Math rev notes8

Review of Quantitative MethodsLectures on Optimization in Economic Theory

Qing Liu

Department of Economics

Chinese University of Hong Kong

September, 2014

Qing (CUHK) Math Camp September, 2014 1 / 56

1. Unconstrained Optimization1.1 One Variable Function

Consider a C2 function f (x) of one variable

Definitions

Local versus global max (or min)

Interior versus boundary max (or min)

Critical point x0 : f ′ (x0) = 0.

First order (necessary) condition

If x0 is an interior max or min of f , then x0 is a critical point

note that the reverse is not ture!

e.g., f (x) = x3

Second order derivative: curvature of the curve

f ′′ (x) > 0: the slope of the curve increases.

f ′′ (x) < 0: the slope of the curve decreases.


1.1 One Variable FunctionSOC

Second order (sufficient) condition

If f ′ (x0) = 0 and f ′′ (x0) < 0, then x0 is a local max of f . (f ′

changes sign from + to −.)

If f ′ (x0) = 0 and f ′′ (x0) > 0, then x0 is a local min of f . (f ′

changes sign from − to +.)

If f ′ (x0) = 0 and f ′′ (x0) = 0, then x0 can be a max, a min,

or neither.

Inflection Point c of a C2 function f

f ′′ (c) = 0 and f ′′ (x) changes sign at c.

For example: f (x) = x3 at x = 0.


1.1 One Variable FunctionExample

Example

f (x) = x3 − 12x2 + 36x + 8

f ′ (x) = 3x2 − 24x + 36 = 0

⇒ x∗ = 2 or 6; (f (2) = 40; f (6) = 8)

f ′′ (x) = 6x − 24

f ′′ (2) < 0 and f ′′ (6) > 0

x∗ = 2 is a local max and x∗ = 6 is a local min.

Exercise:

f (x) = x2 − 4− x4

10+

3x6

1000

1.1 One Variable FunctionGlobal Max and Min

Global max and min

If f is a C2 function defined on an interval I and f ′′ (x) 6= 0

for any x ∈ I, then f has at most one critical point in I. This

critical point is a global min if f ′′ (x) > 0 and a global max if

f ′′ (x) < 0.


1.2. Multivariate Functions

Let f : U → R be a C2 function of n variables defined on a

subset U of Rn:

y = f (x1, x2, ..., xn).

First order condition

If x∗ ∈ Int(U) is a local max or min of f in U, then

Df (x∗) = 0 or∂f

∂xi

(x∗) = 0 for i = 1, ...,n.

We say that x∗ is a critical point of f .


SOC

Suppose that x∗ is a critical point of f .

Hessian

D2f (x∗) =

f11 f12 ... f1n

f21 f22 ... f2n

... ... ... ...fn1 ... ... fnn

.

Sufficient condition:

If D2f (x∗) is negative definite, then x∗ is a local max.

If D2f (x∗) is positive definite, then x∗ is a local min.

If D2f (x∗) is indefinite, then x∗ is neither a local max nor a

local min (saddle point).


SOC: ProofNecessary condition: If x∗ is a local max (min), then

D2f (x∗) is negative (positive) semidefinite.

If f is a concave (convex) function and Df (x∗) = 0, then x∗

is a global max (min).

Proof.By Taylor approximation

f (x∗ + h) ≈ f (x∗) +Df (x∗)h+1

2hT D2f (x∗)h.

Since x∗ is a critical point of f (Df (x∗) = 0),

f (x∗ + h)− f (x∗) ≈ 1

2hT D2f (x∗)h.

If D2f (x∗) is negative definite, then, for all small enough h, the

right-hand side is negative. Thus, f (x∗ + h)− f (x∗) < 0 and x∗

is a local max.

Applications in EconomicsDiscriminating Monopolist

A monopolist sells one product in two distinct and separated

markets. The inverse demand functions are P1 = 50− 5Q1

and P2 = 100− 10Q2 respectively, where Qi is the amount

supplied to market i and Pi is the corresponding price. The

monopolist’s cost function is C (Q) = 90+ 20Q. How to

maximize profit?

The monopolist’s profit function is

π (Q1,Q2) = Q1 (50− 5Q1) +Q2 (100− 10Q2)

− (90+ 20 (Q1 +Q2)) .

