Mathematical Economics:
Lecture 13
Yu Ren
WISE, Xiamen University
November 7, 2012
Chapter 18: Constrained Optimization I
Outline
1
Chapter 18: Constrained Optimization I
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
New Section
Chapter 18:
Constrained
Optimization I
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Definitions
Objective function: f (x1 , x2 , · · · , xn )
Constraint function:
gk (x1 , · · · , xn )≤bk
fk (x1 , · · · , xn ) = ck
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.1
Example 18.1 (Utility Maximization Problem) In
this most basic problem, Xi represents the
amount of commodity i and f (x1 , . . . , xn ), usually
written as U(x1 , . . . , xn ), measures the
individual’s level of utility or satisfaction with
consuming x1 units of good 1, x2 units of good
2, and so on. Let p1 , . . . , pn denote the prices of
the commodities and let I denote the individual’s
income.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.1
The consumer wants to
maximize U(x1 , . . . , xn )
subject to p1 x1 + p2 x2 + · · · + pn xn ≤ I
To be consistent with the general format in (1),
the nonnegativity constraints xi ≥ 0 should be
written as −xi ≤ 0 so that all inequality
constraints are written with ≤ signs.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.2
Example 18.2 (Utility Maximization with
Labor/Leisure Choice) Let U, x1 , . . . , xn ,
p1 , . . . , pn be as in the preceding example. In
addition, let w denote the wage rate, I 0 the
consumer’s nonwage income, l0 hours of labor,
and l1 hours of leisure. The consumer has
I 0 + wl0 dollars to spend and wants to
maximize U(x1 , . . . , xn , l1 )
subject to p1 x1 + p2 x2 + · · · + pn xn ≤ I 0 + wl0 ,
l0 + l1 = 24,
x1 ≥ 0, . . . , xn ≥ 0, l0 ≥ 0, l1 ≥ 0 math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.3
Example 18.3 (Profit Maximization of a
Competitive Firm) Suppose that a firm in a
competitive industry uses n inputs to
manufacture its product. Let y denote the
amount of its output, and let x1 , . . . , xn denote
the amounts of its inputs − all flow concepts.
Let y = f (x1 , . . . , xn ) denote the firm’s
production function, describing the maximal
amount of output that can be produced from
bundle (x1 , . . . , xn ). Let p be the unit price of the
output and let wi denote the cost of input i.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.3
The firm’s goal is to choose (x1 , . . . , xn ) to
maximize its profit
Π(x1 , . . . , xn ) = pf (x1 , . . . , xn ) −
n
X
wi xi
1
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.3
under the constraints
pf (x1 , . . . , xn ) −
n
X
wi xi ≥ 0,
1
g1 (x) ≤ b1 , . . . , gk (x) ≤ bk ,
x1 ≥ 0, . . . , xn ≥ 0.
The first inequality constraints reflects the
requirement that the firm make a nonnegative
profit. The gj −constraints represent constraints
on the availability of the inputs.
Yu Ren
Mathematical Economics: Lecture 13
math
Chapter 18: Constrained Optimization I
Equality Constraints
Two variables and one equality constraint
Setup:
max f (x1 , x2 )
s.t.
h(x1 , x2 ) = c
(2) (3) (4) (5) (6) in Page 414
Figure 18.1
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
NDCQ
NonDegenerate Constraint Qualification:
the rank of Jacobian Dh(x ∗ ) is equal to the
number of the constraints, then x ∗ satisfies
NDCQ.
x ∗ is called a critical point of
h = (h1 , h2 , · · · , hm ) if the rank of the matrix
Dh(x ∗ ) is less than m.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Equality Constraints
