The Economic Approach Applied to Single Variable Optimization

Chapter 5: The Economic Approach Applied to Single Variable Optimization

A Brief Review of the Economic Approach

¥ The Economic Approach can be applied to optimization problems or equilibrium systems. Economic models are abstract, simplified descriptions of the optimization problem or equilibrium system.

¥ As applied to individual decision-makers, the Economic Approach says that agents act as if they optimize. In other words, they do the best they can under given conditions.

¥ The choices agents make are interpreted as being efficient (or optimal or best) from the perspective of the agent.

¥ The often seemingly ridiculous assumption of optimization (or rationality) is used because (1) it organizes variables in a coherent fashion, (2) agents really do face choices and optimization is one way to make decisions, (3) it helps us understand whatÕs going on, and (4) it generates testable predictions.

Did you know that lifeguards are very much aware of the problem they face? They are trained to neither jump right in the water nor to choose the path of least water!1

¥ Economists do not claim that individuals actually use sophisticated mathematics, rather, economists use mathematics to interpret observed variables with the aid of the optimization framework.

Lifeguards are not aware of the mathematics involved in finding the optimal distance to run, then swim. They are conscious of their goal and the fact that either too little or too much running will raise the time it takes to get to the victim. They are trained to use rules of thumb, instinct, experience, and common sense to solve the problem.

¥ Assuming that individuals act as if they optimize enables accurate predictions to be made about the choices observed under varying conditions.

It is the economist who imposes the optimization framework in order to make predictions about the lifeguardÕs changes in behavior (in this case, the distance chosen to run on the sand) as an exogenous variable changes (say, their top running speed).

¥ Often, the optimization framework has provided good predictions of actual decisions. For many analysts, this is the acid test: to them, if a theory makes good predictions, then it doesnÕt matter to them how unrealistic are the assumptions of that theory.

Having come up with a prediction, the economist would observe lifeguards in action in order to see if the predictions matched the empirical data.

1In fact, Wabash College swim coach Gail Pebworth tells us that lifeguards must Òknow their beach.Ó Lifeguards are trained to constantly recalculate the optimal distance as the time of day and season changes on any particular beach. C5Read.pdf 1 The Structure of Optimization Problems

1) Setting Up the Problem

A) Objective function (e.g., utility or profit)

¥ Consumers are utility maximizers ¥ Firms are profit maximizers ¥ Students may be GPA maximizers ¥ The lifeguard is a time minimizer

B) Endogenous variables are things the agent CAN choose; C) Exogenous variables are things the agent CANNOT control.

¥ Consumers choose goods; they cannot control prices ¥ Firms choose output or (maybe) price, they cannot control a competitorÕs moves ¥ Students choose the amount of time they want to study, not the content of the course ¥ The lifeguard chooses the amount of time to run on the sand, not the location of the drowning victim

Objective Endogenous Exogenous Agent Function Variable Variable Prices, Type and Income, Consumer Max utility number of Tastes and goods Preferences Competitive Price, cost Max profit Output Firm conditions Market Monopoly Max profit Output, Price demand, cost conditions Ability, Time Spent course Student Max GPA Studying in content, each course professor's grading scale Top velocity on land and Distance to Lifeguard Min time in water, run on sand location of victim

Sometimes agents have constraints on their choices. That is they face some restriction on the possible values of the endogenous variables. For example, available income restricts the range of options in the consumerÕs choice problem. We will talk about this kind of CONSTRAINED OPTIMIZATION problem later.

C5Read.pdf 2 2) Finding the Optimal Solution

An example of an optimization problem is the lifeguard problem you explored in C4Lab.xls. The optimal value of Distance_in_Sand (how far to run before jumping in and swimming) was found using ExcelÕs Solver. After you have Set Up the Problem, Solver does the work for you and determines the optimal value of the endogenous variable. In earlier work (C3Lab.xls explored the profit maximization problem), we saw how tables and graphs (using either the Direct Method or Method of Marginalism) could be used to find an optimal solution.

Now we will see yet another way to solve optimization problems.

ASIDES: ¥ Note how the steps for these various solution strategies are identical. Try to spot the consistent patterns across the different solution strategies.

¥ There is no one best strategy. The idea is that by exploring these different ways to do the Economic Approach, we learn and understand.

Solving the Lifeguard Problem with the Method of Marginalism via Calculus

Review of the Problem: You are a lifeguard who can run faster than you can swim. You spot a drowning victim. What path should you take to minimize the time necessary to reach the victim? The essential trade-off is that running is faster than swimming, but if you swim all the way, the distance is shortest.

