Math Review for Econ 321

Transcription

1 Math Review for Econ 32 The Derivative and the Marginal Interpretation. The derivative of a function has an important interpretation in economics. It is the marginal change in the value of the function with respect to a marginal change in some variable. By now in your economics courses you have heard phrases like marginal utility, marginal benefit, marginal cost, marginal product of labor, etc. Those concepts are written down formally and logically in calculus through the derivative. Before explaining the derivative and the rules of derivation, lets consider a simple example from microeconomics. The following is a profit maximization problem faced by a monopolist: max Π (0 ) 2 () where Π represents profits, the price is given by 0 and thus represents a downward sloping demand curve. is the quantity that the monopolist will choose to maximize its profits and the cost per unit of produced is 2. In this problem, the monopolist has one choice,. canbechosenandistherefore referred to as an endogenous variable. Note that it is written below max in the above expression indicating it is the variable being chosen to maximize profits. Since 0, intheaboveexpression,then(0 ) is really price times quantity or the total revenue of the firm. 2 is the total cost to the firm. The difference between total revenues and total costs is profits. To find the level of that maximizes profits we take the derivative with respect to. Rewriting slightly by multiplying the through the expression for price we have: max Π (2) Now taking the derivative we get a first-order condition (3) Thederivativemustbeequaltozeroforamaximumofprofits for reasons we will explore below. Now we have one equation with one unknown,. Solve the expression for to get: (4) So the quantity that maximize profits is 4. (The indicates the optimal choice). Indeed if we graph the profit function with on the x-axis and Π on the y-axis, we can see that 4 is the maximizing choice.

2 8 Figure : Profit Function Profits Q* Nowletsgobacktothederivative: Q Quantity Marginal Cost z} { (5) {z } Marginal Benefit There are really two parts to this derivative. The first comes from the revenue portion of the profits: (0 ). That derivative is 0 2 and it represents the marginal revenue, the additional amount of revenue gained by increasing quantity by an infinitesimally small amount. The second part of the derivative is the 2 which comes from the cost portion of profits ( 2 ). That represents the marginal costs, the additional costs incurred by increasing quantity by an infinitesimally small amount. All first-order conditions will have this structure -amarginalbenefit of some kind and a marginal cost of some kind. The profit function is maximized where the marginal benefit exactly equals the marginal cost (something you should recall from Econ 20). 0 2 {z } Marginal Benefit 2 {z} Marginal Cost (6) We can also see this graphically, by graphing the marginal benefit component and the marginal cost component. 2

3 2 Figure 2: Marginal Benefit and Marginal Cost Marginal Revenue Marginal Cost Q: Quantity The downward sloping line is 0 2 and represents the extra revenues the firm gets from producing that level of. The horizontal line at 2 is the marginal cost. At any level of, the additional cost is always 2, so it is a straight flat line. Notice that they intersect precisely at 4, where the marginal benefit equals the marginal cost. To understand this better, think about quantity choices other than at 4. Suppose the monopolist chooses 2. Then the marginal revenue would be (2) 6 and the marginal cost is still 2. Thus, there are extra profits (the difference between the marginal revenue and marginal costs) that the monopolist could be getting. The monopolist is producing too little at 2tomaximize profits. What about 7? The marginal benefits are now: (7) 4. A negative here represents lost revenue and is clearly below the costs at 2. Thus, producing too large a quantity leads to lost profits as well. Compare these results for other levels of to the graph in Figure. Clearly 2or 7does not yield the most profits. The rule is that the function is maximized at the point where the marginal benefits equal the marginal costs. To finish off the example, now that we know the profit maximizing quantity, we can figure out profits: Π 0(4) (4) 2 2(4)6 We can see it visually in figure, but we can also derive it from figure 2. The profits are the difference or the area between the marginal revenue curve and the marginal cost curve. In this case, to figure out the area we have a nice 3

4 simple triangle. The base has length of 4 and the height is 0-2 or 8. So the area is 2 (4)(8) 6.. Practice Problems: Try the following on your own. Find the profit maximizing quantity and graph the marginal revenue and marginal cost.. max Π (20 ) 4 2. max Π (2 ) Formal Definition of the Derivative The derivative of a function tells us the slope of the function at any point. Consider the following graph of an arbitrary function, ( ) Suppose we want to know the slope at the point 0 where the function has the value ( 0). The true slope is drawn in the figure and is a line tangent to the actual curve at that point. The slope represents how the function itself changes, whether it is increasing or decreasing and how fast it is changing. Figure 3: Finding the Slope Second Approximation True Slope y First Approximation f(x0) x0 x0+h2 x x0+h Now, we can get an approximation of the slope by picking some other point away from 0 and then using the point-slope formula from basic algebra. Suppose we add to 0 to get 0+. is just some arbitrary number. Now 4

