Week 5: Calculus and Optimization (Jehle and Reny, Chapter …economics-course.weebly.com/uploads/2/5/7/2/25725158/week_5.pdf · 1 and x 2 in (eq:4) are candidate solutions to this

Week 5: Calculus and Optimization (Jehle andReny, Chapter A2)

Tsun-Feng Chiang

*School of Economics, Henan University, Kaifeng, China

October 10, 2015

Microeconomic Theory Week 5: Calculus and Optimization (Jehle and Reny, Chapter A2)October 10, 2015 1 / 38

Lagrange’s Method

Lagrange’s method is a powerful way to solve constrained optimizationproblems. To see this, let’s reconsider our original problem:

maxx1,x2 f (x1, x2) subject to g(x1, x2) = 0

Suppose we multiply the constraint equation by a new variable, call itλ. If we add it to the objective function, we will have constructed a newfunction, called the Lagrangian function denoted by a script L(·). Thisnew function has three variables instead of two: namely, x1, x2, and λ:

L(x1, x2, λ) ≡ f (x1, x2) + λg(x1, x2)

To find the critical point, we first take the derivatives with respect tothree variables and set the three derivatives to equal zero.

L1 ≡∂L∂x1

=∂f (x∗1 , x

∗2 )

∂x1+ λ∗

∂g(x∗1 , x∗2 )

∂x1= 0 (eq.1)


L2 ≡∂L∂x2

=∂f (x∗1 , x

∗2 )

∂x2+ λ∗

∂g(x∗1 , x∗2 )

∂x2= 0 (eq.2)

∂L∂λ

= g(x∗1 , x∗2 ) = 0 (eq.3)

There are three equations in the three unknowns, x1, x2 and λ.Lagrange’s method asserts that if we can find values x∗1 , x∗2 and λ∗ thatsolve these three equations simultaneously, then we will have a criticalpoints of f (x1, x2) along the constraint g(x1, x2) = 0.

Now we verify why x∗1 , x∗2 and λ∗ make f (x1, x2) as large (or small) aspossible. Consider our contrived function L(·) and take its totaldifferential:

dL =∂L∂x1

dx1 +∂L∂x2

dx2 +∂L∂λ

dλ.

Then substitute the first-order conditions (eq. 1) through (eq. 3) into it,we have


dL =∂f (x∗1 , x

∗2 )

∂x1dx1 +

∂f (x∗1 , x∗2 )

∂x2dx2 + g(x∗1 , x

∗2 )dλ

+ λ∗[∂g(x∗1 , x

∗2 )

∂x1dx1 +

∂g(x∗1 , x∗2 )

∂x2dx2

]= 0

for all dx1, dx2 and dλ. We want to show the equality above impliesthat df = 0 for the dx1 and dx2. From (eq. 3), the equality reduces to

dL =∂f (x∗1 , x

∗2 )

∂x1dx1 +

∂f (x∗1 , x∗2 )

∂x2dx2

+λ∗[∂g(x∗1 , x

∗2 )

∂x1dx1 +

∂g(x∗1 , x∗2 )

∂x2dx2

]= 0

Clearly, if g(x1, x2) must always equal zero, then after x1 and x2change, it must again be equal to zero. Again by (eq. 3), then theequality can be further reduced to


dL =∂f (x∗1 , x

∗2 )

∂x1dx1 +

∂f (x∗1 , x∗2 )

∂x2dx2 = 0

for all dx1 and dx2 satisfying the constraint. This implies that ∂f∂x1

= 0,and ∂f

∂x2= 0. It says that the solution (x∗1 , x

∗2 , λ

∗) to the first-orderconditions for an optimum of Lagrange’s function guarantee that thevalue of the objective function f cannot be increased or decreased forsmall changes in x∗1 and x∗2 that satisfy the constraint. Therefore, wemust be at a maximum or a minimum of the objective function alongthe constraint.