What is the path of The drowning victim is least time? here W A b T E R x is a choice variable for the lifeguard. He or she can set it at x=0 (which means jumping right in) or You are here SAND x=c (which means running down the c beach until he or she is directly across from the victim before jumping in).

Notation:

T = time to reach victim (in minutes) x = distance along sand the lifeguard runs (in yards) b = length of perpendicular connecting victim to beach (in yards c = distance of victim from lifeguard along sand (in yards) vs = velocity of lifeguard running on sand (in yards/minute) vw = velocity of lifeguard swimming in water (in yards/minute)

C5Read.pdf 3 Step 1: Setting Up the Problem

1) What is the Objective Function (Goal)?

minimize the time it takes to reach the victim (T)

¥ recall that dividing distance (in say, yards) by velocity (yd/min) gets you time (min) ¥ total time to victim is the sum of running time and swimming time ¥ once you choose how far to run on the sand, the swimming distance is also determined because the lifeguard makes a bee-line (along the hypotenuse) for the victim

2) What are the Endogenous Variables?

distance to run along the sand (x) before jumping into the water

3) What are the Exogenous Variables?

velocity on sand (vs), velocity in water (vw), position of victim (that is, b and c)

Step 2: Finding the Optimal Solution

The problem can be stated mathematically in this way:

(c - x)2 +b2 x minT= + x vs vw

¥ The variable x, under the Òmin,Ó identifies the endogenous variables in the problem. It is clear that there is only one choice variable in this problem.

¥ The first term on the right-hand side (x/vs) is the time it takes to run on the sand; the second is the time it takes in the water.

¥ We divide distance (in yards or miles or feet) by velocity (in yards per minute or miles per hour or feet per second) to obtain travel time.

C5Read.pdf 4 Before we continue, letÕs quickly review the other methods of solving this problem:

A Review of the Direct Method (using Totals):

Two alternative ways of applying it:

(1) TABLE: For any set of values of the exogenous variables, have Excel calculate the value of the function for different values of x. Choose the value of x that gives the lowest value for the function, total time to victim.

Solving the Lifeguard Problem Using the Direct Method

Vs 300 Vw 100 c 100 b 100

Distance On Total Time Sand Time on Sand Time in Water to Victim (yards) (minutes) (minutes) (minutes) 0 0.0000 1.4142 1.4142 5 0.0167 1.3793 1.3960 10 0.0333 1.3454 1.3787 15 0.0500 1.3124 1.3624 20 0.0667 1.2806 1.3473 25 0.0833 1.2500 1.3333 30 0.1000 1.2207 1.3207 35 0.1167 1.1927 1.3094 40 0.1333 1.1662 1.2995 45 0.1500 1.1413 1.2913 50 0.1667 1.1180 1.2847 55 0.1833 1.0966 1.2799 60 0.2000 1.0770 1.2770 65 0.2167 1.0595 1.2761 70 0.2333 1.0440 1.2774 75 0.2500 1.0308 1.2808 80 0.2667 1.0198 1.2865 85 0.2833 1.0112 1.2945 90 0.3000 1.0050 1.3050 95 0.3167 1.0012 1.3179 100 0.3333 1.0000 1.3333

C5Read.pdf 5 (2) GRAPH: Have Excel graph the value of the function for different levels of x, and choose the x that corresponds to the lowest value of the function, total time.

A Totals Graph of the Lifeguard Problem 1.60

1.40

1.20 min

1.00 Time on Sand (minutes)

0.80 Time in Water (minutes) 0.60 Time to Victim (minutes) 0.40

0.20

0.00 0 102030405060708090100

Distance on Sand (yards)

Although the optimal distance on sand is about 65 yards, that U-shaped curve (or bowl) is pretty flat. That means that mistakes to the right or left (too much or too little running on sand) aren't that big a deal. Unless, of course, that half-second is the difference between life and death!

C5Read.pdf 6 Finding the Optimal Solution with the Method of Marginalism via Calculus

As usual, there are a series of steps that provide a framework for reaching the correct answer. We provide the steps for you and encourage you to refer back to this page whenever you solve an optimization problem with the Method of Marginalism via Calculus.

The steps assume that you have I) SET UP THE PROBLEM and are now ready for II) FINDING THE INITIAL SOLUTION

STEP (1) Set up the problem mathematically; that is, write out the objective function, indicate the choice variable, and indicate what the agent wishes to do (i.e., MAX or MIN).