5 the value of the function at 0+ is ( 0+ ). To find the slope we merely take the ratio of the change in to the change in by comparing the new point to the original point: Slope ( 0+ ) ( 0) 0+ 0 ( 0+ ) ( 0) Visually, looking at the graph is pretty large and thus the slope from the formula above will give us the line labelled "First approximation." Its not a very good estimate. The true slope is much steeper. We can do better by choosing a point closer to our original 0. Let our new try come from adding 2 to 0 to get will be smaller than so that we are closer to the original point. Now the value of the function at 0+ 2 is ( 0+ 2). Again, to findtheslopewetaketheratioofthechangein to the change in : Slope ( 0+ 2) ( 0) ( 0+ 2) ( 0) 2 Looking back at the graph, we would get the line labelled "Second Approximation" which is better than the first,butstillnotaccurate. Nowimagine choosing points closer and closer to 0 such that our value gets smaller and smaller eventually approaching zero. If we look at the slope formula for any level of we have: ( 0+ ) ( 0) Slope which presents a problem because if in the denominator really does go to zero the fraction is undefined. Thus, to get the slope we want we need to take the limit as 0. ( 0+ ) ( 0) Slope lim 0 and that is the formal definition of the derivative. Its really nothing different from the slope formula you used in basic algebra except that the change in is infinitesimally small and effectively zero. It would, however, be a major pain if we had to take the limit of that function every single time we wanted to find the derivative (i.e. find the slope of a function). Fortunately, there are some fairly straightforward rules we can use to find the derivative. Before going on to the rules, one point about notation. There are a number of ways to represent the derivative. If we have a function: ( ) the derivative of with respect to can be written as: or 0 ( ) or ( ) or. They represent the same thing effectively, the derivative. 5

6 3 Rules of Differentiation The following provides all the key rules for taking derivatives. I do not provide proofs for any of these but jump right to what is useful. 3. Constant Rule The derivative of a constant term (i.e. just a number) is zero. we consider some function ( ) where 5,then For example if 0 because does not enter the function at all and therefore does not change. 3.2 Power Rule The derivative of with respect to in a function of the form: is Multiply by the exponent and subtract one from the exponent itself. function is linear (meaning its a straight line) such as: If the 5 then the exponent on is one and the derivative is: 5 since the implied in the exponent after subtracting becomes 0 and anything raised to the power of 0 is one. More generally if the term is multiplied by any constant, wehave: 6

7 Some examples: where is a parameter where and are parameters In the last two examples, when taking the derivative with respect to, wetreat the parameters (numbers represented by letters or Greek letters) as regular numbers. 3.3 Sum-Difference Rule When a function has two components that are additively separable (or subtracted) take the derivatives of each part separately. An additively separable function has the following form: ( )+ ( ) so, to find the derivative with respect to, first take the derivative of ( ) and add the derivative of ( ). In the following example ( ) 2 2 and ( ) The derivative of ( ) is 4 and the derivative of ( ) is 5 2. If the components are linked by a minus sign, the minus sign stays. For example: If the exponent is negative make sure the sign in front changes appropriately. Examples:

8 If the function contains an added constant term, then the constant rule (above) applies to that part of the function. Examples: Product Rule Sometimes the function is complicated and involves two subfunctions that multiply each other. For example, ( +2) 2 +3 where we think of ( ) +2 as the component in the first set of parentheses and ( ) as the component in the second set of parentheses. Now we could algebraically multiply these out to get andthenapplythe sum/difference rules discussed above. However, in many cases we will not be able to do that. In addition, the power rule is often faster and involves fewer steps thereby reducing the chance of making a mistake. For a general function of the form: ( ) ( ) the derivative is: 0 ( ) ( )+ ( ) 0 ( ) Note that here I have used the prime (0) symbol to indicate the derivative of the subfunctions with respect to In words, take the derivative of the ( ) portion and multiply by ( ), thenadd ( ) times the derivative of ( ). Returning to the example above: ( +2) 2 +3 ()( 2 +3)+( +2)(2 ) You should verify that this is the same if you took the derivative directly on Some more examples: (3 + 3)(4 6) Here let ( ) 3 +3and ( ) 4 6 ( 3 2 )(4 6) + (3 + 3)(4 ) 8

9 Simplifying algebraically: ( ) More examples: (0 ) 2 ( ) 2 +(0 ) Quotient Rule 2 3 (6 5 ) ( 4 ) When a function is a fraction or a ratio of two subfunctions: ( ) ( ) it gets a bit messier. The derivative is ( ) 0 ( ) ( ) 0 ( ) [ ( )] 2 Inwords,thederivativehasthesquareofthe ( ) function in the denominator. Inthenumerator,wehavethe function times the derivative of the function minus the function times the derivative of the function. Example: (3) (3 +2) 3 2 ( 3 +) 2 Simplifying using algebra: ( 3 +) ( 3 +) 2 9

10 More examples: + ( + )( ) ( ) ( + ) 2 + ( + ) 2 2 ( + ) ( + 2 )(2 ) ( + 2 ) ( + 2 ) ( + 2 ) (2 ) µ Chain Rule This rule is probably the hardest and the one most frequently forgotten accidentally when trying to take derivatives. The chain rule applies when you can write one subfunction inside another subfunction. Sounds odd, but the general form looks like this: ( ( )) 0 ( ( )) 0 ( ) 0