First order condition:

∂π

∂Q1

= 50− 10Q1 − 20 = 0⇒ Q1 = 3,

∂π

∂Q2

= 100− 20Q2 − 20 = 0⇒ Q2 = 4.


∂π

∂Q1

= 50− 10Q1 − 20 = 0⇒ Q1 = 3,

∂π

∂Q2

= 100− 20Q2 − 20 = 0⇒ Q2 = 4.

Second order condition:

π11 = −10, π22 = −20, π12 = π21 = 0,

the Hessian is negative definite. Thus, the monopolist

maximizes its profit at (Q1 = 3,Q2 = 4).


Applications in Economics

Application: Least Squares Analysis

Suppose that we are interested in the (linear) relationship

between two variables and that we have n observations or

data points: (xi , yi) , i = 1, ...,n. For a given line y = mx + b,

we can measure the vertical distance (|mxi + b− yi |) from

each data point (xi , yi) to the line. Method of least

squares: choose m∗ and b∗ such that the sum of the

squared distances is minimized, that is,

min(m,b)

S (m,b) = ∑n

i=1(mxi + b− yi)

2 .


∂S

∂m= ∑n

i=12 (mxi + b− yi) xi = 0

∂S

∂b= ∑n

i=12 (mxi + b− yi) = 0,


after rearrangement, we obtain(∑i

x2i

)m+

(∑i

xi

)b = ∑i

xiyi(∑i

xi

)m+ n · b = ∑i

yi .

Using Cramer’s rule:

m∗ =n ∑i xiyi − (∑i xi) (∑i yi)

n ∑i x2i − (∑i xi)

2

b∗ =∑i x2

i (∑ni=1 yi)− (∑i xi) (∑i xiyi)

n ∑i x2i − (∑i xi)

2.

Second order condition:

∂2S

∂m2= ∑i

x2i ,

∂2S

∂b2= n

∂2S

∂m∂b=

∂2S

∂b∂m= ∑i

xi

The Hessian is positive semidefinite as n(∑i x2

i

)≥ (∑i xi )

2 .Thus, (m∗,b∗) is the global minimizer (generically).

2. Equality Constraints2.1. Two Variables and One Constraint

Consider the two-variable problem

max(x ,y)

f (x , y)

subject to g(x , y) = c

f (x , y) is the objective function.

g(x , y) is a constraint function


Graphical Illustration

Graphical Illustration

x

y

g (x, y) = c

f (x, y)=kf (x, y)=k’

Note that the slope of the curve f (x , y) at point (x0, y0) is:

− fx (x0,y0)fy (x0,y0)

2.1.1 Necessary Conditions for an OptimumAssuming that f and g are differentiable. At a solution

(x∗, y∗) of the problem, the constraint curve is tangent to a

level curve of f , such that

− fx (x∗, y∗)fy (x∗, y∗)

= −gx (x∗, y∗)gy (x∗, y∗)

,

orfx (x∗, y∗)gx (x∗, y∗)

=fy (x∗, y∗)

gy (x∗, y∗),

assuming that gx (x∗, y∗) 6= 0 and gy (x∗, y∗) 6= 0.

Now introduce a new variable, λ, and let

λ =fx (x∗, y∗)gx (x∗, y∗)

=fy (x∗, y∗)

gy (x∗, y∗). Then,

fx (x∗, y∗)− λgx (x

∗, y∗) = 0

fy (x∗, y∗)− λgy (x

∗, y∗) = 0

Lagrangian Function

Thus the following conditions must be satisfied:


∗, y∗) = 0


∗, y∗) = 0

g(x∗, y∗) = c

Lagrangian Function

L (x , y ,λ) ≡ f (x , y)− λ (g(x , y)− c)

First order conditions

fx (x , y)− λgx (x , y) = 0

fy (x , y)− λgy (x , y) = 0

g(x , y) = c


Lagrangian Method

Theorem Let f and g be C1 functions of two variables.

Suppose that (x∗, y∗) is a solution for

max(x ,y)

f (x , y) subject to g(x , y) = c

Suppose further that (x∗, y∗) is not a critical point of g.