Solution (Theorem 18.1):Suppose (x1∗ , x2∗ ) is not
a critical point of h. Then there is a real number
µ∗ such that (x1∗ , x2∗ , µ∗ )
L(x1 , x2 , µ) ≡ f (x1 , x2 ) − µ(h(x1 , x2 ) − c)
∂f
∂h
(x) − µ
(x) = 0
∂x1
∂x1
∂h
∂f
(x) − µ
(x) = 0
∂x2
∂x2
h(x1 , x2 ) − c = 0
µ is called a Lagrangian multiplier
Yu Ren
Mathematical Economics: Lecture 13
math
Chapter 18: Constrained Optimization I
Example 18.4
Example 18.4 Let’s use Theorem 18.1 to solve
a simple utility maximization problem:
maximize f (x1 , x2 ) = x1 x2
subject to h(x1 , x2 ) ≡ x1 + 4x2 = 16
Since the gradient of h is (1, 4), h has no critical
points and the constraint qualification is
satisfied. From the Lagrangian
L(x1 , x2 , µ) = x1 x2 − µ(x1 + 4x2 − 16),
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.4
and set its partial derivatives equal to zero:
∂L
∂x1
∂L
∂x2
∂L
∂µ
= x2 − µ = 0
= x1 − 4µ = 0
= −(x1 + 4x2 − 16) = 0
we conclude the solution of this system is
x1 = 8, x2 = 2, µ = 2.
Theorem 18.1 states that the only candidate for
a solution is x1 = 8, x2 = 2.
Yu Ren
Mathematical Economics: Lecture 13
math
Chapter 18: Constrained Optimization I
Example 18.5
Example 18.5 Let’s work out a more complex
example:
maximize f (x1 , x2 ) = x12 x2
subject to Ch = {(x1 , x2 ) : 2x12 + x22 = 3}.
To check the constraint qualification, we
compute the critical points of
h(x1 , x2 ) = 2x12 + x22 . The only such critical point
at (x1 , x2 )=(0, 0) − a point which is not in the
constraint set Ch .
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.5
Now, from the Lagrangian
L(x1 , x2 , µ) = x12 x2 − µ(2x12 + x22 − 3),
compute its partial derivatives, and set them
equal to 0:
∂L
∂x1
∂L
∂x2
∂L
∂µ
= 2x1 x2 − 4µx1 = 2x1 (x2 − 2µ) = 0
= x12 − 2µx2 = 0
= −2x12 − x22 + 3 = 0
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.5
The first equation yields
2µ.
√ x1 = 0 or x2 = √
If x1 = 0, therefore (0, 3, 0) and (0, − 3, 0)
are two solutions of the system.
If x1 6= 0, then x2 = 2µ, then
x12 = x22 ⇒ x1 = ±1, x2 = ±1.
If x2 = +1, and µ = 0.5. If x2 = −1, and
µ = −0.5. Then we obtain four more solutions of
the system
(1, 1, 0.5), (-1, -1, -0.5), (1, −1, −0.5), (-1, 1, 0.5) .
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.5
Since
f (1, 1) = f (−1, 1) = 1,
f (1, −1) = f (−1, −1) = −1,
√
√
f (0, 3) = f (0, − 3) = 0.
the max occurs at (1, 1) and (-1, 1). Note that
(1, -1) and (-1, -1) minimize f on Ch .
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Equality Constraints
Several equality constraints
Setup:
max f (x1 , x2 , · · · , xn )
s.t.
h1 (x) = a1 ; · · · , hm (x) = am
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Equality Constraints
Solution (Theorem 18.2): Lagrangian Function:
NDCQ
L(x, µ) ≡ f (x) − µ1 (h1 (x) − a1 ) − µ2 (h2 (x) − a2 )
− · · · − µm (hm (x) − am )
∂L ∗ ∗
∂L ∗ ∗
(x , µ ) = 0, · · · ,
(x , µ ) = 0
∂x1
∂xn
∂L ∗ ∗
∂L ∗ ∗
(x , µ ) = 0, · · · ,
(x , µ ) = 0
∂µ1
∂µn
Yu Ren
Mathematical Economics: Lecture 13
math
Chapter 18: Constrained Optimization I
Example 18.6
Example 18.6 Consider the problem:
maximize f (x, y, z) = xyz
subject to h1 (x, y, z) = x 2 + y 2 = 1
and h2 (x, y , z) = x + z = 1.
First compute the Jacobian matrix of the
constraint functions
2x 2y 0
Dh(x, y, z) =
1 0 1
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.6
Since any point with x=y=0 would violate the
first constraint, all points in the constraint set
satisfy NDCQ. Next from the Lagrangian