11 In words, it says take the derivative of the outer function first and then multiply by the derivative of the inner function. This (hopefully) will become clearer with some examples. Consider the following: The inner function here is ( ) We know how to take that derivative from the basic rules above. To describe the outer function let 2 2 +, then if we rewrite the function we have: 3 and we know how to take the derivative of with respect to Applying the chain rule, we would first take the derivative of 3 then multiply that by the derivative of as follows: 3 2 (4 +) replacing the with its original expression in terms of we have: (4 +) More examples of Chain Rule: Let (2 +) (2 +) where I skipped the step letting equal the inner subfunction 2 + and took the derivative of the outer function directly. (2 +) 2 2

12 In the above example, we will need to use product, quotient and chain rules. Since the entire function is one ratio, we start with the quotient rule, but when we take the derivative of the numerator we will need to apply the product rule becausewehave times (2 +) 2, and then when we take the derivative of (2 +) 2 we need to apply the chain rule. Quotient Rule z } { Product Rule z } { Chain Rule 2 z } { ()(2 +) 2 +( ) (2 +) 2 2 (2 +) 2 (2 ) ( 2 ) 2 2 (2 +) 2 +( ) 2 (2 +) 2 (2 +) 2 (2 ) ( 2 ) 2 The following table summarizes the key rules so far: Rule Function Derivative Constant 0 Linear Power Sum-Difference Product Quotient Chain ( )+ ( ) ( ) ( ) ( ) ( ) ( ( )) 0 ( )+ 0 ( ) 0 ( ) ( )+ ( ) 0 ( ) ( ) 0 ( ) ( ) 0 ( ) [ ( )] 2 0 ( ( )) 0 ( ) 3.7 Practice Problems: Find the derivatives of with respect to for the following

13 (2 ) 5 2. ( 2 9) 2 3 ³ ( ) + 4 Natural logs and the Exponential Function 4. Description Many economic models involve assumptions or results about growth or changes over time in economic variables of interest. This feature is particularly true for macroeconomics. The natural logarithm and the exponential functions are highly convenient for describing, for example, the growth rate of an economy, therateofinflation over the long-run, or the expected increase in the value of an asset. In general, if we have a function of the form: we say that is the logarithm of in base where is some number greater than. Interpreting, we are saying that is the power one needs to raise to, to get the value. The choice of base depends on the situation. Engineering and the sciences will frequently use 2or 0. In economics, we usually stick with a very special base called the natural log, denoted by. is approximately equal to 2.78 and is an irrational number like Π. It happens to be the base that satisfies the following property. If the function is: 3

14 then the derivative is: The derivative is the same as the function!! That makes taking the derivative easy. Moreover what is says is that the slope of the function at any point is the value of the function itself. Below is a graph of : 00 Figure 4: yexp(x) y Note that the function is increasing, and for reasons we shall see below, the percentage increase is constant. Sometimes the exponential function is written as follows: exp( ) instead of as: but they mean exactly the same thing. Again, in words if we raise to the power we get. However, if we have the following function: ln( ) that is effectively the reverse. If we raise to the power of we would get. Below is a graph of the natural log: x 4

15 3 Figure 5: yln(x) 2 y X The natural log also has an easy derivative to work with: ln( ) Note that the derivative could be written as. 4.2 Properties The following are properties and derivatives of the natural log and exponential function you should be comfortable with using. 5

16 ln( ) ln() 0 ln ln() ln +ln ln( )ln ln ln ln ln ln[ ( )] ( ) 0 ( ) ( ) 0 ( ) ( ) Note that in the derivatives, when the natural log contains a function of ( ) we need to apply chain rule where ln is the outer function and ( ) is the inner function. For example: and: ln(3 ) 3 (3) ln(3 2 + ) (6 +) When is raised to a power that is a function of we again need to apply chain rule where exp is the outer function and ( ) is the inner function. For example: and 2 2 (2) (6 +) 6

17 A simple example may help illustrate their usefulness. Consider a bottle of wine which you can purchase today for dollars. The value of the bottle of wine is expected to grow at a constant rate of 3% per year. How much will that bottle of wine be worth in 5 years? The exponential function provides a way to findout. Letthebottleofwineatsometime be ( ), where is the length of time from the purchase of the bottle of wine. Let (0) be the initial value of the bottle of wine (the dollars you need to spend on it.) Then: ( ) (0) 0 03 That formula gives the value of the bottle of wine after years if it grows at a rate of 3%. Suppose the initial value is $00, how much is it worth after 5years? ( ) (00) 0 03(5) Plugging that into your calculator (make sure you know how to do that) we get: ( ) (00) 0 03(5) $6 8 Note that the exponential function works for continuous compounding - where interest is added at all points in time. If we want to calculate growth with discrete compounding we use a different formula: ( ) (0)( + ) where is the growth rate and is again the length of time. The difference here is that "interest" is only added at the end of the period. For most calculations the two formulas give very similar results unless the numbers are quite large. Under the continuous compounding formula using the exponential function, the value will be larger. To illustrate, suppose you are choosing between two banks where you will place your savings of $,000. Bank A offers an interest rate of 2% and continuous compounding - meaning interest is added to your account every fraction of every second. Bank B offers an interest rate of 2.0% with annual compounding - meaning you get your interest added to your account at the end of each year. If you plan on holding your savings in the account for 3 years, which bank has the better offer? If you choose bank A, the value of your savings account after 3 years would be: At bank B you would have: (3) ($000) 0 02(3) $ (3) ($000)( ) 3 $ So bank A pays slightly more, even though the interest rate offered by bank B is slightly higher. The reason is that bank A is adding interest throughout the entire year and thus you are also earning interest on the interest. With Bank B, you do not get the interest payment until the end of the year, so up until then you would not earn interest on the interest. 7