Then, there is a real number λ∗ such that (x∗, y∗,λ∗) is a

critical point of the Lagrangian function

L (x , y ,λ) ≡ f (x , y)− λ (g(x , y)− c) .

In other words, at (x∗, y∗,λ∗)

∂L

∂x=

∂L

∂y=

∂L

∂λ= 0.


Lagrangian Method

Remarks

The theorem holds whether we are maximizing or minimizing

f with the same constraint.

No restrictions on the sign of the multiplier

reduce the constrained problem to unconstrained problem


Example 1Utility Maximization with Budget Constraint

max(x ,y){xy} subject to pxx + pyy = I

The Lagrangian is

L (x , y ,λ) ≡ xy − λ (pxx + pyy − I)

The first-order conditions are

∂L

∂x= y − λpx = 0

∂L

∂y= x − λpy = 0

∂L

∂λ= pxx + pyy − I = 0

The unique solution is: (x∗, y∗,λ∗) =(

I2px, I

2py, I

2px py

).

Example 2

Example 2

Consider the problem

max(x ,y)

{x2y

}subject to 2x2 + y2 = 3

The Lagrangian is

L (x , y ,λ) ≡ x2y − λ(

2x2 + y2 − 3)

The first-order conditions are

∂L

∂x= 2x(y − 2λ) = 0

∂L

∂y= x2 − 2λy = 0

∂L

∂λ= 2x2 + y2 − 3 = 0

Analysis: Either x = 0 or y = 2λ

If x = 0, y = ±√

3, and λ = 0

If y = 2λ, x = ±2λ, thus x = ±1

If x = 1,y = 1 and λ = 1

2, or y = −1 and λ = −1

2

If x = −1,y = 1 and λ = 1

2, or y = −1 and λ = −1

2

The first-order conditions have six solutions:

1 (x , y ,λ) = (0,√

3,0), f (x , y) = 0.2 (x , y ,λ) = (0,−

√3,0), f (x , y) = 0.

3 (x , y ,λ) = (1,1, 12), f (x , y) = 1.

4 (x , y ,λ) = (1,−1,− 12), f (x , y) = −1.

5 (x , y ,λ) = (−1,1, 12), f (x , y) = 1.

6 (x , y ,λ) = (−1,−1,− 12), f (x , y) = −1.

We conclude that the problem has two solutions,

(x , y) = (1,1) and (x , y) = (−1,1).

2.1.2. Sufficient Conditions for a Local

Optimum

Consider the two-variable problem

max(x ,y)

f (x , y) subject to g(x , y) = c

Let h be implicitly defined by g(x ,h(x)) = c. Then the

problem is:

maxx

f (x ,h(x))

Define F (x) = f (x ,h(x)). Then

F ′(x) = fx (x ,h(x)) + fy (x ,h(x))h′(x)

Let x∗ be a critical point of F (i.e. F ′(x∗) = 0). A sufficient

condition for x∗ to be a local maximizer of F is that

F ′′(x∗) < 0. We have

F ′′(x∗) = fxx (x∗,h(x∗)) + 2fxy (x

∗,h(x∗))h′(x∗)

+fyy (x∗,h(x∗))

(h′(x∗)

)2+ fy (x

∗,h(x∗))h′′(x∗)

Now, since g(x ,h(x)) = c for all x , we have

gx (x ,h(x)) + gy (x ,h(x))h′(x) = 0

h′(x) = −gx (x ,h(x))

gy (x ,h(x))

Using this expression we can find h′′(x∗), and substitute it

into the expression for F ′′(x∗):

F ′′(x∗) = −|H (x∗, y∗,λ∗)|

(gy (x∗, y∗))2

Bordered Hessian of the Lagrangian

H (x∗, y∗,λ∗)

=

0 gx (x∗, y∗) gy (x∗, y∗)gx (x∗, y∗) Z ∗xx Z ∗xy

gy (x∗, y∗) Z ∗yx Z ∗yy

where

Z ∗xx = fxx (x∗, y∗)− λ∗gxx (x

∗, y∗)

Z ∗yy = fyy (x∗, y∗)− λ∗gyy (x

∗, y∗)

Z ∗xy = Z ∗yx = fxy (x∗, y∗)− λ∗gxy (x

∗, y∗)


Generalization

Theorem Consider the problems

max(x ,y)

f (x , y) subject to g(x , y) = c; or

min(x ,y)

f (x , y) subject to g(x , y) = c.