STEP (2) Take the derivative of the objective function with respect to the choice variable.

STEP (3) Set the derivative of the objective function equal to zero. The value of the endogenous variable that satisfies this condition (called a Òfirst-order conditionÓ by mathematicians) is unique (for the simple problems we will be dealing with in this course); we indicate this by sticking a star or asterisk (*) on the choice variable.

This is a crucial point. A choice variable, say X, can be set by the agent to many different values. One (and only one) of these values is the optimal (or best or efficient) solution for the problem at hand. This special value, X*, is denoted by the asterisk convention.

STEP (4) Solve for the optimal value of the endogenous variable; that is, rearrange the equation (or first-order condition) in Step 3 so that you have X* by itself on the left-hand side and only exogenous variables on the right. This is called a reduced form: it tells you the optimal value of the endogenous variable for any set of values for the exogenous variables.

STEP (5) Plug the optimal value back into the objective function. This will give you what is called Òthe maximum value functionÓ. You can think of it as a sort of reduced form of the objective function: it gives the maximum value (or minimum value, in the case of, for example, the lifeguard problem) of the objective function as a function of the parameters of the problem.

Let's put these steps to work in solving the lifeguard problem. We repeat that you should always follow these steps carefully when solving an optimization problem with the Method of Marginalism via Calculus.

C5Read.pdf 7 We have already Set Up the Problem , so we are prepared to apply the steps discussed on the previous page.

STEP (1) Set up the problem mathematically; that is, write out the objective function, indicate the choice variable, and indicate what the agent wishes to do (i.e., MAX or MIN).

x ()c − x 2 + b2 minT = + x vs vw

This can be rewritten in the following form which is easier to work with:

11 1 minT =+x []()cx −2 + b2 2 x vswv

STEP (2) Take the derivative of the objective function with respect to the choice variable.2

− 1 dT 1 1 1 2 = + []()c − x + b 2 2 ()− 2()c − x) dx vs vw 2

This derivative is a little complicated. We had to use the chain rule to get it. DonÕt worry about that right now.

This expression can be evaluated at any value of x. The crucial idea is that thereÕs one special value of x, the optimal value, that can be found when the derived expression is equal to zero. Thus, the next step is to set the derivative equal to zero.

STEP (3) Set the derivative of the objective function equal to zero. The value of the endogenous variable that satisfies this condition (called a Òfirst-order conditionÓ by mathematicians) is unique; we indicate this by sticking a star or asterisk (*) on the choice variable.

− 1 dT 111 2 2 =+[]()cx −**+ b2 ()−−20() cx) = dx vsw v 2

Many students don't see the need to separate steps 2 and 3. After all, the only difference is that Step 3 carries an "=0" and a few stars ("*"). But what a difference that is! You can take a derivative of any function at any time. Doing so will always report how the function changes. Setting the derivative equal to zero is a special step in the application of the derivative to the solving of optimization problems. The appendix at the end tries to emphasize this crucial point.

When you set the derivative of the objective function equal to zero and put asterisks on the choice variables, you have made an important claim. By putting the Ò*Ó superscript on the choice variable, you are saying that this is the optimal value. In essence, you have found the answer to the problem. All that remains is to present the solution in a nicer form.

2 We offer a helpful appendix at the end of this reading on the "derivative." Go to it now if you would like to learn more about the derivative before continuing through this example. C5Read.pdf 8 STEP (4) Solve for the optimal value of the endogenous variable; that is, rearrange the equation (or first-order condition) in Step 3 so that you have the optimal value of the e variable by itself on the left-hand side and only exogenous variables on the right. This is called a reduced form: it tells you the optimal value of the endogenous variable for any set of values for the exogenous variables.

Applying this step to the equation in step 3 and solving for x* you obtain (after much tedium): v x* = c − w b 2 − 2 vs vw

This is the answer. It tells you the optimal value of x* (distance on sand) for any value of the exogenous variables. All that remains is to substitute this optimal value into the objective function in order to find the minimum time to the victim.

v c < w b v 2 − v 2 The expression above holds only for x* ≥ 0. If s w , then x* = 0. It doesnÕt make any sense for the lifeguard to run away from the x*=0 b victim! One intuitive explanation for the corner solution is that if b is much higher than c, then running is not at all worth it and the c lifeguard should jump right in.3

2 b v w Š b v w ‘ 2 c − Œ “ + b 2 − 2 Œ 2 − 2 “ ∗ v s v w ‹ v s v w ’ T = + v s v w

Now, most people would say, "Yecch! What a mess!!" or maybe even, "What on earth is that thing?!?" But, actually, although admittedly seemingly complicated, it's merely an expression that says, "The absolute minimum (i.e., fastest) time to the drowning victim

depends on b, c, vw, and vs." Given any configuration of those four exogenous variables, the expression above will immediately reveal T*, the minimum time to the victim. That's powerful.