18 Reconsider the growth rate formula again, but lets think of it as representing the growth of an entire economy where ( ) is GDP per capita and the average growth rate of the economy is. Then to find GDP per capita after years we would write: ( ) (0) So if our fictitious economy started with a GDP per capita of $5,000 and grew at an average rate of 2.5%, then after 30 years the GDP per capita would be: ( ) ($5 000) (0 025)(30) $0 585 Referring back to the general growth formula above, ( ) (0),takethe derivative of with respect to time. What do we get? ( ) (0) ( ) (0) ( ) ( ) Hm, can we make any sense of that? Well, is the change in GDP per capita with time. If we divided by ( ), we have an expression very, very close to the basic growth rate formula. That is, when you calculate the growth rate of GDP between two years you use the following: growth rate which gives you the percentage change between year and year. Well, ( ) is effectively equivalent to,howmuchdid change over time - but for a marginally small amount of time. Thus if we divide by ( ) we get a percentage back, i.e. the growth rate of ( ). ( ) (0) ( ) ( ) (0) ( ) ( ) ( ) But since ( ) (0) we can replace the ( ) in the denominator on the right-hand side and then almost everything cancels out: ( ) ( ) (0) ( ) (0) which just says that the growth rate of the economy is. Another way to see this relationship is by taking the natural log of both sides of the growth equation: ( ) (0) ln( ( )) ln (0) ln( ( )) ln( (0)) + ln[ ] ln( ( )) ln( (0)) + ln( ( )) ln( (0)) 8

19 The growth rate is the difference in the natural logs divided by the length of time. If we consider the length of time to be one (e.g. one year, one month), then the growth rate is just the difference in the natural logs or effectively the slope of the natural log of. Consider the 2nd to last line above: ln( ( )) ln( (0)) +. How does ln( ( )) change with? taking a derivative note that the right-hand side is a simple linear function. ln( ( )) ln( (0)) + ln( ( )) and this is perhaps easier to see graphically. The following figure shows the following function: ( ) (0) where the values are on the left-side y-axis and (0) is 00 and is 5% (or 0.05). The straight dashed line is the graph of ln( ( )) ln( (0)) + for the same numbers but uses the axis on the right-side (so that both can be seen more easily). 400 Figure 6: Growth in Levels and Natural Logs y(t) ln(y(t)) y(t) Level of y 200 Scale on Primary Axis 6 ln(y(t)) in natural logs Scale on Secondary axis 5 50 Slope is exactly g Time t 9

20 4.3 Practice Problems Find the derivatives of with respect to for the following ln(7 5 ) 6. ln( ) 7. 5ln( +) ln 9. Suppose that the value of a bottle of wine after yearsisgivenby: ( ) 00 2( ) Compute that value of the bottle of wine after 2 years, 5 years, and 0 years. 0. From the wine bottle problem in #9, take the derivative of with respect to time,. What is the growth rate of the value?. Take the natural lns of the value equation in problem #9. Graph the ln of the value against time. 2. Repeat #9-# for the following: 5 Partial Differentiation ( ) 00 2 The examples above going through the derivatives were all for single-variable calculus. Frequently though we are working with multi-variate calculus. Fortunately, if you can do single-variable calculus then doing multivariate calculus is a pretty simple step. When we say single-variable calculus we mean a function like the following: where is a function of just one other variable. The derivative of with respect to is: 6 20

21 However,supposewereplacethe8 above with, which is also a variable: Now we are in the world of multi-variate calculus. changes with both and. So how do we take derivatives? Pretty much that same way we did before. What we do, however, is take a "partial" derivative. That is, if we want to know how changes with, then we don t care about and we treat it just like any other number. The only immediate difference is that instead of writing we write where the script sign indicates a partial derivative. For the example above we have: 6 which means the following: The change in with respect to is 6 OR the slope of along the x-axis is 6, holding constant. We can also take the derivative with respect to to get: where the term 3 2 is treated as just a constant and therefore its derivative with respect to is zero. Consider the following example: Taking derivatives with respect to and we would get: The interpretation of each is as follows: shows how changes with holding constant, i.e. it is the slope of the function along the x-axis. shows how changes with holding constant, i.e. it is the slope of the function along the z-axis. Here s a standard application in economics. Suppose that the total output (GDP) of an economy is given by the following production function: 2 2 where is total GDP, is productivity (or technology), is the capital stock (the stock of machinery, factories, equipment, etc.) and isthesizeofthelabor force. The production function tells us how much output ( )wegetforthe inputs ( and ) that we have and the level of productivity (the efficiency with which the economy transforms inputs into outputs). In this problem we have output depending on 3 variables:,, and. Soifwewanttoknowhow 2