L(x, y , z, µ1 , µ2 ) = xyz−µ1 (x 2 +y 2 −1)−µ2 (x+z−1),
and set its first partial derivatives equal to 0:
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.6
∂L
∂x
∂L
∂y
∂L
∂z
∂L
∂µ2
∂L
∂µ1
= yz − 2µ1 x − µ2 = 0
= xz − 2µ1 y = 0
= xy − µ2 = 0
= 1 − x2 − y2 = 0
=1−x −z =0
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.6
Solve the second and third equations for µ1 and
µ2 in terms of x, y , and z and plug these into
the first equation to obtain
y 2 z − x 2 z − xy 2 = 0
Then, solve the fourth equation for y 2 in terms of
x 2 and the last equation for z in terms of x, then
(1 − x 2 )(1 − x) − x 2 (1 − x) − x(1 − x 2 ) = 0
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.6
√
so x = 16 (−1 ± 13), approximately -0.7676 and
0.4343. We obtain the four solution candidates
x ≃ 0.4343, y ≃ ±0.9008, z ≃ 0.5657;
x ≃ −.07676, y ≃ ±0.6409, z ≃ 1.7676;
Evaluate the objective functions at these four
points, we find that the maximizer is
x ≃ −.07676, y ≃ −0.6409, z ≃ 1.7676
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Inequality Constraints
Figure 18.4 Figure 18.5
One inequality constraint:
Setup:
max f (x, y )
s.t.
g(x, y ) ≤ b
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Inequality Constraints
∗
∗
Solution (Theorem 18.3): If ∂g
∂x (x , y ) 6= 0 or
∂g
∗
∗
∂y (x , y ) 6= 0 Lagrangian Function:
L(x, y , λ) ≡ f (x, y ) − λ[g(x, y ) − b]
∂L ∗ ∗ ∗
(x , y , λ ) = 0
∂x
∂L ∗ ∗ ∗
(x , y , λ ) = 0
∂y
λ∗ [g(x ∗ , y ∗ ) − b] = 0
λ∗ ≥ 0
g(x ∗ , y ∗ ) ≤ b
Yu Ren
Mathematical Economics: Lecture 13
math
Chapter 18: Constrained Optimization I
Example 18.7
Example 18.7 Consider the problem:
maximize f (x, y ) = xy
subject to g(x, y) = x 2 + y 2 ≤ 1
The only critical point of g occurs at the origin −
far away from the boundary of the constraint set
x 2 + y 2 = 1. So the constraint qualification will
be satisfied any candidate for a solution. From
the lagrangian
L(x, y , λ) = xy − λ(x 2 + y 2 − 1),
and write out the first order conditions described math
in Theorem 18.3:
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.7
∂L
∂L
∂x = y − 2λx = 0, ∂y = x − 2λy = 0,
2
2
2
2
λ(x + y − 1) = 0, x + y ≤ 1 , λ ≥ 0
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.7
The first two equations yield
λ=
y
x
=
or x 2 = y 2 .
2x
2y
If λ = 0, then x = y = 0, which is a candidate for
a solution.
If λ 6= 0, then x 2 + y 2 − 1 = 0 ⇒ x 2 = y 2 = 12 ⇒
x = ± √12 , x = ± √12 , then we find the following
four candidates:
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.7
x = + √12 , y = + √12 , λ = + 21
x = − √12 , y = − √12 , λ = + 12
x = + √12 , y = − √12 , λ = − 21
x = − √12 , y = + √12 , λ = − 21
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.7
We disregard the last two candidates since they
involve a negative multiplier. Plugging the left
three candidates into the object function, we find
that
x = √12 , y = √12 and x = − √12 , y = − √12
are the solutions of our original problem. The
two points with the negative multipliers are the
solutions of the problem of minimizing xy on the
constraint set x 2 + y 2 ≤ 1.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Inequality Constraints
Several Inequality Constraint:
Setup:
max f (x)
s.t.
g1 (x) ≤ b1 , · · · , gk (x) ≤ bk
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Inequality Constraints
Solution (Theorem 18.4): NDCQ condition
Assume the first k0 constraints are binding at x ∗
and the last k − k0 are not binding. NDCQ holds
if the rank of
 ∂g1 ∗