Suppose that (x∗, y∗,λ∗) satisfies the first order conditions


∗, y∗) = 0


∗, y∗) = 0

g(x∗, y∗) = c

1 If |H (x∗, y∗,λ∗)| > 0, then (x∗, y∗) is a local maximizer of f

subject to g(x , y) = c.2 If |H (x∗, y∗,λ∗)| < 0, then (x∗, y∗) is a local minimizer of f

subject to g(x , y) = c

Example

Utility maximization revisited

max(x ,y){xy} subject to pxx + pyy = I

The Lagrangian is: L (x , y ,λ) ≡ xy − λ (pxx + pyy − I)The unique solution is

(x∗, y∗,λ∗) =

(I

2px,

I

2py,

I

2pxpy

).

Bordered Hessian of the Lagrangian

H (x∗, y∗,λ∗) =

0 px py

px 0 1

py 1 0

.|H (x∗, y∗,λ∗)| = 2pxpy > 0, so

(I

2px, I

2py

)is indeed a local

maximizer

2.2. Generalization

The Lagrangian method can easily be generalized to a

problem of the form

maxx

f (x) subject to gj(x) = cj for j = 1, ...,m

with n variables (x = (x1, ...xn)) and m constraints.

The Lagrangian for this problem is

L (x,λ) ≡ f (x)−m

∑j=1

λj

(gj(x)− cj

),

that is, one Lagrange multiplier for each constraint

NDCQNondegenerate Constraint Qualification (NDCQ)

With two variables and one constraint, we require(∂g

∂x1

(x∗) ,∂g

∂x2

(x∗)

)6= (0,0) .

With n variables and one constraint, we generalize it as(∂g

∂x1

(x∗) , ...,∂g

∂xn(x∗)

)6= (0, ...,0) .

With n variables and m constraints, we further generalize it

as the Jacobian matrix

Dg (x∗) =

∂g1

∂x1(x∗) ... ∂g1

∂xn(x∗)

∂g2

∂x1(x∗) ... ∂g2

∂xn(x∗)

... ... ...∂gm

∂x1(x∗) ... ∂gm

∂xn(x∗)

has a rank of m

2.2.1 First Order Conditions

Theorem Let f and g1, ...,gm be C1 functions of n variables.

Consider the problem of maximizing (or minimizing) f (x) on

the constraint set:

Cg ≡{(x1, ..., xn) : gj(x) = cj for j = 1, ...,m

}.

Suppose that x∗ ∈ Cg is a (local) max or min of f on Cg .Suppose further that x∗ satisfies condition NDCQ above.

Then, there exist λ∗ = (λ∗1, ...,λ∗m) such that (x∗,λ∗) is a

critical point of the Lagrangian

L (x,λ) ≡ f (x)−m

∑j=1

λj

(gj(x)− cj

)In other words, at (x∗,λ∗)

∂L

∂xi

= 0, i = 1,2, ...,n,

∂L

∂λj

= 0, j = 1,2, ...,m.

2.2.2. Second Order ConditionsBordered Hessian

H ≡(

0 Dg (x∗)Dg (x∗)T D2

x L (x∗,λ∗)

)where

Dg (x∗) =

∂g1

∂x1(x∗) ... ∂g1

∂xn(x∗)

∂g2

∂x1(x∗) ... ∂g2

∂xn(x∗)

... ... ...∂gm

∂x1(x∗) ... ∂gm

∂xn(x∗)

and

D2xL (x∗,λ∗) =

∂2L

∂x21

(x∗) ... ∂2L∂x1∂xn

(x∗)

∂2L∂x2∂x1

(x∗) ... ∂2L

∂x22

(x∗)

... ... ...∂2L

∂xn∂x1(x∗) ... ∂2L

∂x2n(x∗)

.

SOC

Theorem Let f and g1, ...,gm be C2 functions on Rn.Consider the problem of maximizing (or minimizing) f (x) on

the constraint set


}.