22 output changes with capital, we take the partial derivative of with respect to, treating and as constant numbers. We get: In words, shows the change in output (GDP) with a change in capital stock holding productivity and labor constant. That is more commonly referred to as the marginal product of capital. If we take the derivative with respect to labor we would have: which is the marginal product of labor, i.e. the additional amount of output we get with an additional amount of labor. Now the term partial again indicates holding all other variables constant. There is another way to take the derivative where we let all the variables change andthisiscalledthetotal derivative. Taking the total derivative follows the same rules for differentiation, but we now need to go through the function and take the partial derivative for each variable separately and then add these together. Return to our first example: The total derivative would be: where the 6 is the partial from taking the derivative with respect to and is the partial from taking the derivative with respect to (as above). However, notice that the left-hand side is and not a ratio like. represents the change in but its now dependent on changes in two variables: and. The changes in those variables are represented by and, respectively. The total change in is given by the slope of the function along the x-axis (6 ) timesthe change in,, plus the slope of the function along the z-axis () times the changes in,. Notice that, if we hold constant that means doesn t change and 0. Our total derivative would become: and we can divide by to get: (0) 6 6 which is the same as the partial derivative of with respect to. 22

23 Consider the production function example above and suppose we want the total derivative to see how changes with, and simultaneously. Taking the total derivative we have: 2 2 ³ µ µ In words, the total change in GDP is given by the change in productivity () times the marginal product of plus the marginal product of capital times the change in capital plus the marginal product of labor times the change in labor. 5. Practice Problems: Find the partial and total derivatives of the following functions: ln( ) 4. ln 5. ( is a parameter and you do not need to take derivatives on it). 6. ( + ),(,, and are parameters and you do not need to take derivatives on them). 6 Implicit Function Theorem In all the derivatives we have been taking so far, we have been using explicit functions. By explicit we mean that the variable is isolated on the left-hand side entirely by itself and the also does not appear on the right-hand side. That is, has been isolated and expressed in terms of. For example, is an explicit function. However, is not explicit because is not isolated. You could, of course, merely add to both sides of the equation and make it explicit. In some cases that is not possible such as the following:

24 Such functions are called because there still exists a relationship between and even though we can t isolate the variable. With an implicit function, we can still find,howthe variable changes with, i.e. the slope. It turns out there is a pretty simple rule for finding this derivative call the implicit function rule. To apply the implicit function rule there are two steps: First write the equation such that it is equal to zero. In the above example, we would write it as: For notational convenience, we shall call everything on the left-hand side ( ). So, the first step is really finding ( ) 0. The second step involves finding the derivatives of ( ) with respect to both and. The derivative of with respect to is: ( ) ( ) (7) In words, the derivative of with respect to is the negative of the ratio of the derivative of ( ) with respect to to the derivative of ( ) with respect to. Let s do a few examples. Suppose we have: which is an explicit function and from our basic rules we know how to do this: 6 Suppose we write it as an implicit function: If we apply the implicit function rule, we better get the same result. ( ) ( ) 6 6 (8) where the numerator, 6, isthederivativeof with respect to and the denominator,, isthederivativeof with respect to. Now let s try this on the more complicated implicit function we had from above: which, as a reminder, we must rewrite such that it is equal to zero: Applying the implicit function rule we have: ( ) 4 +2 ( ) (9) 24

25 The interpretation here is exactly the same as before, the derivative shows us the marginal change in that occurs because of a marginal change in. In some ways this is even easier to intuitively understand with an implicit function. Again reconsider the original equation when set equal to zero: Suppose we increase by a small amount. How much must change in order to keep the above equation true and equal to zero? By. There is an alternative, though a bit longer, way to findthederivativeofan implicit function. We can use total derivatives. I ll do that here on the example above to illustrate. Starting from: we do not have to arrange it such that it equals zero. We could, but with total derivatives it does not matter. Taking the total derivative with respect to both and we get: Remember we are looking for the ratio. We can get there from the above by simply using algebra and solving. First, arrange the equation such that all the items are on the left-hand side and all the items are on the right-hand side Factor out the and : ( 2 +4 ) Now divide both sides by and by the factor multiplying, : Same answer as before. 6. Practice Problems (0) Find from the following using the implicit function rule: () (2) (3) 25

26 (4) (5) 7 Second Derivatives, Increasing, Decreasing, Concave, Convex functions When we take a derivative, it reveals important information about a function which may not be obvious from looking at the function directly. The sign of the derivative tells us whether the function is increasing or decreasing. For example, the function for any value of other than zero. That means the slope goes upward which also means the function is increasing, larger values of give a larger values of. If the derivative is negative the slope is downward and the function is decreasing as in the following example: In the monopolist problem at the beginning of this review, we had the following: and the derivative we got was. max Π (6) Π (7) That derivative could be positive or negative depending on the value of. If 4, then the derivative is positive and it means the function is increasing over the interval [0 4). However, if is greater than 4, the derivative is negative and the function is decreasing. Compare that with Figure. At exactly 4, thederivativeiszeroandhencetheslopeiszero. Similarly consider the production function from above and the derivative with respect to :