∂g1
∗
(x
)
·
·
·
(x
)
∂x
∂xn
 1 ..

..
..
.
.
.


∂gk0
∗
∂x1 (x )
···
∂gk0
∗
∂xn (x )
is k0
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Inequality Constraints
Solution (Theorem 18.4): Lagrangian Function:
L(x1 , · · · , xn , λ1 , · · · , λk ) ≡
f (x) − λ1 [g1 (x) − b1 ] − · · · − λ1 [gk (x) − bk ]
∂L ∗ ∗
∂L ∗ ∗
(x , λ ) = 0, · · · ,
(x , λ ) = 0
∂x1
∂xn
λ∗1 [g1 (x ∗ , y ∗ ) − b1 ] = 0 · · · λ∗k [gk (x ∗ , y ∗ ) − bk ] = 0
λ∗1 ≥ 0 · · · λ∗k ≥ 0
g1 (x ∗ ) ≤ b1 · · · gk (x ∗ ) ≤ bk
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.8
Example 18.8 Consider again the standard
utility maximization problem of Example 18.1.
We continue to ignore the nonnegativity
constraints but do not force the budget
constraint to be binding in the statement of the
problem.
maximize U(x1 , x2 )
subject to p1 x1 + p2 x2 ≤ I
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.8
We assume that for each commodity bundle
(x1 , x2 ),
∂U
∂U
(x1 , x2 ) > 0 or x = ∂x
(x1 , x2 ) > 0
x = ∂x
1
2
This is a version of the usual monotonicity or
nonsatiation assumption. It states that the
commodities under study are goods in that
increasing consumption increases utility. Since
the usual constraint qualification is satisfied, so
from the lagrangian
L(x1 , x2 , λ) = U(x1 , x2 ) − λ(p1 x1 + p2 x2 − I)
Yu Ren
Mathematical Economics: Lecture 13
math
Chapter 18: Constrained Optimization I
Example 18.8
and compute its x1 − and x2 − critical points:
∂L
∂x1 (x1 , x2 )
∂L
∂x2 (x1 , x2 )
∂U
= ∂x
(x1 , x2 ) − λp1 = 0,
1
∂U
= ∂x
(x1 , x2 ) − λp2 = 0.
1
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.8
At the maximizer, the multiplier λ cannot be
∂U
∂U
zero; otherwise both ∂x
and ∂x
would be zero
1
2
− a contradiction to our monotonicity
assumption. Since
λ > 0 and λ(p1 x1 + p2 x2 − I) = 0,
it follows that p1 x1 + p2 x2 = I; the consumer will
spend all available income and we can treat the
budget constraint as an equality constraint.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.9
Example 18.9 Consider the problem:
max f (x, y , z) = xyz
s.t x + y + z ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.9
we rewrite the three nonnegatitivity constraints
as
−x ≤ 0, −y ≤ 0, and −z ≤ 0.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.9
The Jacobian of the constraint functions is


1 1 1
 −1 0 0 


 0 −1 0 
0 0 −1
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.9
Since the NDCQ holds at any solution
candidate. From the lagrangian
L(x, y , z, λ1 , λ2 , λ3 , λ4 ) = xyz − λ1 (x + y + z − 1)
−λ2 (−x) − λ3 (−y) − λ4 (−z).
we can rewrite it more aesthetically as
L(x, y , z, λ1 , λ2 , λ3 , λ4 ) = xyz − λ1 (x + y + z − 1)
+ λ2 x + λ3 y + λ4 z.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.9
according to Theorem 18.4:
(1)
(2)
(3)
∂L
∂x = yz − λ1 + λ2 = 0,
∂L
∂y = xz − λ1 + λ3 = 0,
∂L
∂z = xy − λ1 + λ4 = 0,
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.9
(4) λ1 (x + y + z − 1) = 0, (5) λ2 x = 0,
(6)
λ3 y = 0,
(7) λ4 z = 0,
(8)
λ1 ≥ 0,
(9) λ2 ≥ 0,
(10)
λ3 ≥ 0,
(11) λ4 ≥ 0,
(12)
x + y + z − 1 ≤ 1,
(13) x ≥ 0,
(14)
y ≥ 0,
(15) z ≥ 0.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.9
Rewrite conditions 1, 2, and 3, without minus
signs, as
λ1 = yz + λ2 = xz + λ3 = xy + λ4
If λ1 = 0, then yz = xz = xy = 0 and
λ1 = λ2 = λ3 = λ4 = 0
If λ1 > 0, suppose x = 0, then
λ1 = λ3 = λ4 > 0 ⇒ y = z = 0− a contradiction
to x + y + z = 1, so x > 0. Similarly
y > 0, z > 0 ⇒ λ2 = λ3 = λ4 = 0 and
1
yz = xz = xy = 13 . Since f ( 13 , 31 , 13 ) = 27
> 0, is
the solution of the constraint maximization
problem.
Yu Ren
Mathematical Economics: Lecture 13
math
Chapter 18: Constrained Optimization I
Mixed Constraints
Setup:
max f (x)
s.t.
g1 (x) ≤ b1 , · · · , gk (x) ≤ bk
h1 (x) = c1 , · · · , hm (x) = cm
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Mixed Constraints
Solution (Theorem 18.5): NDCQ condition
Assume the first k0 constraints are binding at x ∗
and the last k − k0 are not binding.

