Finance 30210: Managerial Economics Optimization

Functions Optimization deals with functions. A function is simply a mapping from one space to another. (that is, a set of instructions describing how to get from one location to another) is a function Is the range Is the domain

For example For Domain Range

For 05 Domain 5 20 Range X =3 Y =14

05 Domain 5 20 Range Optimization involves finding the maximum value for y over an allowable domain. Here, the optimum occurs at x = 5 (y = 20)

What is the solution to this optimization problem? 105 There is no optimum because f(x) is discontinuous at x = 5

06 12 What is the solution to this optimization problem? There is no optimum because the domain is open (that is, the maximum occurs at x = 6, but x = 6 is NOT in the domain!)

0 12 What is the solution to this optimization problem? There is no optimum because the domain is unbounded (x is allowed to become arbitrarily large)

The Weierstrass theorem provides sufficient conditions for an optimum to exist, the conditions are as follows: The domain foris closed and bounded is continuous over the domain of

Derivatives Formally, the derivative ofis defined as follows: All you need to remember is the derivative represents a slope (a rate of change) Actually, to be more accurate, the derivative represents a trajectory

Useful derivatives Exponents Logarithms Products Composites Linear Functions

Practice Makes Perfect…

Unconstrained maximization Strictly speaking, no problem is truly unconstrained. However, sometimes the constraints don’t “bite” (the constraints don’t influence the maximum) First Order Necessary Conditions Ifis a solution to the optimization problem then or

An Example Suppose that your company owns a corporate jet. Your annual expenses are as follows:  You pay your flight crew (pilot, co-pilot, and navigator a combined annual salary of $500,000.  Annual insurance costs on the jet are $250,000  Fuel/Supplies cost $1,500 per flight hour  Per hour maintenance costs on the jet are proportional to the number of hours flown per year. Maintenance costs (per flight hour) = 1.5(Annual Flight Hours) If you would like to minimize the hourly cost of your jet, how many hours should you use it per year?

An Example Let x = Number of Flight Hours First Order Necessary Conditions

An Example Hourly Cost ($) Annual Flight Hours

How can we be sure we are at a minimum? Ifis a solution to the maximization problem then Ifis a solution to the minimization problem then Secondary Order Necessary Conditions

Note: The second derivative is the rate of change of the first derivative Slope is increasingSlope is decreasing

An Example Let x = Number of Flight Hours First Order Necessary Conditions Second Order Necessary Conditions For X>0

Multiple Variables Suppose you know that demand for your product depends on the price that you set and the level of advertising expenditures. Choose the level of advertising AND price to maximize sales

Partial Derivatives When you have functions of multiple variables, a partial derivative is the derivative with respect to one variable, holding everything else constant First Order Necessary Conditions

(1) (2) (1) (2) Multiple Variables

Again, how can we be sure we are at a maximum?

its generally sufficient to see if all the second derivatives are negative… Recall, the second order condition requires that

Practice Questions 1) Suppose that profits are a function of quantity produced and can be written as Find the quantity that maximizes profits 2) Suppose that costs are a function of two inputs and can be written as Find the quantities of the two inputs to minimize costs

Constrained optimizations attempt to maximize/minimize a function subject to a series of restrictions on the allowable domain To solve these types of problems, we set up the lagrangian Function to be maximized Constraint(s) Multiplier

To solve these types of problems, we set up the lagrangian We know that at the maximum…

Once you have set up the lagrangian, take the derivatives and set them equal to zero First Order Necessary Conditions Now, we have the “Multiplier” conditions…

Constrained Optimization Example: Suppose you sell two products ( X and Y ). Your profits as a function of sales of X and Y are as follows: Your production capacity is equal to 100 total units. Choose X and Y to maximize profits subject to your capacity constraints.

Constrained Optimization The first step is to create a Lagrangian Objective Function Constraint Multiplier

Constrained Optimization First Order Necessary Conditions “Multiplier” conditions

Constrained Optimization:

The Multiplier Lambda indicates the marginal value of relaxing the constraint. In this case, suppose that our capacity increased to 101 units of total production. Assuming we respond optimally, our profits increase by $5

Another Example Suppose that you are able to produce output using capital (k) and labor (l) according to the following process: The prices of capital and labor areandrespectively. Union agreements obligate you to use at least one unit of labor. Assuming you need to produce units of output, how would you choose capital and labor to minimize costs?

Minimizations need a minor adjustment… To solve these types of problems, we set up the lagrangian A negative sign instead of a positive sign!!

Just as in the previous problem, we set up the lagrangian. This time we have two constraints. Doesn’t necessarily hold with equality Will hold with equality Non-Binding Constraints

First Order Necessary Conditions

Case #1: Constraint is non-binding

First Order Necessary Conditions Case #2: Constraint is binding

Constraint is Non-Binding Constraint is Binding

Try this one… You have the choice between buying apples and bananas. You utility (enjoyment) from eating apples and bananas can be written as: The prices of Apples and Bananas are given byand Maximize your utility assuming that you have $100 available to spend

(Income Constraint) (Objective) (You can’t eat negative apples/bananas!!) Objective Non-Negative Consumption Constraint Income Constraint

First Order Necessary Conditions We can eliminate some of the multiplier conditions with a little reasoning… 1.You will always spend all your income 2.You will always consume a positive amount of apples

Case #1: Constraint is non-binding First Order Necessary Conditions

Case #1: Constraint is binding First Order Necessary Conditions

Constraint is Non-Binding Constraint is Binding