27 logically,,, and are all positive values. Thus, the entire derivative is always positive. That means that output,, is always rising with. Beyond the first derivative, the second derivative also carries important information about the shape of the function. It tells us how the slope itself is changing over the function. Specifically when the second derivative is negative, it means the slope is decreasing. We call functions of that type concave. If the second derivative is positive, the slope is increasing and the function is convex. Figures 7 and 8 below show concave and convex functions. 5 Figure 7: Convex and Concave Functions Convex concave

28 7 Figure 8: Convex and Concave Functions 6 Convex 5 concave In both diagrams, notice that for the convex functions (solid line), the slope is always increasing, whereas for the concave functions (dashed line) the slope is always decreasing. From the monopolist problem the derivative we got was: Π Takingthesecondderivativetellsuswhetherwehaveaconcavefunctionora convex function. 2 Π 2 2 whichisalways-2andthereforewehaveaconcavefunction(seefigure). Similarly consider the production function from above and the derivative with respect to : is always negative and therefore concave. 28

29 From our earlier examples, we take the second derivatives and they will tell us whether the functions are concave or convex which is positive whenever 0. Thus the function is convex for positive values of, but concave for negative values of Has the opposite result. The function is negative whenever 0 and therefore is concave, but is convex when 0. The Figure below shows both 6 3 and 6 3. Note the ranges that are convex and concave. Figure 9: Functions with both Concave and Concave Regions Convex Convex y 6x^3 y 6x^ Concave Concave Concavity has a straightforward economics interpretation - diminishing marginal returns. In the context of the production function example, it means that capital exhibits diminishing marginal returns. In other words, output rises 29

30 with more capital (as we found from the first derivative) but the amount of the increase gets smaller and smaller with more capital. Convexity is frequently associated with cost functions. Consider the following cost function: The second derivative is always 6 and therefore always positive. Thus, the marginal costs (which is the first derivative 6 4) are always increasing in quantity. Try graphing that cost function and convince yourself that it is indeed convex. Whether a function is convex or concave is really important for the behavior of economic variables. To see this, let s look at a simplified version of the production function where we eliminate and focus only on : Now, output is given by productivity,, which must be positive, and capital,. The exponent is the parameter. The value of determines the shape of the function and how output responds to adding physical capital. Take the first derivative of with respect to and we get: Since we expect capital to increase output, it makes sense that 0. That will mean that the first derivative is always positive and we have a positive marginal product of capital. However, the function could still be concave or convex. Taking the second derivative we have: 2 ( ) 2 2 where the sign depends on whether is greater than, less than, or equal to. The following graph shows the difference: 30

31 4 Figure 0: Convex, Linear, and Concave Production Functions Y a2 a a/ K (Capital) The curve labelled (for alpha) 2, is convex. The slope is increasing. In this case, it says that we are getting increasing returns for each additional unit of capital. The straight line ( ) shows constant returns, each additional unit of capital yields the same increase in output. The final line ( 2) is concave and shows diminishing returns to capital. 7. Practice Problems: Find the second derivatives of the following and figure out whether they are concave or convex functions (or have ranges of both): ln 4. 4 For the following production functions are they concave or convex and what does that depend on? If the answer is possibly both, over what ranges are the functions convex or concave? ( + ) 3

32 Optimization We now get to the primary use of calculus in economics and that is optimization. That is, maximizing or minimizing some objective. In your principles classes, you certainly have come across this notion, though probably without the calculus representation. Consumers maximize utility, firms maximize profits, or firms minimize costs, governments try to minimize dead weight losses, or the central bank may try to minimize the costs of inflation and unemployment. In many cases, our economic agents are trying to optimize some objective with a constraint of some form. Consumers have a budget constraint, firms have a technological constraint, governments are constrained by the resources available. In essence constrained optimization is the formal expression of the basic definition of economics: How do people makes decisions with unlimited wants when resources are scarce? To begin, let us return to the monopolist example at the very beginning. max Π (0 ) 2 where Π represents profits, the price is given by 0 and thus represents a downward sloping demand curve. is the quantity that the monopolist will choose to maximize its profits and the cost per unit of produced is 2. In this problem, the monopolist has one choice. canbechosenandistherefore referred to as an endogenous variable. Note that it is written below max in the above expression indicating it is the variable being chosen to maximize profits. Since 0, intheaboveexpressionthen(0 ) is really price times quantity or the total revenue of the firm. 2 is the total cost to the firm. The difference between total revenues and total costs is profits. To find the level of that maximizes profits we take the derivative with respect to. Rewriting slightly by multiplying the through the expression for price we have: max Π Now taking the derivative we get: Π Recall Figure. The profits reach a maximum at 4. Notice that at exactly that point, if we draw a tangent line to the function the tangent line will be a flat horizontal line with slope equal to zero. Therefore, the derivative must be equal to zero for a maximum of profits. That is, at the maximum, the slope is positive to the left, but negative to the right (which is another way of saying 32