Example A2.8Let’s consider a problem of the sort we’ve been discussing and applyLagrange’s method to solve it. Suppose our problem is to

maxx1,x2 − ax21 − bx2

2 subject to x1 + x2 − 1 = 0

where a > 0 and b > 0. First, from the Lagrangian,


L(x1, x2, λ) ≡ −ax21 − bx2

2 + λ(x1 + x2 − 1)

Then set all of its first-order partials equal to zero:

∂L∂x1

= −2ax∗1 + λ∗ = 0

∂L∂x2

= −2bx∗2 + λ∗ = 0

∂L∂λ

= x∗1 + x∗2 − 1 = 0

Solve the system, we get,

x∗1 =b

a + b, x∗2 =

aa + b

, λ∗ =2ab

a + b. (eq.4)

Only x∗1 and x∗2 in (eq.4) are candidate solutions to this problem. Theadditional bit of information we have acquired, the value of theLagrangian multiplier there, is only "incidental." We may obtain thevalue the objective function achieves along the constraint by


substituting the values for x1 and x2 into the objective function:

y∗ = −a(

ba + b

)2

− b(

aa + b

)2

=−(ab2 + ba2)

(a + b)2

From the first-order conditions alone, we are unable to tell whether thisis a maximum or a minimum value of the objective function subject tothe constraint.

The Lagrangian method is quite capable of addressing a much broaderclass of problems than we have yet considered. Lagrange’s method"works" for functions with any number of variables, and in problemswith any number of constraints, as long as the number of constraints isless than the number of variables being chosen.Suppose we have a function of n variables and we face m constraints,where m < n. Our problem is


maxx1,···,xn f (x1, · · · , xn) subject to g1(x1, · · · , xn) = 0

g2(x1, · · · , xn) = 0...gm(x1, · · · , xn) = 0.

To solve this, from the Lagrangian by multiplying each constraintequation g j by a different Lagrangian multiplier λj and adding them allto the objective function f . For x = (x1, · · · , xn) and Λ = (λ1, · · · , λm),we obtain the function of n + m variables

L(x,Λ) = f (x) +m∑

j=1

λjg j(x).

The first-order conditions again require that all partial derivatives of Lbe equal to zero at optimum. Because L has n + m variables, there willbe of n + m equations determining the n + m variables x∗ and Λ∗:


∂L∂xi

=∂f (x∗)∂xi

+m∑

j=1

λ∗j∂g j(x∗)∂xi

= 0, i = 1, · · · ,n

∂L∂λi

= g j(x∗) = 0, j = 1, · · · ,m

In principle, these can be solve for the n + m values, x∗ and Λ∗. Underthe condition that ∆g j(x∗) = (g1(x∗), · · · ,gm(x∗)) are linearlyindependent, all solution vectors x∗ will then be candidates for thesolution to the constrained optimization problem. The followingtheorem restates Lagrange’s method for constrained optimization

Theorem A2.16 Lagrange’s Theorem

Let f (x) and g j(x), j = 1, · · · ,m, be continuously differentiablereal-valued functions over some domain D ⊂ Rn. Let x∗ be an interiorpoint of D and suppose that x∗ is an optimum (maximum or minimum)of f subject to the constraints, g j(x∗) = 0, j = 1, · · · ,m. If the gradientvector ∆g j(x∗) , j = 1, · · · ,m, are linearly independent, then there exist


Theorem A2.16 (Continued)m unique number λ∗j , j = 1, · · · ,m, such that

∂L∂xi

=∂f (x∗)∂xi

+m∑

j=1


= 0, i = 1, · · · ,n

The conditions characterizing the Lagrangian solution to constrainedoptimization problems have a geometrical interpretation. We againconsider the problem with two-variables and a constrain. We canrepresent the objective function geometrically by its level sets,L(y0) ≡ {(x1, x2)|f (x1, x2) = y0} for some y0 in the range. Bydefinition, all points in the set must satisfy the equation

f (x1, x2) = y0

If we change x1 and x2 and are remain on that level set, dx1 and dx2must be such as to leave the value f unchanged at y0. Therefore bytotally differentiating each side of the equation, they must satisfy


∂f (x1, x2)