3 Mathematically, whatÕs happening is that minus x is being treated as a negative number in the T expression, when, in fact, we should be using the absolute value of x. Minus x indicates traveling in the opposite direction, but such a move cannot possibly result in negative running time! Including |x| in the T expression doesnÕt seem worth the effort since we know the lifeguard will never run away from the victim.

C5Read.pdf 9 ASIDE ON MATH: We realize that cranking through a series of equations is probably not your idea of a great time. That's why we hid a lot of the work. It turns out that the lifeguard problem is interesting, but a bit cumbersome. Keeping the variables straight and not making any mistakes requires a little patience. WeÕre teaching you the Economic Approach, not mathematics.

LetÕs test out our answers to x* and T*: c = distance on sand = 100 yards b = distance on water = 100 yards

vs = velocity on sand = 300 yards/minute

vw = velocity on water = 100 yards/minute ⋅ 100∗ 100 b vw =−100 ≈ 64. 6yards x* =c - 22− 2 2 1/2 300 100 (vs -vw) This is the same answer that Excel's Solver gave us, remember?

To find out what the minimum time, T*, is, we plug the values of the exogenous variables into our Step 5 maximum value function The bar at the end tells you to b v b v 2 evaluate the w ŒŠ ŒŠ w “‘ “‘ 2 expression at the c − Œ c − Œ c − “ “ + b 2 − 2 Œ Œ 2 − 2 “ “ values given. ∗ v s v w ‹ ‹ v s v w ’ ’ T = + = v v b 100 s w c = 100 v = 300 s v = 100 w

2 100•100 Š Š 100•100 ‘ ‘ 2 100 − Œ 100 − Œ 100 − “ “ + 100 2 2 Œ Œ 2 2 “ “ 300 − 100 ‹ ‹ 300 − 100 ’ ’ = + 300 100

= 1 . 28 minutes

This, of course, perfectly agrees with our solution from the Solver (in C4Lab.xls).

To be sure, working with complicated expressions with several variables is not easy, but neither is it so fantastically complicated that it is beyond the reach of all but the most brilliant mathematicians. All we have done is found precisely where the objective function has a flat spot (where the first derivative expression is equal to zero) and then worked with the equation to isolate the optimal value of the endogenous variable, x*, on the left-hand side. The power of the resulting reduced-form expression should be evidentÑany set of values for the four exogenous variables can be immediately plugged in and x* can be calculated. The same holds true for the maximum value function, T*. It may be a messy expression, but once we have it, it is quite powerful.

C5Read.pdf 10 In fact, the ability to quickly find how x* and T* change given a shock (a change in a single exogenous variable, holding all others constant), is the main reason why calculus is so prevalent in economics.

To see a demonstration of this, read on!

Step 3: Comparative Statics

In general, economists are less interested in finding a single answer to an optimization problem than in seeing how the optimal choice of an agent changes when the environment in which she makes that choice changes. This is called comparative statics.

The exercise of comparative statics allows you to make predictions (either qualitative or quantitative) about the real world. These predictions can be tested to see if your model of observed behavior is any good. The test is simple: Is the prediction accurate?

Comparative statics takes the following form: Initial Shock Recalculate New Compare

Shock means change one of the exogenous variables while holding all of the others constant. Recalculate New means find the NEW optimal value (for example, by using the Solver).

TWO EXAMPLES:

Example 1: Suppose we solve the firmÕs profit maximization problem and determine its optimal output Q* as a function of the exogenous variables including output price. We may be interested in finding out how the optimal choice changes when the price of the product changes or when production inputs become more expensive.

Example 2: In the lifeguard problem, we may be interested in how the optimal choice of distance to run on sand changes when the location of the swimmer changes or when you get a lifeguard with a different top swimming or running speed.

