Multivariable and Constrained Optimization

Mathematical Economics

Vilen Lipatov

Fall 2014

Outline

I multi-variable optimization

I constrained optimizationReading: Sydsaeter and Hammond, chapters 13-14

Unconstrained Optimization

I Objective: Find the maximum of the function z = f (x , y).

I That is, find a pair (x∗, y∗) such that f (x∗, y∗) ≥ f (x , y)holds for all pairs (x , y).

I All pairs of numbers (x , y) (including (x∗, y∗) ) must be inthe domain of the function.

I For the minimum, analogously, we require f (x∗, y∗) ≤ f (x , y).

I In the optimization context, f (x , y) is called the objectivefunction.

I Show that the function

z = f (x , y) = −(x − 2)2 − (y − 3)2

has a maximum at (x∗, y∗) = (2, 3).

First-order conditions

I If we keep the variable y at the optimal value y∗ and varyonly x , then the function of one variable

g(x) = f (x , y∗)

must have a maximum at x∗:

dg

dx(x∗) = 0.

I Thus we obtain the first-order conditions necessary for amaximum:

∂f

∂x(x∗, y∗) = 0,

∂f

∂y(x∗, y∗) = 0.

I These conditions determine the stationary points of f (x , y).

Exercise

I Consider a firm with a production function Y = F (K , L),where Y is the number of units produced, K is the capitalinvested and L is the labor input.

I The profit function is given byΠ(K , L) = PF (K , L)− rK − wL, where P is the price of oneunit of output, r is the cost of capital (the interest rate), andw is the wage rate.

I Find the first-order conditions for the profit maximum.

Exercise

I A monopolist with total cost function C (Q) = Q2 sells herproduct in two different countries.

I When she wants to sell QA units of the good in country A,she cannot charge the price higher than PA = 22− 3QA foreach unit.

I When she wants to sell QB units of the good in country B,she cannot charge the price higher than PB = 34− 4QB .

I How much should the monopolist sell in the two countries inorder to maximize profits?

SOC: a maximum

(0, 0) is a stationary point of f (x , y) = −x2 − y2.

SOC: a minimum

(0, 0) is a stationary point of f (x , y) = x2 + y2.

SOC: a saddle point

(0, 0) is a stationary point of f (x , y) = x2 − y2.

Types of extrema and Hessian: a Theorem

I Suppose that the function y = f (x , y) has a stationary pointat (x∗, y∗).

I Consider the determinant of the Hessian matrix

detH =

∣∣∣∣∣ ∂2f∂x2

∂2f∂y∂x

∂2f∂x∂y

∂2f∂y2

∣∣∣∣∣ =∂2f

∂x2∂2f

∂y2−(∂2f

∂x∂y

)2

.

I If detH < 0 at (x∗, y∗), then (x∗, y∗) is a saddle point.I If detH > 0 at (x∗, y∗), then (x∗, y∗) is a maximum or a

minimum.I When detH > 0, the signs of ∂2f /∂x2 and ∂2f /∂y2 are the

same.I If both signs are positive, then (x∗, y∗) is a minimum.I If both signs are negative, then (x∗, y∗) is a maximum.

Exercises

I Classify the stationary points of f (x , y) = −x2 − y2.

I Classify the stationary points of f (x , y) = x2 + y2.

I Classify the stationary points of f (x , y) = x2 − y2.

Constrained optimization

y ≥ 1

Constrained optimization

I Objective: Find the maximum of the function z = f (x , y)subject to the inequality constraints

g1(x , y) ≥ 0,

g2(x , y) ≥ 0,

...

gK (x , y) ≥ 0.

I That is, find a pair (x∗, y∗) satisfying the constraints, suchthat f (x∗, y∗) ≥ f (x , y) holds for all pairs (x , y) that alsosatisfy the constraints.

I All pairs of numbers (x , y) (including (x∗, y∗) ) must be inthe domain of the function.

I For the minimum, we require f (x∗, y∗) ≤ f (x , y).

I Just as in unconstrained case, f (x , y) is called objectivefunction.

Geometry of constraints

Budget set

Typical examples

I A consumer maximizes her utility subject to a budgetconstraint.

I A producer minimizes costs subject to the constraint that acertain amount is produced.

I Asymmetric information setting: An insurer selects aninsurance contract that maximizes profits subject to theconstraint that the insurance is valuable for the consumer(participation constraint) and that the consumer has anincentive to be careful (incentive compatibility constraint).