∂x1dx1 +

∂f (x1, x2)

∂x2dx2 = 0 (eq.5)

This must hold at any point along any level set of the function. Bysolving (eq. 5) for dx2/dx1, the slope of the level set through (x1, x2)will be

dx2

dx1

∣∣∣∣along L(y0)

= (−1)f1(x1, x2)

f2(x1, x2)(eq.6)

The notation |along · · · is used to remind you of the very particular sortof changes dx1 and dx2 we are considering. Thus, as depicted inFigure A2.6 (see the next slide), the slope of the level set through anypoint (x1, x2) is given by the (negative) ratio of first-order derivatives off at (x1, x2).We can think of the constraint function, too, as a kind of level set. It isthe set of all (x1, x2) such that

g(x1, x2) = 0


Figure A2.6 (left): The slope of a level set ; Figure A2.7(right): The slope of a

constraint


Just as before, we can derive the slop of this constraint set at any pointalong it by totally differentiating both sides of the equation andrearranging terms,

dx2

dx1

∣∣∣∣along g(·)=0

= (−1)g1(x1, x2)

g2(x1, x2)(eq.7)

Figure A2.7 (see the last slide) illustrates a constraint function and itsslope. Recall the first-order conditions for a critical point of theLagrangian function given in (eq. 1) and through (eq. 3). Rearrangethese three equations, we obtain

∂f (x∗1 , x∗2 )

∂x1= −λ∗

∂g(x∗1 , x∗2 )

∂x1

∂f (x∗1 , x∗2 )

∂x2= −λ∗

∂g(x∗1 , x∗2 )

∂x2

g(x∗1 , x∗2 ) = 0


Suppose λ∗ 6= 0. Then dividing the first of the equations by the secondto eliminates λ∗ altogether and leaves us with just two conditions todetermine the two variables x∗1 and x∗2 :

Condition 1:f1(x∗1 , x

∗2 )

f2(x∗1 , x∗2 )

=g1(x∗1 , x

∗2 )

g2(x∗1 , x∗2 )

(eq.8)

Condition 2: g(x∗1 , x∗2 ) = 0 (eq.9)

Multiply both side of (eq. 8) by −1, then compare it with (eq. 6) and(eq. 7), we can see Condition 1 says that the solution values of x1 andx2 will be at point where the slope of the level set for the objectivefunction and the slope of the level set for the constraint are equal.Condition 2 tells us the solution values must also be on the level set ofthe constraint equation. A point that is on the constraint and where theslope of the constraint are equal is, by definition, a point of tangencybetween the constraint and the level set. See Figure A2.8 on the nextslide as a visual example.


Figure A2.8: The first-order conditions for a solution to Lagrange’s problem

identify a point of tangency between a level set of the objective function and

the constraint


Second-Order Conditions (Equality Constraints)

To know that we have a maximum (minimum), all we really need is thatthe second differential of the objective function at the point that solvesthe first-order conditions is decreasing (increasing) along theconstraint. Now we only show the case of maximization.With two variables and one constraint, the conditions for this can beeasily derived by exploiting the interdependence between x1 and x2arising from the constraint. Suppose we arbitrarily view x1 as free totake any value and think of x2(x1) as the required value of x2 imposedby the constraint. We then can think of the constraint as the identity

g(x1, x2(x1)) = 0

Here, we view the constraint equation as defining x2 as an implicitfunction of x1. Differentiating with respect to x1 and solving for dx2/dx1,

dx2

dx1=−g1

g2(eq.10)


Letting y = f (x1, x2(x1)) be the value of the objective function subjectto the constraint, we get y as a function of the single variable x1.Differentiate with respect to x1, we get dy/dx1 = f1 + f2(dx2/dx1).Substituting from (eq. 10) gives

dydx1

= f1 − f2g1

g2(eq.11)

Differentiating again, remembering always that x2 is a function of x1,and that the fi and gi all depend on x1 both directly and through itsinfluence on x2, we obtain the second derivative,

d2ydx2

1=f11 + f12

dx2

dx1−[f21 + f22

dx2

dx1

]g1

g2

− f2

{g2[g11 + g12(dx2/dx1)]− g1[g21 + g22(dx2/dx1)]

g22

}(eq.12)