Note: It is VERY important that you change only one exogenous variable at a time and observe its effect on the optimal choice of the endogenous variable. If you change more than one parameter (say, cost of labor and price in the profit max problem or lifeguard water velocity and position of the victim in the lifeguard problem), it is extremely difficult to disentangle the effects in order to say how one variable alters the optimal choice. This is the heart of the difficulty facing the econometrician.

C5Read.pdf 11 TWO WAYS OF DOING COMPARATIVE STATICS:

The size of the shock, a finite distance (say a change in Price from $3/unit to $4/unit) versus an infinitesimally small shock, determines two alternative ways of doing comparative statics.

As we saw in our study of the derivative in C3Lab.xls, as the size of the change gets smaller and smaller, the finite distance method approaches the method based on the calculus. You will see what we mean when we work through an example.

Comparative Statics Method 1: Method of Actual Comparison

In this method, you simply redo the problem with the new value of the parameter in question. In the lifeguard problem, you were asked to run the Solver twice for different running velocities of the lifeguard (300 yards/minute and 250 yards/minute). LetÕs compare your results:

X*(vw=100, c=b=100, vs=300) = 64.64 yards for a minimum time of 1.28 minutes X*(vw=100, c=b=100, vs=250) = 56.36 yards for a minimum time of 1.32 minutes

If you run Solver again with a running velocity of 200 yards/minute (an even slower lifeguard) you will obtain:

X*(vw=100, c=b=100, vs=200) = 42.26 yards for a minimum time of 1.37 minutes

These data can be graphed in order to be understood more easily. The idea is that X*, the optimal distance to run on the sand, depends upon the lifeguard's foot speed. Thus, the

graph is plotted with X* on the vertical axis and vs on the horizontal axis. It looks like this:

Top_Velocity_on_Sand Optimal_Distance_on_Sand 150 10.56 200 42.26 250 56.36 300 64.64

Comparative Statics Presentation Graph

80.00

60.00

40.00

20.00

0.00 0 100 200 300 400 Top_Velocity_on_Sand Optimal_Distance_on_Sand

Notes: ¥ the relationship between distance run on sand to sand velocity is a positive one: the faster the lifeguard, the longer she should run on sand ¥ the relationship appears to be non-linear: in other words, a one yard/minute increase in running speed does not uniformly increase optimal distance. It appears that the slope is decreasingÑthe higher the lifeguard velocity on sand, the less a one yard/minute increase in speed will increase the time she should run on sand.

C5Read.pdf 12 Presentation Graphs:

The graph of x*=Ä(vs|b, c, vw) (read "x star as a function of vs holding constant b, c, and vw") is called a Òpresentation graphÓ because it depicts how the optimal value of an endogenous variable changes as an exogenous variable changes, ceteris paribus.

The presentation graph on the previous page, reproduced below for ease of comparison, was constructed by TRACKING the optimal distance on sand for a given top velocity on sand as shown below:

1.420 A Totals Graph of the Lifeguard Problem 1.400 vs = 25 0

1.380

1.360

1.340 vs = 300 1.320

1.300

1.280

1.260 X (Distance on Sand) 0102030405060708090100 64.64 56.36

Comparat ive Stat ics Present at ion Graph O p 80.00 64.64 ti 60.00 56.36 m al 40.00 _ Di 20.00 st 0.00 a 0 100 200 300 400 n Top_Velocity_on_Sand c

How to Read a Presentation Graph:

One important question to ask in order to understand a presentation graph is, ÒWhat does it mean to be off the curve?Ó

In the underlying graph, the lifeguard is certainly free to run other than 64.64 yards when top velocity in sand is 300 yards/min. The point is that we will probably NOT observe the lifeguard doing this because points above or below the curve yield higher time to victim. C5Read.pdf 13 In other words, this presentation graph depicts the connected minimum points of time to victim given various values of top velocity in sand.

"Connected minimum points" is a difficult, but extremely important concept. The totals graph shows two alternative lifeguards and their optimal distance on sand values; while the presentation graph shows only the optimal solution for any lifeguard with a given

configuration of exogenous variables (vs, b, c, vw).

"Why is this so important?" Because there are essentially two kinds of graphs in economicsÑunderlying graphs and presentation graphs. Understand how presentation graphs are created and read and you overcome a major hurdle to the understanding of economics.

We could also have created a presentation graph of T*=Ä(velocity in sand). It would have similar properties to the x*=Ä(velocity of sand) presentation graph. Instead of tracking

X*=Ä(vs|b, c, vw), we would plot T*=Ä(vs|b, c, vw). WeÕll examine this later.