33 the function is concave - the slope is continuously getting smaller). Therefore we set the derivative equal to zero: Π andthistypeofexpressionisreferredtoasafirst-order condition (FOC) for an optimal solution. The first-order condition requires that the derivative of the function be equal to zero - for a minimum or amaximum. Thus,the condition is necessary for a maximum or a minimum - but it is not sufficient. The solution to that equation, from before, was 4. It is possible, if we do not know what the function looks like, that 4represents a maximum or it could be a minimum. The way we check is to obtain another condition that is sufficient to tell us whether its a maximum or minimum. This condition is where the second derivative comes into play. We look at the second derivative to determine whether the function is concave (which ensures a maximum) or convex (which ensures a minimum). When checking an optimization problem werefertothisasthesecond-order condition. The second derivative is: 2 Π 2 2 which is negative. That means the function is concave and the point we found must be a maximum. Consider the cost function: min where a firm is trying to minimizes its costs here which we can set equal to zero as the first-order condition. That means the slope of the function will be zero at the level of that satisfies this equation. Solving we get 4. Still,wecannotbesurewehaveaminimumora maximum until we check the second-order condition. For a minimum we need a positive second derivative. We get: which is clearly positive. Therefore, the function is convex and we found the minimum point. First-order necessary condition Second-order sufficient condition Maximum 0 2 0, Concave 2 Minimum 0 2 0, Convex 2 33

34 Look back at Figure 9 for the functions 6 3 and 6 3. Notice also, that for both functions, the slope is equal to zero at 0. Thus, the first-order condition for a minimum or maximum would be satisfied. However, the second-derivative at 0is also equal to zero, so the sufficient condition for an extreme point is not satisfied. In fact, we have an inflection point - a point where the function goes from convex to concave (or vice versa). (There are further tests for extreme points using higher order derivatives when the second derivative is zero, but we shall not go into those here). 8. Practice Problems: Be sure to check your second- Maximize or minimize the following functions. order conditions.. max Π (20 ) 4 2. max Π (2 ) max Π max Π (where and are exogenous parameters and is the wage rate on labor, ) 9 Optimization with Constraints As mentioned above, our models frequently involve maximizing or minimizing afunctionbutsubjecttoaconstraint. Therearetwomethodsfordealingwith a constraint when optimizing a function: ) Substitute the constraint into the objective function (the function you are maximizing or minimizing); or 2) Use a Lagrangian. We will start with the former. 9. Substituting the Constraint into the Objective Function To explain constrained optimization, we will use a simple example from microeconomics. Consider a single consumer, Jane, who gets utility,, (happiness) from consuming two goods: (cars) and (pizza). The utility of this consumer is given by the following: max The consumer wants to be as happy as possible, i.e. get the highest value of possible. Right now there is no constraint. So if Jane could choose the amount of cars and pizza to consume, she would simply choose an infinite amount of both cars and pizza and that would give her an infinite level of utility. No calculus needed. However, suppose our consumer has a budget of 32 dollars 34

35 and cars cost $4 and pizza costs $2 (its just a silly example, don t worry about the unrealistic prices). Thus, the constraint is: which simply states that the amount spent on cars, 4, plus the amount spent on pizza, 2, must be equal to the income level our consumer has available, 32. Now the problem faced by our consumer would be written as: max which reads - the consumer maximizes utility by choosing the amounts and, subject to () the constraint. Now Jane can t choose an infinite amount of cars and pizza because she has to live within her budget. But how do we solve this problem for the utility-maximizing choices of and? If we use the budget constraint and solve it algebraically for either or, we can put it into the utility function. Suppose we solve it for. We get the following: which says that the amount of Jane will buy is equal to 6 minus 2 times the amount of she buys. With that we can replace the in the objective function: max 2 +(6 2 ) 2. The constraint has been accounted for here because as increases, will decrease. Moreover, notice that our original problem was a multi-variate problem - we would have needed two derivatives: one for and another for. Now our problem has been reduced to choosing just and has become a single-variable problem. To maximize utility we know that the derivative with respect to must be equal to zero, that is the necessary first-order condition. Taking the derivative we have: (6 2 ) 2 ( 2) 0 Simplifying a bit: 2 2 (6 2 ) 2 0 Recall from the very start of this handout that each first order condition has two parts: a marginal benefit and a marginal cost. In the above example that decomposition would be: Marginal Benefit z } { Marginal Cost z } { (6 2 ) 2 0

36 The marginal benefit portion is 2 2, which is the marginal utility of. That is, it is the marginal increase in utility that comes from an infinitesimally small increase in consuming more (cars). The marginal cost portion, notice it has a negative sign on it, represents a reduction in utility. (6 2 ) 2 shows how utility decreases by choosing more, because consuming more means that Jane must consume less (pizza). To "solve" the problem and find the optimal choices of and, wesolvethe first-order condition for algebraically: 2 2 (6 2 ) (6 2 ) (6 2 ) Since the optimal choice for is 8/3, we can now use the budget constraint to find the optimal choice of : 6 2 µ Just to be sure that the solution ( )( ) really is a maximum, we should check the second-order condition of our function. Recall that for a maximum we need a negative second-derivative so that the function is concave. 2 2 (6 2 ) (6 2 ) 3 2 ( 2) (6 2 ) 3 2 For any values of or that are positive, the second derivative is always negative and thus the sufficient condition for a maximum is satisfied. 9.2 Lagrangian multiplier The second method for solving a constrained optimization problem is to set up a Lagrangian. In a Lagrangian we add the constraint to the objective function, 36