 ∂g1 ∗
∂g1
∗
(x
)
·
·
·
(x
)
∂x
∂xn

 1 ..
..
..


.
.
.

 ∂gk ∗
∂g
k0
∗ 
 0 (x ) · · ·
(x
)
∂x
∂x

r
 ∂h11 (x ∗ ) · · · ∂hn1 (x ∗ )  =?

 ∂x1
∂xn


.
.
.
..
..
..


∂hm
∂hm
∗
∗
∂x1 (x ) · · ·
∂xn (x )
Yu Ren
Mathematical Economics: Lecture 13
math
Chapter 18: Constrained Optimization I
Mixed Constraints
Solution (Theorem 18.5): Lagrangian Function:
L(x1 , · · · , xn , λ1 , · · · , λk , µ1 , · · · , µm ) ≡
f (x) − λ1 [g1 (x) − b1 ] − · · · − λk [gk (x) − bk ]
−µ1 [h1 (x) − c1 ] − · · · − µm [hm (x) − cm ]
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Mixed Constraints
∂L ∗ ∗ ∗
∂L ∗ ∗ ∗
(x , λ , µ ) = 0 · · ·
(x , λ , µ ) = 0
∂x1
∂xn
λ∗1 [g1 (x ∗ , y ∗ ) − b1 ] = 0 · · · λ∗k [gk (x ∗ , y ∗ ) − bk ] = 0
h1 (x ∗ ) = c1 · · · hm (x ∗ ) = cm
λ∗1 ≥ 0 · · · λ∗k ≥ 0
g1 (x ∗ ) ≤ b1 · · · gk (x ∗ ) ≤ bk
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.10
Example 18.10 Consider the problem:
maximize x − y 2
subject to x 2 + y 2 = 4, x ≥ 0, y ≥ 0.
Checking the NDCQ, first note that the gradient
of x 2 + y 2 is zero only at the origin, a point
which is not in the constraint set.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.10
If either nonnegativity constraint is binding, then
the solution candidates is (2, 0) and (0, 2). In
both cases, the corresponding 2X2 Jacobian
matrix of constraints has rank two. Therefore,
the NDCQ will automatically be satisfied. From
the lagrangian
L = x − y 2 − µ(x 2 + y 2 − 4) + λ1 x + λ2 y .
The first order conditions become:
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.10
(1)
(2)
(3)
∂L
∂x = 1 − 2µx + λ1 = 0,
∂L
∂y = −2y − 2µy + λ2 = 0,
2
2
x + y − 4 = 0,
(4) λ1 x = 0,
(6) λ1 ≥ 0,
(8) x ≥ 0
(5) λ2 y = 0,
(7) λ2 ≥ 0,
(9) y ≥ 0.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.10
Write 1 as 1 + λ1 = 2µx.Since
λ1 ≥ 0, 1 + λ1 > 0. Therefore,
µ > 0, x > 0, λ1 = 0. Write 2 as 2y(1 + µ) = λ2 .
Since 1 + λ1 > 0, either both y and λ2 are zero
or both are positive. By 5, both cannot be
positive. Therefore, λ2 = y = 0. Now, x = 2 by 3
and 8, λ1 = 0 by 4, and µ = 1/4 by 1. This leads
to the solution
(x, y, µ, λ1 , λ2 ) = (2, 0, 1/4, 0, 0).
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.11
Example 18.11 Consider the problem:
minimize f (x, y) = 2y − x 2
subject to x 2 + y 2 ≤ 1, x ≥ 0, y ≥ 0.
From the lagrangian
L(x, y , λ1 , λ2 , λ3 ) = 2y − x 2
− λ1 (−x 2 − y 2 + 1) − λ2 x − λ3 y .
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.11
The first order conditions are:
∂L
= −2x + 2λ1 x − λ2 = 0,
∂x
∂L
= 2 + 2λ1 y − λ3 = 0,
∂y
λ1 (−x 2 − y 2 + 1) = 0,
λ2 x = 0,
λ3 y = 0
λ1 , λ2 , λ3 ≥ 0,
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.11
∂L
rewrite the ∂x
= 0 equations, then
2x + λ2 = 2λ1 x, 2 + 2λ1 y = λ3 . Since
λ3 ≥ 2 > 0, we conclude that y = 0 and λ3 = 2.
Next, if x + 0 ⇒ λ1 = 0, y = 0 and λ2 = 0, thus
f (0, 0) = 0.
If x > 0 ⇒ λ2 = 0, λ1 = 1, y = 0, x = 1, thus
f (1, 0) = −1.
We conclude that (x, y) = (1, 0) minimizes
f (x, y ) = 2y − x 2 on the constraint set.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.12
Example 18.12 In this framework, the
Kuhn-Tucker Lagrangian for the usual utility
maximization problem of Example 18.1 would
be:
e 1 , x2 , λ) = U(x1 , x2 ) − λ(p1 x1 + p2 x2 − I),
L(x
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.12
The first order conditions are:
∂U
∂x1 − λp1 ≤ 0,
∂U
x1 ( ∂x
− λp1 ) = 0,
1
∂U
∂x2 − λp2 ≤ 0,
∂U
x2 ( ∂x
− λp2 ) = 0,
2
e
∂L
∂λ = I − p1 x1 − p2 x2 ≥ 0,
e
∂L
λ ∂λ = λ(I − p1 x1 − p2 x2 ) = 0
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.13
Example 18.13 Consider the problem:
minimize f (x, y ) = x 2 + x + 4y 2
subject to 2x + 2y ≤ 1, x ≥ 0, y ≥ 0.
The Jacobian of the constraint functions is