Comparative Statics Method 2: Method of the Reduced Form

The Method of Actual Comparison works really well when you have access to Excel. The Solver is quick, once youÕre accustomed to it, and you can very easily and quickly generate a list of the different values of x* when you vary one of the parameters. (WeÕll learn an easy way to do this in the next lab using something called the Comparative Statics Wizard, which saves the values of x* and the different parameter values for you to work with later).

Another method for generating comparative statics, i.e., to answer the question, ÒWhat happens when the environment changes?Ó is to use the mathematical expression for the reduced form. LetÕs talk about the reduced form in a Q&A format.

What is a reduced form?

The reduced form is an expression for the optimal value of the endogenous variable(s) in terms of the exogenous variables.

How do I get a reduced form?

You get it by solving the optimization problem mathematically. For example, in the lifeguard problem, we took the derivative of the objective function with respect to x, set it

equal to zero, and solved for x. The equation X*=Ä(vs, b, c, vw) is the reduced form. When we substituted it into the objective function (step 5), we got a reduced form of T*=Ä(exogenous variables), which has a special name, the maximum value function.

C5Read.pdf 14 What does the reduced form tell me?

It tells you how the optimal value of an endogenous variable is related to the exogenous variables. From here, predictions are easy.

What is the advantage of the Method of the Reduced Form over the Method of Actual Comparison (with Excel's Solver)?

Once you know the reduced form, you donÕt have to re-solve the problem: the reduced form tells you everything you want to know about how the optimal value of an endogenous variable changes when an exogenous variable changes.

So I should always use the Method of the Reduced Form over the Method of Actual Comparison (with Excel's Solver)?

No. Sometimes the math to get the reduced form can be pretty cumbersome. It's pretty easy to make algebra mistakes and other errors. In fact, sometimes, it can be impossible (and so the problem is said to have "no analytical solution" or "no closed form"). In these tough cases, the Method of Actual Comparison is the only way to go.

Do the Method of the Reduced Form and the Method of Actual Comparison (with Excel's Solver) always give the same answer?

Ouch! That's a tough question! The answer is "It depends." If the reduced form is linear in the exogenous variable under consideration, then the Method of Actual Comparison (with Excel's Solver) and the Method of the Reduced Form give the exact same answer. But, f the reduced form is non-linear in the exogenous variable under consideration, then the Method of Actual Comparison (with Excel's Solver) and the Method of the Reduced Form give different answers. We will explore this in much detail.

TWO EXAMPLES OF THE METHOD OF THE REDUCED FORM:

Example 1: We saw that the reduced form for x* was b v x ∗ = c − w 2 − 2 v s v w

Suppose that you are interested in seeing how x* varies when vs changes, other things equal. If you plug in numbers for all the OTHER exogenous variables, your reduced form becomes 100∗ 100 x* =−100 2 − 2 vs 100

This tells you how x* varies for different values of vs, other things equal. You can see by looking at the expression that x* is not a linear function of vs, just as the presentation graph we drew above suggested. If you were really interested, you could take the derivative of

x* with respect to vs and find out exactly what the slope looks like and how it changes when vs changes.

Note how the presentation graph is a graph of the reduced form, x*=Ä(vs|b, c, vw).

C5Read.pdf 15 Example 2: Suppose that you were interested on how x* changed when the victim is further out from the beach (that is, when b increases)? Substituting for all of the other parameters, we obtain:

b x* =−100 283.

¥ Clearly x* is negatively related to b: the further out from the beach the victim is, the less time you should run on the sand.

¥ Furthermore, the relationship is linear: every yard further from the beach reduces optimal distance on sand by 1/2.83 = 0.35 yard.

¥ You can plug in different values for b to get the corresponding x*. For example:

When b = 100 (and vs = 300, vw = 100, c = 100), then x* = 64.64. When b = 200 (and vs = 300, vw = 100, c = 100), then x* = 29.29.

Thus, increasing b by 100 yards leads to approximately a 35 yard decrease (64.64 - 29.29 _ 35) in x*.

A presentation graph would be:

X* as a Function of b

40 x* 20

0 50 150 250 b (yards)

So, in this case, how are the Method of Actual Comparison and the Method of the Reduced Form related to each other?

In this case, they are the same because x* is linear in b. However, x* is non-linear in vs so the two methods of comparative statics would give different answers. This is a really difficult idea and we will continue discussing it in the next few chapters.