37 but multiply the constraint by a new variable called the Lagrangian multiplier, commonly represented by the lower-case Greek letter lambda,. The workings of a Lagrangian are probably easiest to understand through an example and we will use the same example from above with Jane choosing cars and pizza. Here is the basic problem again: max Writing this problem as a Lagrangian is as follows: L Objective Function z } { Constraint z } { + [ ] The script letter, L, simply indicates that this representation is a Lagrangian. The Lagrangian is equal to the objective function (the function we are trying to maximize or minimize) plus times the constraint. Notice that the constraint has been rearranged such that the constraint is equal to zero. You must always do this with the constraint. It means that we are adding effectivelyzerotothe objective function and therefore not changing its value. The multiplier,, has a particularly useful interpretation and will be discussed in more detail below. Notice that we are maximizing this utility function ( )choosing two variables, and. Therefore we will need first order conditions for both of them. Taking the derivatives of the Lagrangian with respect to and respectively we have: L L Marginal Benefit z } { 2 2 Marginal Benefit z } { 2 2 Marginal Cost z} { (4) 0 Marginal Cost z} { (2) 0 Both first-order conditions must be equal to zero for a maximum. Technically here, we have 2 equations and 3 unknowns:,, and. Therefore we will need a third equation. However, the third equation just comes from taking the derivative of L with respect to. Doing so merely reproduces the constraint: L Since it reproduces the constraint, which we had from before, this step is often omitted, but we do need the constraint to solve the problem. OK. Look at the derivatives L L and. They, too, have the marginal benefit and marginal cost components. In the first, the marginal benefit of consuming a little bit more of, is that is, the marginal utility of (cars). Similarly in the second equation, 2 2 is the marginal utility of 37

38 pizza. The marginal cost components, however, are expressed in terms of the multiplier and the price of the good. If the price of one of the goods increases, it means the marginal cost will increase. Before we explore the meaning of the, lets solve this problem out using these equations. Rewrite the two first-order conditions such that we have marginal benefits equals marginal costs as follows: 2 2 (4) (2) 0 become: Divide the first condition by the second condition: Using algebra: Notice that the lambda divides out and then solving for in terms of we get: Now we need to use the third condition (the budget constraint) to solve. Replace in the budget constraint with 4 and we have: (4 ) Same as what we had above. Now that we have we can find. We found above that 4, so 4(8 3) 32 3 Same answer again. 38

39 What about lambda? Well, we could use either of the first-order conditions to get a value for it. Let us use the condition on. 2 2 (4) We now know that 8 3, so 8(8 3) 2 6(2 3) 2 That does not look very intuitive. However, the interpretation is the marginal utility of wealth which means the marginal increase our agent would get in utility from an infinitesimally small increase in her budget. Right now she has $32 dollars to spend. The multiplier tells us how fast utility will rise if we increase her wealth by a very small amount. A more mathematical way of saying the same thing would be: how much does the objective function increase when we relax the constraint just slightly. We shall do two more examples that will better illustrate the meaning of. Consider a more general form of the consumer problem choosing between two goods and where the utility function is: and the budget constraint is: ln +ln + where is wealth (or income), is the price of good and is the price of good. Thus we have: max ln +ln + In Lagrangian form the problem becomes: L The first order conditions are: Objective Function z } { ln +ln Constraint z } { + [ ] L L Marginal Benefit z} { Marginal Benefit z} { Marginal Cost z} { 0 Marginal Cost z} { 0 39

40 Combining the two equations and solving for in terms of we have: Substituting into the budget constraint: + µ Then is: + µ We can use either of the first-order conditions to find a value for. (2 ) 2 So the marginal utility of wealth is 2. To see this more clearly, lets go back to the original utility function: ln +ln 40

41 and replace the and with our solutions 2 and 2 : µ µ ln +ln 2 2 The marginal utility of wealth,, can be found directly from this expression by taking the derivative of with respect to : µ µ µ + 2 µ Thus: 2 which means that the added utility our consumer would get from an infinitesimally small increase in (a relaxing of the budget constraint) is 2 or, the Lagrangian multiplier. 9.3 Two-Period Consumer Model One long last example, more closely related to macroeconomics and consumer choice between consumption and savings. We will use this model extensively throughout the course. Consider a single consumer who lives for 2 periods and is choosing how much to spend on consumption in the first period,, and how much to spend on consumption in the second period, 2, and how much to save,. The consumer has the following utility function over consumption today and in the future: ln + ln 2 is a parameter and takes a value between zero and one, 0. is called the subjective discount factor and represents the empirical observation that people place more weight on consumption today than on consumption in the future (due to risk, uncertainty, and impatience among other things). The consumer receives income in the first period and income 2 in the second period. The consumer can also borrow or save at the real interest rate. The consumer has a budget constraint in each period. The first period budget constraint is: + which simply says the consumer uses first period income to consume and to save. Note that can be negative which means the consumer is borrowing against future income. In the second period, the budget constraint is: 2 +(+ ) 2 4