Exercise: Utility maximization

I A consumer maximizes her utility

u(x , y) = (x + 1)(y + 1)

subject to her budget constraint

pxx + pyy ≤ B

and the non-negativity constraints

x ≥ 0, y ≥ 0.

I Find the optimal amount of goods consumed and the optimalvalue of utility that consumer attains.

Exercise: Cost minimization

I A producer in a perfectly competitive market has a productionY = F (K , L) = K 1/6L1/2, where Y is the number of unitsproduced, K is the capital invested and L is the labor input.

I She wants to minimize costs subject to producing at least Y0

units.

I The optimization problem is equivalent to

max(−rK − wL)

subject toF (K , L) ≥ Y0,

K ≥ 0, L ≥ 0.

I Find the optimal amount of inputs and total costs ofproduction.

Exercise: swimming

I A swimmer who is currently at the coordinates (a, b) wants toswim along the shortest route to the square island with cornerpoints (−1, 1), (1,−1), (−1,−1), (1, 1).

I Instead of minimizing the distance, we can maximize thenegative of the square of the distance

−(x − a)2 − (y − b)2,

subject to the constraints

−1 ≤ x ≤ 1,

−1 ≤ y ≤ 1.

The island

The swimmer

Binding constraints

I A constraint is binding at the optimum if it holds withequality in the optimum.

I Here, only two of the four constraints are binding. Allnon-binding constraints can be ignored, because the optimalsolution does not change, if we leave them out.

The Lagrangian approach

I The Lagrangian approach transfers a constrained optimizationproblem into

I an unconstrained optimization problem andI a pricing problem.

I The new function to be optimized is called the Lagrangian.

I For each constraint, a shadow price is introduced, called aLagrange multiplier.

I In the new unconstrained optimization problem a constraintcan be violated, but only at a cost.

I The pricing problem is to find shadow prices for theconstraints such that the solutions to the new and the originaloptimization problem are identical.

I L(x , y) = f (x , y) +λ1g1(x , y) +λ2g2(x , y) + ...+λKgK (x , y).

The method

1. Make an informed guess about which constraints are bindingat the optimum.

2. Suppose there are k∗ such constraints. Set the Lagrangemultipliers for all other k − k∗ constraints to zero, i.e. ignorethese constraints.

3. Solve the first-order conditions

∂L/∂x = 0, ∂L/∂y = 0,

together with the conditions that the k∗ constraints hold withequality. Note that we obtain a system of k∗ + 2 equations forthe same number of unknowns.

4. Check whether the solution is indeed an unconstrainedoptimum of the Lagrangian. This may be difficult.

5. Check that the Lagrange multipliers are all non-negative andthat the solution (x∗, y∗) satisfies all the constraints.

6. If 4) or 5) are violated, start again at 1) with a new guess.

The constrained optimum theorem

I Suppose we are given numbers λ1, λ2, ...λK and a pair ofnumbers (x∗, y∗) such that

I λ1, λ2, ...λK , i.e. Lagrange multipliers are nonnegative;I (x∗, y∗) satisfies all the constraints, i.e.

gk(x∗, y∗) ≥ 0,∀k = 1, 2, ..,K ;I (x∗, y∗) is an unconstrained maximum of the Lagrangian L;I The complementary slackness conditions λkgk(x∗, y∗) = 0 are

satisfied, i.e. either the kth Lagrange multiplier is zero or thekth constraint binds.

I Then (x∗, y∗) is a maximum for the constrained maximizationproblem.

Constrained optimum: comments

I The Lagrangian approach does not immediately tell us, whichconstraints are binding in the optimum. We have to start withan informed guess using all problem-specific information.

I We write down the Lagrangian assuming that only certainconstraints bind.

I We solve the system of simultaneous equations consisting ofthe FOCs and the binding constraints.

I We check if the solution satisfies the other constraints andthat Lagrange multipliers are nonnegative.

I We check if the solution found is an unconstrained optimumof the Lagrangian.

Single binding constraint

I Suppose only constraint g1 is binding.

I The FOCs are then

∂f

∂x= −λ1

∂g1∂x

,

∂f

∂y= −λ1

∂g1∂y

.

I We can get rid of λ1 to obtain the following system to solve:

∂f /∂x

∂f /∂y=∂g1/∂x

∂g1/∂y,

g1(x , y) = 0.