x
gets farther away, f becomes larger, or smaller, so the critical
points are “obviously” maxes, or mins.
(Example later)

Analytic conditions
Recall Equation 16.13, cf. Section 18.4

For the two-variable constrained optimization problem, we
have (look in the book if you want the gory details):

Lλλ Lλx Lλy
0 gx gy
d 2f
fxy − λgxy = Lxλ Lxx Lxy
− λgxx
= gx fxx
dx 2
Ly λ Lyx Lyy
g − λgyx gyy − λgyy
gy fyx

The critical point is a local max if this determinant is
negative, and a local min if this is positive.
The matrix on the right is the Hessian of the Lagrangian. But
there is still a distinction between this and the unconstrained
case. The constrained extrema are critical points of the
Lagrangian, not extrema.
Don’t worry too much about this!

Problem 17.1.10

Problem

Solution
The Lagrange equations are

Aax a−1 y b z c = λp
Abx a y b−1 z c = λq
Acx a y b z c−1 = λr

We rule out any solution with x, y , z, or λ equal to 0 (they will
minimize f , not maximize it).

Dividing the ﬁrst two equations gives
ay p bp
= =⇒ y = x
bx q aq
Dividing the ﬁrst and last equations gives
az p cp
= =⇒ z = x
cx r ar
Plugging these into the equation of constraint gives
bp cp m a
px + x + x = m =⇒ x =
a a p a+b+c

General method for more than one constraint

If we are optimizing f (x1 , . . . , xn ) subject to gj (x1 , . . . , xn ) ≡ 0,
j = 1, . . . , m we need multiple lambdas for them. The new
Lagrangian is
m
L (x1 , . . . , xn ) = f (x1 , . . . , xn ) − λj gj (x1 , . . . , xn )
j=1

∂L ∂L
The conditions are that = 0 and = 0 for all i and j. In
∂xi ∂λj
other words,
∂f ∂g1 ∂gm
+ · · · + λm
= λ1 (all i)
∂xi ∂xi ∂xi
gj (x1 , . . . , xn ) = 0 (all j)

Example
Find the minimum distance between the curves xy = 1 and
x + 2y = 1.

Example
Find the minimum distance between the curves xy = 1 and
x + 2y = 1.
Reframing this, we can minimize

f (x, y , u, v ) = (x − u)2 + (y − v )2

subject to the constraints

xy − 1 = 0 u + 2v = 1.

•
•
•
• xy = 1
•
•
• •
x + 2y = 1

The Lagrangian is

L = (x − u)2 + (y − v )2 − λ(xy − 1) − µ(u + 2v − 1)

So the Lagrangian equations are

2(x − u) = λy −2(x − u) = µ
2(y − v ) = λx −2(y − v ) = 2µ

Dividing the two λ equations and the two µ equations gives
x −u x −u
y 1
= =.
y −v y −v
x 2
Since the left-hand-sides are the same, we have 2y =√ Since
x.
√ 1 1
√ , or x = − 2, y = − √
xy = 1, we can say either x = 2, y = 2 2

√ 1
Suppose x = 2, y = √. Then
2
√ √
2−u 1 32
=⇒ 2u − v =
=
1 2 2
√ −v
2

This along with u + 2v = 1 gives
√ √
1 1
4−3 2
u= 1+3 2 v=
5 10
√ 1
If we instead choose x = − 2, y = − √2 , we get

√
1 1 3
2+ √
1−3 2
u= v=
5 5 2

√
1
9−4 2
5

• •
xy = 1
•
•

x + 2y = 1
√
1
9+4 2
5

Because f gets larger as x, y , u, and v get larger, the absolute
minimum is the smaller of these two critical values. So the
√
1
minimum distance is 5 9 − 4 2 .

Lesson 28: Lagrange Multipliers II

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