Second-order necessary conditions for a maximum in one variablerequire that this second derivative be less than or equal to zero at the


point where the first-order conditions are satisfied. Sufficientconditions require that the inequalities hold strictly at the point. Thefirst-order conditions (eq. 1) and (eq. 2) require that f1 = −λg1 andf2 = −λg2. Young’s theorem tells us that f12 = f21 and g12 = g21.Substituting from these and rearranging, we can express (eq. 12) as

d2ydx2

1=

1(g2)2

[(f11 + λg11)(g2)2 − 2(f12 + λg12)g1g2 + (f22 + λg22)(g1)2

](eq.13)

The second-order partial derivatives of L (Taking the derivatives of (eq.1) and (eq. 2) with respect to x1 and x2) would be,

L11 = f11 + λg11; L12 = f12 + λg12; L22 = f22 + λg22 (eq.14)

It is clear that the terms involving λ inside the brackets in (eq. 13) arejust the second-order partials of the Lagrangian function. Suppose weform the symmetric matrix


H̄ ≡

0 g1 g2g1 L11 L12g2 L21 L22

The matrix is called the bordered Hessian of the Lagrangian function,because it involves the second-order partials of L bordered by thefirst-order partials of the constraint equation and a zero. If we take itsdeterminant, we see that

D̄ ≡

∣∣∣∣∣∣0 g1 g2g1 L11 L12g2 L21 L22

∣∣∣∣∣∣ = −[L11(g2)2 − 2L12g1g2 + L22(g1)2] (eq.15)

This is just the value in the big bracket in (eq. 13) multiplying by -1. Bycombining (eq. 13), (eq. 14) and (eq. 15), the second derivative of theobjective function subject to the constraint can be written in term of thedeterminant of the bordered Hessian of the Lagrangian function as


d2ydx2

1=

(−1)

(g2)2 D̄ (eq.16)

We are now in a position to state a sufficient condition for thetwo-variable, one-constraint problem.

Theorem A2.17 A Sufficient Condition for a Local Optimum inthe Two-Variable, One-Constraint Optimization ProblemIf (x∗1 , x

∗2 , λ

∗) solves the first-order conditions (eq. 1) through (eq. 3),and if D̄ > 0 (< 0), when evaluated at (x∗1 , x

∗2 , λ

∗), then (x∗1 , x∗2 ) is a

local maximum (minimum) of f (x1, x2) subject to the constraintg(x1, x2) = 0.

Example A2.9Consider whether the critical point we obtain in Example A2.8 is aminimum or a maximum. It is easy to see that L11 = −2a,L12 = L21 = 0, and L22 = −2b. From the constraint equation,


Example A2.9 (Continued)g1 = 1 and g2 = 1. Constructing the bordered Hessian, its determinantwill be

D̄ ≡

∣∣∣∣∣∣0 1 11 −2a 01 0 −2b

∣∣∣∣∣∣ = 2(a + b) > 0

Because here, D̄ > 0 for all value of x1, x2 and λ, by Theorem A2.17,the critical point must be a maximum subject to the constraint.

We can extend the problem of two-variable and one constraint to ageneral problem of n-variable and m constraints (m < n), thesecond-order sufficient conditions again tell us we will have amaximum (minimum) if the second differential of the objective functionis less than zero (great than zero) at the point where the first-orderconditions are satisfied. In the multivariable, multiconstraint case, thebordered Hessian is again formed by bordering the matrix ofsecond-order partials of L by all the


first-order partials of the constraints and enough zeros to form asymmetric matrix. The test for definiteness then involves checking thesign pattern on the appropriate principal minors of this bordered matrix:

H̄ =

0 · · · 0 g11 · · · g1

n...

. . ....

.... . .

...0 · · · 0 gm

1 · · · gmn

g11 · · · gm

1 L11 · · · L1n...

. . ....

.... . .