C5Read.pdf 16 SUMMARY

Optimization problems have three steps: I. Setting up the problem (identifying the objective function and the exogenous and endogenous variables) II. Solving the problem III. Generating comparative statics.

¥ The solution method, Step 2, can be either direct using totals (graphs or tables) or can rely on changes using marginals. Economists generally use marginals. ExcelÕs Solver is a yet another alternative.

¥ The Direct Method and Marginalism can be done using pencil and paper and your own math skills or by using Excel. Excel is much faster.

¥ Comparative statics, Step 3, involves changing one of the exogenous variables in the problem and observing how the optimal choice changes. You can use either the Method of Direct Comparison or the Method of the Reduced Form.

¥ If the reduced form is linear, the two methods of doing comparative statics give the same answer; if not, they don't.

The material in this reading is difficult. Please do not panic. We will be using Excel to put into practice the lessons stated here. You will have several opportunities to learn these concepts. Good luck!

Take the time to read the appendix and return to it as needed for help in understanding the role of derivatives in optimization and comparative statics.

C5Read.pdf 17 Appendix: Some Information About Derivatives

What is a derivative?

A derivative is a mathematical expression that tells you how a function changes when the endogenous variable (or one of the endogenous variables, if there are several) changes a little bit. Graphically, it is the slope of the function at that particular value of the endogenous variable. A function can change in two waysÑlinearly and non-linearly.

¥ Linear functions have a constant slope and, therefore, a constant value for the derivative. If the derivative of a function is 0.3, for example, this tells you that every time the x variable goes up by 1, the y variable goes up by 0.3. This is true regardless of whether you increase x from 1 to 2 or from 3.000 to 3.001. An example of a linear function of x is y = 0.7 + 0.3x; its derivative with respect to x is 0.3. Notice how x does not appear in the derivative. A mathematician would say, ÒIn this case, y is linear in x.Ó

y dy/dx

1.6 1.3 0.3 1.0 .7

123 x x This is a picture of a linear This is a picture of the derivative function of x of a linear function of x

¥ Non-linear functions have a changing slope and, therefore, a derivative that takes on different values at different values of x. If the derivative is increasing in x, that means that the slope is getting larger as x gets larger. If the derivative is decreasing in x, that means that the slope of the function is getting smaller as you move out along the x axis. An example of a non-linear function of x is y = x2; its derivative, dy/dx, is 2x. Notice how x appears in the expression for the derivative. A mathematician would say, ÒIn this case, y is non- linear in x.Ó

y dy/dx

123 x x

This is a picture of a This is a picture of the derivative non-linear function of x. of that non- linear function of x. Note that the slope is Note that the value of the slope is increasing as x increases. increasing as x increases.

Keep clear in your mind the difference between the function itself and the slope of the function.

C5Read.pdf 18 Some Uses of Derivatives in Economics

(1) Solving Optimization Problems

An optimization problem typically requires you to find the value of an endogenous variable (or variables) that maximizes or minimizes a particular objective function. We can use derivatives to find that value (via the Method of Marginalism using calculus) because of the following mathematical fact:

at the maximum or minimum value of functions that we will deal with in this course, the slope is zero (that is, the value of the derivative is zero). So all we need to do is find the derivative and then solve for the value of the endogenous variable that makes the derivative equal to zero.

The equation that you make when you set the first derivative equal to zero is called the first-order condition. The first-order condition (FOC) is different from the derivative because the derivative by itself is not equal to anythingÑyou can plug in any value of x and the derivative will pump out an answer that tells you whether and by how much the function is rising or falling at that point.

When you solve the first -order condition for the optimal value of the endogenous variable, the expression that you get when you is called the reduced form.

Below are some graphs that show how the derivative can be used to solve an optimization problem:

At the At the y maximum, y minimum, the slope = the slope = dy/dx = 0 dy/dx = 0

x* x x* x Above are pictures of the functions; below are pictures of their derivatives. To the left of x* To the left of x* y the derivative dy/dx the derivative is positive but is negative and falling decreasing; to the (getting less negative); right of x* to the right of x* the derivative is the derivative negative. is positive and increasing. x x x* x*

The two graphs on the left show that if you want to maximize something by choosing x, find the value of x where the derivative is zero; the two graphs on the right show that if you want to minimize something by choosing x, find the value of x where the derivative is zero.