2 2
 −1 0 
0 −1
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.13
At most two constraints can be binding at the
same time, and any 2X2 submatrix of the
Jacobian has rank two. Therefore, the NDCQ
will hold at any solution candidate. From the
lagrangian
L(x, y , λ1 , λ2 , λ3 ) = x 2 + x + 4y 2
− λ1 (2x + 2y − 1) + λ2 x + λ3 y .
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.13
The first order conditions are:
∂L
= −2x + 1 − 2λ1 + λ2 = 0,
∂x
∂L
= 8y − 2λ1 + λ3 = 0,
∂y
λ1 (2x + 2y − 1) = 0, λ2 x = 0, λ3 y = 0,
λ1 ≥ 0, λ2 ≥ 0 , λ3 ≥ 0,
2x + 2y ≤ 1, x ≥ 0, y ≥ 0.
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.13
Since 2λ1 ≥ 1 > 0 implies that 2x + 2y = 1 is
binding.
Look at the first case
λ2 > 0 ⇒ x = 0, y = 0.5, λ3 = 0, λ1 = 2, λ2 = 3.
So the assumption λ2 > 0 leads to the candidate
(x, y , λ1 , λ2 , λ3 ) = (0, 0.5, 2, 3, 0),
try the opposite case: λ2 = 0, then we find
2x + 1 + λ2 = 2λ1 and 2 = 10y + λ3 .
math
Yu Ren
Mathematical Economics: Lecture 13
Chapter 18: Constrained Optimization I
Example 18.13
So this leads to the conclusion that either y = 0
or λ3 = 0, and we get two candidates:
(x, y, λ1 , λ2 , λ3 ) = (0.5, 0, 1, 0, 2) or
(x, y, λ1 , λ2 , λ3 ) = (0.3, 0.2, 0.8, 0, 0) by
evaluating the objective function at each of
these three candidates, we find that the
constrained maximum occurs at the point
x = 0, y = 0.5, where λ1 = 2, λ2 = 3 and λ3 = 0.
math
Yu Ren
Mathematical Economics: Lecture 13