Suppose that (x∗,λ∗) is a critical point of the Lagrangian

L (x,λ) ≡ f (x)−m

∑j=1

λj

(gj(x)− cj

).

1 If the Hessian of L with respect to x at (x∗,λ∗) , D2xL (x∗,λ∗) ,

is negative definite on the linear constraint set

{v : Dg (x∗) v = 0} ; that is

v 6= 0 and Dg (x∗) v = 0

⇒ vT(

D2xL (x∗,λ∗)

)v < 0,

then, x∗ is a strict local constrained max of f on Cg

(cont.)

1 If the Hessian of L...2 If the Hessian of L with respect to x at (x∗,λ∗) , D2

xL (x∗,λ∗) ,is positive definite on the linear constraint set

{v : Dg (x∗) v = 0} ; that is

v 6= 0 and Dg (x∗) v = 0

⇒ vT(

D2xL (x∗,λ∗)

)v > 0,

then, x∗ is a strict local constrained min of f on Cg .

Remember that:

Negative definite: if the last (n−m) leading principle

minors of the bordered Hessian H alternate in sign, with the

sign of the determinant of H the same as the sign of (−1)n .Positive definite: if the last (n−m) leading principle minors

of the bordered Hessian H have the same sign as the sign of

(−1)m .

2.2.3. Summary

Consider the problem of maximizing f (x) on the constraint

set


}.

Suppose that (x∗,λ∗) is a critical point of the Lagrangian

(i.e., first order condition)

L (x,λ) ≡ f (x)−m

∑j=1

λj

(gj(x)− cj

).

If the second order condition is satisfied at (x∗,λ∗), then x∗

is a strict local constrained max.

If L is concave, in particular, if f is concave and λjgj is

convex for j = 1, ...,m, then x∗ is a global constrained max.

If f is strictly quasiconcave and Cg is convex, then x∗ is a

unique global constrained max

3. Inequality Constraints3.1. Example: Failure of Lagrange Principle

Consider the consumer optimization problem with log-linear

utility function

max (x1 + ln x2) s.t.

{p1x1 + p2x2 ≤ I

x1 ≥ 0, x2 ≥ 0(3.1)

The Lagrangian is

L (x1, x2,λ) = x1 + ln x2 − λ (p1x1 + p2x2 − I) ,

and the FONCs set forth by Lagrange Theorem are∂L∂x1= 1− λp1 = 0

∂L∂x2= 1

x2− λp2 = 0

p1x1 + p2x2 = I

⇒ λ =1

p1

, x2 =p1

p2

, x1 =I − p1

p1

.

Example: Failure of Lagrange Principle

solution:

λ =1

p1

, x2 =p1

p2

, x1 =I − p1

p1

If p1 > I, then the above solution is not feasible. But notice

that the problem (3.1) certainly has an optimal solution

Therefore, if p1 > I, this optimal solution does not satisfy the

above Lagrange condition.

In fact, as we will see soon, if p1 > I, the optimal solution is:

x1 = 0, x2 = I/p2 (corner solution) with Lagrange multiplier

λ = 1/I. Hence at the optimum, ∂L/∂x2 = 0 while

∂L/∂x1 < 0.

Remark:

The failure demonstrated below is due to the fact that

optimum is reached at the boundary of the set of feasible

points, i.e. the "interior " requirement of Lagrange Theorem

is violated

3.2. Kuhn-Tucker Conditions

Many models in economics are naturally formulated as

optimization problems with inequality constraints.

Consider, for example, a consumer’s choice problem.

maxx

u(x) subject to p · x ≤ w and x ≥ 0.

Consider a problem of the firm

maxx

f (x) s.t. gj (x) ≤ cj for j = 1, ...,m.

where f and gj for j = 1, ...,m are functions of n variables,

x = (x1, ..., xn), and cj for j = 1, ...,m are constants


Case with One Constraint

Consider the case with one constraint

maxx

f (x) subject to g (x) ≤ c,

which has a solution x∗. There are two possibilities:

The constraint is binding: g (x∗) = c.The constraint is NOT binding: g (x∗) < c.

Define the Lagrangian function as before

L(x,λ) = f (x)− λ(g(x)− c).