...g1

n · · · gmn Ln1 · · · Lnn

Its principal minors are the determinants of submatrices obtained bymoving down the principal diagonal. The n −m principal minors ofinterest here are those beginning with the (2m + 1)-st and ending withthe (n + m)-th, i.e. the determinant of H̄. That is, the principal minors


D̄k =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

0 · · · 0 g11 · · · g1

k...

. . ....

.... . .

...0 · · · 0 gm

1 · · · gmk

g11 · · · gm

1 L11 · · · L1k...

. . ....

.... . .

...g1

k · · · gmk Lk1 · · · Lkk

∣∣∣∣∣∣∣∣∣∣∣∣∣∣k = m + 1, · · · ,n, (eq.17)

We can summarize the sufficient conditions for optima in the generalcase with the following theorem.

Theorem A2.18 Sufficient Conditions for Local Optima withEquality ConstraintsLet the objective function be f (x) and the m < n constraints beg j(x) = 0, j = 1, · · · ,m. Let the Lagrangian be given by (A2.14). Let(x∗,Λ∗) solve the first-order conditions in (A2.15). Then


Theorem A2.18 (Continued)1. x∗ is a local maximum of f (x) subject to the constraints if the n −mprincipal minors in (eq. 17) alternate in sign beginning with positiveD̄m+1 > 0, D̄m+2 < 0, · · ·, when evaluated at (x∗,Λ∗).2. x∗ is a local minimum of f (x) subject to the constraints if the n −mprincipal minors in (eq. 17) are all negative D̄m+1 < 0, D̄m+2 < 0, · · ·,when evaluated at (x∗,Λ∗).


Optimization with Inequality Constraints

In many economic applications, we are faced with maximizing orminimizing something subject to constraints that involve inequalities.The Lagrangian analysis must be modified to accommodate problemsinvolving this and more complicated kinds of inequality restrictions onthe choice variables. First we begin with the simplest possibleproblem: maximizing a function of one variable subject to anonnegativity constraint on the choice variable. Formally,

maxx f (x) subject to x ≥ 0

As before, we are interested in deriving conditions to characterize thesolution x∗. If we consider the problem carefully, keeping in mind thatthe relevant region for solutions is the nonnegative real line, it seemsthat any one of three possibilities, depicted in Figure A2.9 (see thenext slide), can happen.


Figure A2.9 Three possibilities for maximization with nonnegativity constraints:

(left) case 1, constraint is binding; (center) case 2, constraint is binding but

irrelevant; and (left) case 3, constraint is not binding.

Case 1. x∗ = 0, and f ′(x∗) < 0Case 2. x∗ = 0, and f ′(x∗) = 0Case 3. x∗ > 0, and f ′(x∗) = 0


In each case, multiply the two conditions together and notice that theproduct will always be zero. Thus, in all three cases, x∗[f ′(x∗)] = 0.However, from Figure A2.9 Case 3, we see the x̃ = 0 also satisfiesx̃ [f ′(x̃)] = 0 although x̃ does not give a maximum of the function in thefeasible region. We can rule out this unwanted possibility by simplyrequiring the function to be nonincreasing as we increase x .All together, we have identified three conditions that characterize thesolution to the simple maximization problem with nonnegativityconstraints. If x∗ solves this problem, then all three of the followingmust hold:Condition 1. f ′(x∗) ≤ 0 (nonincreasing)Condition 2. x∗[f ′(x∗)] = 0Condition 3. x∗ ≥ 0In case 3, x̃ is ruled out because f (x̃) > 0, which violates Condition 1.Similarly, for the a minimum of f (x) subject to x ≥ 0, Condition 1 for x∗

should be altered as Condition 1. f ′(x∗) ≥ 0 (nondecreasing). Theother two conditions are the same.


Example A2.10Consider the problem

maxx f (x) = 6− x2 − 4x subject to x ≥ 0

Differentiating with respect to x , we get f ′(x) = −2x − 4. There are twocritical points 0 and -2. The latter one violates Condition 3 so -2 isruled out as an optima. x = 0 obviously satisfies Condition 2 andCondition 3. And furthermore, f ′(0) = −4 ≤ 0 which satisfy Condition1 for a maximum. So the solution must be x∗ = 0.

