Mathematics 2 for Business Schools Topic 7: Application of Integration to Economics. Building Competence. Crossing Borders.

Transcription

1 Mathematics 2 for Business Schools Topic 7: Application of Integration to Economics Building Competence. Crossing Borders. Spring Semester 2017

2 Learning objectives After finishing this section you should be able to interpret the integration of marginal functions such as marginal revenue, marginal costs, and marginal profit. calculate the consumer and the producer surplus using integration. explain the definition of an ordinary differential equation. solve ordinary differential equations for exponential growth functions with predefined elasticity using the method of the separation of the variables and integration. 2

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5 Integration of the marginal revenue function For a given marginal revenue function E x, there is exact one revenue function E x, which can be found using the fundamental theorem as follows: and consequently 0 x E t dt = E x E(0) E x = 0 x E t dt + E(0) = 0 Revenue is the product of price and quantity. The revenue at quantity 0 must therefore be 0. Example: E x = x E x = 3

6 Revenue E x and the area under the marginal revenue function E(x) = 34x 10 3 x2 E (x) = E x 4

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8 Integration of the marginal cost function For a given marginal cost function K x, there is exactly one total variable cost function Kvar x, which can be found using the fundamental theorem as follows: and consequently Example: 0 x K t dt = K x K(0) K x = 0 x K t dt + K(0) total variable cost fixed cost K (x) = 1.2x 2 7.2x Kvar x = The fixed cost, i.e. the integration constant, cannot be calculated, but must be otherwise obtained. 5

9 Total variable cost Kvar(x) as the area under the marginal cost function K(x) = 0.4x 3 3.6x x K (x) = total variable cost fixed cost K x 6

10 Contribution margin as area between the marginal revenue and the marginal cost functions E x = x K (x) = 1.2x 2 7.2x E K The area between the marginal revenue function and the marginal cost function is the profit before fixed costs, the so called contribution margin (= contribution to fixed cost and profit). x 7

11 Contribution margin and profit Since fixed cost K fix = K(0) cannot be found by integration, profit cannot be calculated form marginal revenue and marginal cost alone. We can only calculate the the contribution margin G D (x): G D x = When, in addition, fixed cost is known, the profit function can be found: G x = 0 0 x x Example: K fix = 25.2, therefore K(x) = 0.4x 3 3.6x x E(x) = 34x 10 3 x2 G x = E t K t E t dt = E x Kvar x K fix 0 x dt = E x Kvar x K t dt K 0 8

12 Profit maximum Summary The area under the marginal revenue function corresponds to the total revenue. The area under the marginal cost function corresponds to the total variable cost. The area between the marginal revenue and the marginal cost function corresponds to the contribution margin (contribution to fixed cost and profit). Maximizing profit is equivalent to maximizing contribution since the fixed cost is constant. At the profit maximum: marginal revenue = marginal cost Bear in mind! 9

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14 Consumer surplus Let p N (x) be a demand function and (x 0 ; p 0 ) an arbitrary point on it, then x 0 p N x dx 0 p 0 x 0 x 0 p N x dx p 0 x 0 0 is the amount the consumers would be willing to pay, is the amount the consumers actually pay, and is the so called consumer surplus K R at (x 0 ; p 0 ). 10

15 Consumer surplus Example Actual market price: p 0 = 15 GE/ME Demand function: p n x = 1 20 x2 3x + 40 and x < 20 ME Consumer surplus: K R = 11

16 Producer surplus Let p A (x) be a supply function and (x 0 ; p 0 ) an arbitrary point on it, then p 0 x 0 x 0 p A x dx is the amount the producers actually receive, is the amount the producers would be willing to accept, and 0 p 0 x 0 x 0 p A x dx 0 the producer surplus P R in (x 0 ; p 0 ). 12

17 Producer surplus Example Actual market price: p 0 = 32 GE/ME Supply function: p A x = 0.25x + 20 Producer surplus: P R = 13

18 Ordinary differential equation Introduction (I) Find a function f with the characteristic that the local rate of change f (x) is identical to its function value f(x). Thus: f (x) = f x What is its equation? 14

19 Ordinary differential equation Introduction (II) Approach: But: f x = e x and f (x) = e x f x = C e x with an arbitrary real constant C is also a solution of f (x) = f x (*) Conclusion: The equation (*) has an indefinite number of solutions if C is not given. Example: f(x) goes through the point (0; 1): 1 = C e 0 C = 1 and f x = e x 15

20 Definitions with regard to differential equations An equation relating a function, its derivatives, and the independent variable in any arbitrary way is called a differential equation. The solution of a differential equation is a function that together with its derivatives satisfies the differential equation. A differential equation is called ordinary if it contains one or several derivatives of an unknown function f that has only one independent variable, and partial if f has more than one independent variable. The abbreviation of ordinary differential equation is ODE. 16

21 Solution approach for many ordinary differential equations One family of ordinary differential equations that is common in economics can be solved using the method which is called separation of the variables. Such ODEs are called separable. The solution approach is: 1. Restate the differential equation using the Leibniz notation 2. Separate, i.e., restate the differential equation such that all terms containing the dependent variable are on the left-hand side and all terms containing the independent variable on the right-hand side of the equality sign: dy = dx 3. Integrate both sides of the equation. 4. Substitute and solve. Example: y x = 0.02 y(x) 17

22 Example of an ODE Find the (general) solution of y x = 0.02 y(x) Solution approach: 1. Restate 2. Separate 3. Integrate 4. Substitute and solve 18

23 General and special solution of a differential equation A differential equation has an indefinite number of solutions as long as the integration constant is not specified. The set of all solutions of a differential equation is called the general solution. The special or particular solution is the element of the general solution that satisfies an additional condition, for example a solution that goes through a specified point. Find the special solution of y x = 0.02 y(x) if y 0 = 300. The general solution is y x = C e 0.02x. 19

24 continuous General discrete Example Application (I) Continuous compounding Yearly interest 30 = = Principal times yearly interest rate Monthly interest = Principal times monthly interest rate 2.5 = Interest of period K ΔK Δt = = Principal times interest rate of period K i t = K i ΔK lim t 0 Δt = K i K = K i K t = K 0 e i t 20

25 Application (II) Exponential growth For a growth process, the time t is generally the independent variable. The rate of growth per time period of a population B is its derivative: B t = db dt Exponential growth is based on the assumption that a population (or any other quantity) B grows at a rate that is proportional to the size of the population. This leads to the ordinary differential equation B t = i B(t) having the solution B t = B 0 e i t with i the continuous growth rate and B 0 the initial size of the population. 21

26 Application (III) Functions with given elasticities The point elasticity is by definition a differential equation: Example 1: ε f,x = f (x) f(x) x Find the family of functions f that have a constant elasticity ε f,x = k, so called isoelastic functions. 22

27 Application (III) Functions with given elasticities The point elasticity is by definition a differential equation: Example 2: ε f,x = f (x) f(x) x Find the function f whose elasticity is x, i.e., ε f,x = x 23