If g(x∗) = c and the constraint satisfies a regularity

condition, then ∂L∂xi(x∗,λ∗) = 0 for all i . In this case, it must

be that λ ≥ 0. To see this, suppose that λ < 0, then a small

decrease in c raises the value of f . In other words, moving

x∗ inside the constraint set raises the value of f ,

contradicting the fact that x∗ is a local max.

Complementary Slackness Condition

If g(x∗) < c, then ∂f∂xi(x∗) = 0 for all i . In this case, the value

of λ does not enter the conditions, so we can choose any

value for it. Given the interpretation of λ, setting λ = 0

makes sense. Under this assumption we have∂f∂xi(x) = ∂L

∂xi(x,λ) for all x, so that ∂L

∂xi(x∗,λ∗) = 0 for all i .

We now combine the two cases by writing the conditions as

∂L

∂xi

(x∗,λ∗) = 0 for i = 1, ...,n

λ∗ ≥ 0, g(x∗) ≤ c

λ∗ [g(x∗)− c] = 0

Such a condition in which one of the two inequalities must

be binding is called a complementary slackness

condition.


Kuhn-Tucker Conditions

Definition The Kuhn-Tucker conditions for the problem

maxx

f (x) s.t. gj (x) ≤ cj for j = 1, ...,m.

are

∂L

∂xi

(x,λ) = 0 for i = 1, ...,n

λj ≥ 0, gj (x) ≤ cj

λj

[gj (x)− cj

]= 0, for j = 1, ...,m,

where

L(x,λ) = f (x)−m

∑j=1

λj(gj(x)− cj).


Example

Example: Consider the problem

max(x ,y)

[−(x − 4)2 − (y − 4)2]

subject to x + y ≤ 4 and x + 3y ≤ 9

Lagrangian:

L (x , y ,λ1,λ2) = [−(x − 4)2 − (y − 4)2]− λ1 (x + y − 4)

−λ2 (x + 3y − 9) .

The Kuhn-Tucker conditions are

−2(x − 4)− λ1 − λ2 = 0

−2(y − 4)− λ1 − 3λ2 = 0

x + y ≤ 4,λ1 ≥ 0

x + 3y ≤ 9,λ2 ≥ 0

λ1 (x + y − 4) = 0 and λ2 (x + 3y − 9) = 0

Analysis: There are four cases.1 None of the constraints is binding.

In this case, λ1 = λ2 = 0, then x = y = 4, the constraints

are not satisfied.2 Both constraints are binding.

In this case, x = 32, y = 5

2, then λ1 = 6, λ2 = −1 < 0.

3 Constraint 1 is binding, 2 is not.In this case, λ2 = 0, we have

2(x − 4) + λ1 = 0

2(y − 4) + λ1 = 0

x + y − 4 = 0

We have x = y = 2, λ1 = 4. f (2,2) = −8.4 Constraint 2 is binding, 1 is not.In this case, λ1 = 0, we have

2(x − 4) + λ2 = 0

2(y − 4) + 3λ2 = 0

x + 3y − 9 = 0

We have x = 3310, y = 19

10, but then constraint 1 is violated.

Thus, (2,2) is the solution

Failure of Kuhn-Tucker Conditions

It is not difficult to imagine that Kuhn-Tucker conditions are

not sufficient for a local constrained max. Without an extra

regularity condition, they are not even necessary conditions


max(x ,y){x} s.t. y − (1− x)3 ≤ 0; y ≥ 0

The solution is clearly (1,0). The Lagrangian is

L(x ,λ1,λ2) = x − λ1(y − (1− x)3) + λ2y .


1− 3λ1 (1− x)2 = 0

−λ1 + λ2 = 0

y − (1− x)3 ≤ 0,λ1 ≥ 0,

and λ1(y − (1− x)3) = 0

−y ≤ 0,λ2 ≥ 0, and λ2y = 0.

The Kuhn-Tucker conditions have no solution. From the

last condition, either λ2 = 0 or y = 0. If λ2 = 0 then λ1 = 0

from the second condition, so that no value of x is

compatible with the first condition. If y = 0 then from the

third condition either λ1 = 0 or x = 1, both of which are

incompatible with the first condition.