In the multivariable case, the three conditions must hold for eachvariable separately, with the function’s partial derivatives beingsubstituted for the single derivative. The theorem is straightforward.

Theorem A2.19 Necessary Conditions for Optima ofReal-Valued Functions Subject to Nonnegativity ConstraintsLet f (x) be continuously differentiable.


Theorem A2.19 (Continued)If x∗ maximizes (minimizes) f (x) subject to x ≥ 0, then x∗ satisfies

1

∂f (x∗)∂xi

≤ 0(≥ 0), i = 1, · · · ,n

2

x∗i

[∂f (x∗)∂xi

]= 0, i = 1, · · · ,n

3

x∗i ≥ 0, i = 1, · · · ,n

We now take up a two-variable optimization problem in which the onlyconstraint is given by the inequality g(x1, x2) ≥ 0. Formally, ourproblem is

maxx1,x2 f (x1, x2) subject to g(x1, x2) ≥ 0

To solve this problem, we’ll convert the problem to one with equalityMicroeconomic Theory Week 5: Calculus and Optimization (Jehle and Reny, Chapter A2)October 10, 2015 29 / 38

constraints and nonnegativity constraints, then apply what we know.The constraint requires that g be nonnegative. Suppose we introducea new variable, call it z, that we define as the amount by which gexceeds zero. Because g must be nonnegative, z, by definition. Wenow can rewrite the single inequality constraint as two now conditions,g(x1, x2)− z = 0 and z ≥ 0. By introducing the new variable z, wehave converted an inequality constraint into an equality constraint andone more nonegativity constraint. Then we can rewrite the problem as

maxx1,x2,z f (x1, x2) subject to g(x1, x2)−z = 0z ≥ 0

First considering the equality constraint, we form an Lagrangianfunction,

L(x1, x2, z, λ) ≡ f (x1, x2) + λ[g(x1, x2)− z].

With Lagrange’s theorem in mind, we wish to maximize L in thevariable x1, x2,z, subject to z ≥ 0. Because x1 and x2 areunconstrained, and z is constrained to be nonnegative, the first-order


conditions for the variables are given by combining Theorem A2.16and A2.19 as follows,

L1 ≡ f1 + λg1 = 0 (i)L2 ≡ f2 + λg2 = 0 (ii)Lz ≡ −λ ≤ 0 (iii)

zLz ≡ z(−λ) = 0 (iv)

z ≥ 0 (v)

where (i) and (ii) are from Theorem A2.16, (iii) through (v) are fromTheorem A2.19. The first-order condition on the choice of λ is simply,

Lλ ≡ g(x1, x2)− z = 0. (vi)

From (iv) we know z = g(x1, x2). Plug this into (iv) and (v) to replacez, and rewrite (iii) as λ ≥ 0. Then conditions (i) through (vi) can berewritten as an economical set of conditions called Kuhn-Tuckerconditions :


f1 + λg1 = 0 (i)f2 + λg2 = 0 (ii)

λg(x1, x2) = 0 (iii ′)λ ≥ 0, g(x1, x2) ≥ 0 (iv ′)

Conditions (i) through (iv ′) are the necessary conditions formaximization problems with inequality constraints. For minimizationproblems, it is λ ≤ 0 in (iv ′). For now, we will simply collect theseresults together in the following extension of Lagrange’s theorem(Theorem A2.16).