C5Read.pdf 19 A Few Words About The Second Derivative The second derivative is simply the derivative of the first derivative. It tells you the slope of the function that describes the slope of the function itself. For example, if a function has a constant slope, we saw that its first derivative is a constant value (like 0.3 in the first example). The second derivative of the function tells you how the slope changes when x changes ... Well, since the slope is unchanging, the second derivative would be zero. Second derivatives are useful for the following reason: when you find the value of the endogenous variable that makes the first derivative equal to zero, the point that you have located could be either a maximum or a minimum (see the set of pictures above). If you want to be sure which one you have found, you can check out the second derivative. If you plug in the proposed optimal value of x into the second derivative and you find that the value of the second derivative is negative, this tells you that the slope of the function is declining when you move a little past x*. If the slope is declining after you leave x*, it must mean that x* was the top of the function, i.e., that x* gave you the maximum value of the function. If on the other hand you plug x* into the second derivative and you find that its value is positive, this tells you that if you move a little past x* the slope starts increasing. This must mean that x* gave you the minimum value of the function. So, the second-order condition for a maximum is that the value of the second derivative at x* is negative; the second-order condition for a minimum is that the value of the second derivative is positive at x*. (This is for single-variable optimization problems; the second-order conditions for multi-variable problems are more complicated.) The above is merely by way of explanation. Our primary focus is not mathematics and we will not work any problems where the second-order conditions do not hold. You should understand, however, what a second derivative is and how it is used.

(2) Exploring Comparative Statics

Once you have used derivatives to solve for the reduced form of an endogenous variable, you have an expression for the optimal or best value of the choice variable as a function solely of the exogenous (or environmental) variables. Economists are interested in how the optimal choice changes when one of those environmental variables changes. To explore this, you can take the derivative of the optimal value (x*) with respect to the parameter you are interested in. Once you get that expression for the derivative, you donÕt set it equal to zero and solve for anything since you are not trying to find the optimal value of anything Ñthat part is already done. What you are interested in is TELLING A STORY ABOUT HOW THE OPTIMAL VALUE IS RELATED TO THE PARAMETERS OF THE PROBLEM. Basically, you should look at your derivative and ask yourself two questions: ¥ Is the relationship between the exogenous variable and x* always positive, always negative, or does it depend? ¥ Is the relationship between the exogenous variable and x* linear or non-linear? If the exogenous variable does not appear in the expression for the derivative, then the relationship is LINEAR. If the exogenous variable DOES appear in the expression for the derivative, then the slope depends on the value of the exogenous variable and the relationship is NON-LINEAR.

Conclusion Economists use derivatives for (1) Solving Optimization Problems and (2) Exploring the Comparative Statics properties of a reduced form. These are two different uses of the derivative and should not be confused. In particular, the derivative of a reduced form with respect to an exogenous variable is not a second derivative. It is merely another, different first derivative of a function. This common confusion can be avoided if you understand exactly where you are in the Economic Approach.

C5Read.pdf 20 Rules for Taking Derivatives

Let x be the variable; a, n, and e are constants (e = 2.7183....) General Rule Example of its Application d (x) = 1 dx d (ax) = a d (4x) = 4 dx dx d (a) = 0 d (4) = 0 dx dx d (xa) = axa-1 d (x4) = 4x3 dx dx

2 2-1 d (f(x))a) = af(x)a-1⋅f'(x) d (2x2) = 2(2x2) ⋅ 4x = 16x3 dx dx d f(x)⋅g(x) = f(x)⋅g'(x) + g(x)⋅f'(x) d (4x2+3)⋅(x3+ 2x2) = (4x2+3)⋅(3x2+ 4x)+ (x3+ 2x2)⋅8x dx dx

f(x) g(x)⋅f'(x) - f(x)⋅g'(x) 2 (4x + 10)⋅(4x + 3) - (2x2+ 3x)⋅4 d = d 2x + 3x = dx g(x) g(x)2 dx 4x + 10 4x + 10 2 d ln(x) = 1 d 4 ln(x) = 4 dx x dx x

... and Some Other Math Reminders Laws of Exponents

Rule Example xa⋅xb =axa+b x3⋅x4 =x7 (xy)a =xa⋅ya (xy)4 =x4⋅y4 b 1/2 (xa ) =xab (x3 ) =x3/2 x0 = 1 30 = 1 b xa/b =xa x3/2 = x3

-a -2 1 x = 1 x = xa x2 Quadratic Formula -b ± b2 - 4ac If ax2 + bx + c = 0 and a≠0, then x = 2a

C5Read.pdf 21