NDCQ: Suppose that k out of m inequality constraints are

binding at x∗, then the NDCQ requires that at x∗, the rank of

the Jacobian matrix of the binding constraints is k .

If at some x∗ in the constraint set Cg , the NDCQ is NOT

satisfied, then x∗ is a possible maximizer. No need to

see whether or not the Kuhn-Tucker conditions are also

satisfied at x∗.

If at some x∗ in the constraint set Cg , the NDCQ is

satisfied but the Kuhn-Tucker conditions are not

satisfied, then x∗ is NOT a possible maximizer.

3.3. Nonnegativity Constraints

Many of the optimization problems in economic theory have

nonnegativity constraints on the variables. For example, a

consumer chooses a bundle x of goods to maximize her

utility u(x) subject to her budget constraint p · x ≤ I and the

condition xi ≥ 0. The general form of such a problem is

maxx

f (x) subject to

gj (x) ≤ cj for j = 1, ...,m

xi ≥ 0 for i = 1, ...,n

This problem is a special case of the general maximization

problem with inequality constraints.

The Lagrangian is

L(x,λ, ν) = f (x)−m

∑j=1

λj(gj(x)− cj)−n

∑i=1

νi (−xi) .




∂L

∂xi

(x,λ, ν) = 0 for i = 1, ...,n

λj ≥ 0, gj(x) ≤ cj

λj

[gj(x)− cj

]= 0, for j = 1, ...,m,

νi ≥ 0, xi ≥ 0, and νixi = 0, for all i .

In this way we have to work with n+m Lagrange multipliers,

which can be difficult if n is large.


Modified LagrangianNote that

∂L

∂xi

=∂f

∂xi

−m

∑j=1

λj

∂gj

∂xi

+ νi , thus

∂f

∂xi

−m

∑j=1

λj

∂gj

∂xi

≤ 0

xi

[∂f

∂xi

−m

∑j=1

λj

∂gj

∂xi

]= 0

It allows us to simplify the calculations as follows.

Modified Lagrangian

L̃ (x,λ) = f (x)−m

∑j=1

λj(gj(x)− cj).

Note that this Lagrangian does not include the nonnegativity

constraints explicitly.

Modified Lagrangian

Notice the difference:

1 Modified Lagrangian

L̃ (x,λ) = f (x)−m

∑j=1

λj(gj(x)− cj).

2 previously:

L(x,λ, ν) = f (x)−m

∑j=1

λj(gj(x)− cj)−n

∑i=1

νi (−xi) .



Then we give the Kuhn-Tucker conditions for the modified

Lagrangian:

∂L̃

∂xi

(x,λ) ≤ 0, xi ≥ 0,

and xi

∂L̃

∂xi

(x,λ) = 0, for i = 1, ...,n

λj ≥ 0, gj(x) ≤ cj

λj

[gj(x)− cj

]= 0, for j = 1, ...,m.

If (x,λ,v) satisfies the original Kuhn-Tucker conditions, then

(x,λ) satisfies the conditions for the modified Lagrangian,

and if (x,λ) satisfies the conditions for the modified

Lagrangian then we can find numbers (ν1, ..., νn) such that

(x,λ,v) satisfies the original set of conditions.



Remark:

This result means that in any problem for which the original

Kuhn-Tucker conditions may be used, we may alternatively

use the conditions for the modified Lagrangian. For most

problems in which the variables are constrained to be

nonnegative, the Kuhn-Tucker conditions for the modified

Lagrangian are easier to work with than the conditions for

the original Lagrangian.


Example


max(x ,y)

xy subject to

x + y ≤ 2, x ≥ 0 and y ≥ 0

The modified Lagrangian is

L̃ (x , y ,λ) = xy − λ (x + y − 2) ,

and the Kuhn-Tucker conditions are

y − λ ≤ 0, x ≥ 0, and x(y − λ) = 0

x − λ ≤ 0, y ≥ 0, and y(x − λ) = 0

λ ≥ 0, x + y ≤ 2, and λ(x + y − 2) = 0.