Theorem A2.20 (Kuhn-Tucker) Necessary Conditions for Optimaof Real-Valued Functions Subject to Inequality Constraints

Let f (x) and g j(x), j = 1, · · · ,m, be continuously differentiablereal-valuable functions over some domain D ⊂ Rn. Let x∗ be aninterior point of D and suppose that x∗ is an optimum (maximum or


Theorem A2.20 (Continued)

minimum) of f subject to the constraints, g j(x∗) ≥ 0, j = 1, · · · ,m.If the gradient vector ∆g j(x∗) associated with all binding constraintsare linearly independent, then there exists a unique vector Λ∗ such that(x∗,Λ∗) satisfies the Kuhn-Tucker conditions:

∂L(x∗,Λ∗)∂xi

≡ ∂f (x∗)∂xi

+m∑

j=1


= 0 i = 1, · · · ,n

λ∗j g j(x∗) = 0, g j(x∗) ≥ 0 j = 1, · · · ,m

Furthermore, the vector Λ∗ is nonnegative if x∗ is a maximum, andnonpositive if it is a minimum.


Value Functions

We often encounter optimization problems like

maxx f (x,a) s.t . g(x,a) = 0 and x ≥ 0 (problem 1)

where x is a vector of control variables, and a ≡ (a1, · · · ,am) is avector of parameters that may enter the objective function, theconstraint, or both. Clearly, the solutions to this problem will depend insome way on the vector of parameters, a. Suppose for each vector a,the solution is unique and denoted by x(a).We can define a new function, M(a), that gives the value achieved bythe objective function when x is chosen to maximize f subject to theconstraints. M(a) is called a maximum-value function and is definedformally as follows:

M(a) ≡ maxxf (x,a) s.t . g(x,a) = 0 and x ≥ 0

if we evaluate f (x,a) at the optimal solutions to this problem x(a), thenof course the value of the function will be as great as possible subject


to the constraint. We therefore could have defined M(a) equivalently as

M(a) ≡ f (x(a),a)

Now we want to know how the maximum value of a function changesas change one or more parameters of the problem. Envelopetheorem is a very powerful theorem that can be used to analyze thebehavior of the maximum value function as its parameters change.

Theorem A2.21 The Envelop TheoremConsider the (problem 1) and suppose the objective function andconstraint are continuously differentiable in a. For each a, let x(a)� 0uniquely solve (problem1) and assume that it is also continuouslydifferentiable in the parameter a. Let L(x,a, λ) be the problem’sassociated Lagrangian function and let (x(a), λ(a)) solve theKuhn-Tucker conditions in Theorem A2.20. Finally, let M(a) be theproblem’s associated maximum-value function. Then the Enveloptheorem states that


Theorem A2.21 (Continued)∂M(a)

∂aj=

∂L∂aj

∣∣∣∣x(a),λ(a)

j = 1, · · · ,m

where the right-hand side denotes the partial derivative of theLagrangian function with respect to the parameter aj evaluated at thepoint (x(a), λ(a)).

The theorem says that the total effect on the optimized value of theobjective function when a parameter changes can be deduced simplyby taking the partial of the problem’s Lagrangian with respect to theparameter and then evaluating that derivative at the solution to theoriginal problem’s first order Kuhn-Tucker conditions. The theoremapplies regardless of the number of constraints.

Example A2.11Suppose we have the following maximization problem


Example A2.11 (continued)

maxx1,x2 x1x2 s.t . a− 2x1 − 4x2 = 0

The Lagrangian function is

L = x1x2 + λ[a− 2x1 − 4x2]

with first-order conditions:

L1 = x2 − 2λ = 0L2 = x1 − 4λ = 0

Lλ = a− 2x1 − 4x2 = 0

These can be solved to find x1 = a/4, x2 = a/8 and λ(a) = a/16. Wecan calculate the maximum value:

M(a) = x1(a)x2(a) = (a/4)(a/8) =a2

32


Example A2.11 (continued)Take the derivative with respect to a,

dM(a)

da=

a16.

To verify the Envelope theorem, simply differentiate the Lagrangian forthe maximization problem with respect to the parameter

dLda

= λ

and then evaluate that derivative at the solution to the first-orderconditions. But we had known that λ = a/16. This gives us

dLda

=a16

=dM(a)

da

thus verify the theorem.


Documents

Week 5: Calculus and Optimization (Jehle and Reny, Chapter …economics-course.weebly.com/uploads/2/5/7/2/25725158/week_5.pdf · 1 and x 2 in (eq:4) are candidate solutions to this