Analysis

1 If x > 0 then from the first set of conditions we have y = λ.

If y = 0 in this case then λ = 0, so that the second set of

conditions implies x ≤ 0, contradicting x > 0. Hence y > 0,

and thus x = λ, so that x = y = λ = 1.2 If x = 0 then if y > 0 we have λ = 0 from the second set of

conditions, so that the first condition contradicts y > 0. Thus

y = 0 and hence λ = 0 from the third set of conditions.

We conclude (as before) that there are two solutions of the

Kuhn-Tucker conditions, in this case (x , y ,λ) = (1,1,1) and

(0,0,0). Since the value of the objective function at (1,1) is

greater than the value of the objective function at (0,0), the

solution of the problem is (1,1)

3.4. Mixed Constraints

Consider a problem of the form

maxx

f (x) subject to

h1 (x) = c1, ...,hm (x) = cm.

g1 (x) ≤ b1, ...,gk (x) ≤ bk ;

where f , g1, ...,gk , h1, ...,hm are C1 functions of n variables;

b1, ...,bk , c1, ..., cm are constants.

The Lagrangian is

L(x,λ, µ) = f (x)−k

∑j=1

λj(gj(x)− bj)−m

∑j=1

µj

(hj (x)− cj

).

FOC

The first order conditions are

∂L

∂xi

(x,λ, µ) = 0 for i = 1, ...,n

λ1 ≥ 0, ..., λk ≥ 0

g1 (x) ≤ b1, ...,gk (x) ≤ bk

λ1 [g1 (x)− b1] = 0, ...,λk [gk (x)− bk ] = 0,

h1 (x) = c1, ...,hm (x) = cm.

NDCQ: Without loss of generality, suppose that the first k0

inequality constraints are binding at x∗, and the other

inequality constraints are not binding. Then, the NDCQ

requires that the rank of the Jacobian matrix of the m

equality constraints and the k0 binding inequality constraints

is k0 +m


Log-Linear Example Revisited

Let us now apply Kuhn-Tucker necessary conditions to

problem (3.1).

max (x1 + ln x2)

s.t. p1x1 + p2x2 ≤ I

x1 ≥ 0, x2 ≥ 0

where the Lagrangian is given by

L (x1, x2,λ) = x1 + ln x2 − λ (p1x1 + p2x2 − I) .

Kuhn-Tucker Conditions for this problem areLx1≤ 0, x1 ≥ 0 with "CS"

Lx2≤ 0, x2 ≥ 0 with "CS"

p1x1 + p2x2 ≤ I,λ ≥ 0 with "CS"


or

1 ≤ λp1, x1 ≥ 0 with "CS" (3.7)

1

x2

≤ λp2, x2 ≥ 0 with "CS" (3.8)

p1x1 + p2x2 ≤ I,λ ≥ 0 with "CS" (3.9)

According to (3.7), λ > 0. Hence "CS" in (3.9) implies that

p1x1 + p2x2 = I (3.10)

If x2 = 0 then by (3.8) λ = +∞. But then "CS" in (3.7) would

imply x1 = 0, which violates "CS" in (3.9) (0 < I and λ > 0).

The contradiction shows that we must have x2 > 0. Then

"CS" in (3.8) implies 1x2= λp2. Substitute this into (3.10), we

obtain:

p1x1 +1

λ= I (3.11)


we obtain: p1x1 +1λ = I

1 Case 1: Check whether x1 = 0 is consistent with

Kuhn-Tucker conditions. Substitute x1 = 0 in (3.11): λ = 1I.

Then (3.7) becomes: 1 ≤ p1I. This is consistent with "CS" if

p1 ≥ I. Thus, if p1 ≥ I, then x1 = 0,λ = 1I

is consistent with

Kuhn-Tucker conditions. And in this case, x2 =I

p2.

Hence, we have shown that x1 = 0, x2 =I

p2is the point

satisfying Kuhn-Tucker conditions when p1 ≥ I.

2 Case 2: Now let p1 < I. Then the above arguements imply

that x1 = 0 would violate (3.7). Therefore, we must have

x1 > 0. But then by "CS" in (3.7), 1 = λp1, so λ = 1/p1. It

follows that x2 = p1/p2 and x1 =I−p1

p1. Note this solution

satisfies Kuhn-Tucker conditions with λ = 1/p1, if p1 ≤ I


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