BEE1024 Mathematics for Economists

Transcription

1 BEE1024 Mathematics for Economists Dieter and Jack Rogers and Juliette Stephenson Department of Economics, University of Exeter February 1st 2007

2 1 Objective 2 Isoquants 3

3 Objective. The lecture should enable you to calculate partial derivatives and marginal rates of substitution.

4 Objective. Partial di erentiation! partial analysis in economics The lecture should enable you to calculate partial derivatives and marginal rates of substitution.

5 Objective. Partial di erentiation! partial analysis in economics level curves! indi erence curves or isoquants The lecture should enable you to calculate partial derivatives and marginal rates of substitution.

6 Objective. Partial di erentiation! partial analysis in economics level curves! indi erence curves or isoquants slopes of tangents to level curves! marginal rates of substitution The lecture should enable you to calculate partial derivatives and marginal rates of substitution.

7 A function or simply z = f (x, y) z (x, y) in two independent variables with one dependent variable assigns to each pair (x, y) of (decimal) numbers from a certain domain D in the two-dimensional plane a number z = f (x, y). x and y are hereby the independent variables z is the dependent variable.

8 Outline Objective 1 Objective 2 Isoquants 3

9 The graph of f is the surface in 3-dimensional space consisting of all points (x, y, f (x, y)) with (x, y) in D. z = f (x, y) = x 3 3x 2 y x y Exercise: Evaluate z = f (2, 1), z = f (3, 0), z = f (4, 4), z = f (4, 4)

10 Outline Objective 1 Objective 2 Isoquants 3

11 Q = 6p K p L = K 1 6 L 1 2 capital K 0, labour L 0, output Q K L

12 Outline Objective 1 Objective 2 Isoquants 3

13 Example: pro t function Assume that the rm is a price taker in the product market and in both factor markets. P is the price of output

14 Example: pro t function Assume that the rm is a price taker in the product market and in both factor markets. P is the price of output r the interest rate (= the price of capital)

15 Example: pro t function Assume that the rm is a price taker in the product market and in both factor markets. P is the price of output r the interest rate (= the price of capital) w the wage rate (= the price of labour)

16 Example: pro t function Assume that the rm is a price taker in the product market and in both factor markets. P is the price of output r the interest rate (= the price of capital) w the wage rate (= the price of labour) total pro t of this rm: Π (K, L) = PQ rk wl = PK 1 6 L 1 2 rk wl

17 P = 12, r = 1, w = 3: Π (K, L) = PK 1 6 L 1 2 rk wl = 12K 1 6 L 1 2 K 3L K L 15 20

18 P = 12, r = 1, w = 3: Π (K, L) = PK 1 6 L 1 2 rk wl = 12K 1 6 L 1 2 K 3L K L Pro ts is maximized at K = L = 8.

19 Outline Objective 1 Objective 2 Isoquants 3

20 Objective The level curve of the function z = f (x, y) for the level c is the solution set to the equation where c is a given constant. f (x, y) = c

21 Objective The level curve of the function z = f (x, y) for the level c is the solution set to the equation where c is a given constant. f (x, y) = c Geometrically, a level curve is obtained by intersecting the graph of f with a horizontal plane z = c and then projecting into the (x, y)-plane. This is illustrated on the next page for the cubic polynomial discussed above:

22 y x 1 1 y 1 x compare: topographic map

23 Isoquants Objective In the case of a production function the level curves are called isoquants. An isoquant shows for a given output level capital-labour combinations which yield the same output K K L L 20

24 Finally, the linear function z = 3x + 4y has the graph and the level curves: y x y x The level curves of a linear function form a family of parallel lines: c = 3x + 4y 4y = c 3x y = c 4 slope 3 4, variable intercept c y

25 Exercise: Describe the isoquant of the production function Q = KL for the quantity Q = 2. Exercise: Describe the isoquant of the production function for the quantity Q = 4. Q = p KL

26 Remark: The exercises illustrate the following general principle: If h (z) is an increasing (or decreasing) function in one variable, then the composite function h (f (x, y)) has the same level curves as the given function f (x, y). However, they correspond to di erent levels L L K K Q = KL Q = p KL

27 Objective Exercise: What is the derivative of z (x) = a 3 x 2 with respect to x when a is a given constant? Exercise: What is the derivative of z (y) = y 3 b 2 with respect to y when b is a given constant?

28 Consider function z = f (x, y). x y = y 0, vary only x: z = F (x) = f (x, y 0 ). The derivative of this function F (x) at x = x 0 is called the partial derivative of f with respect to x and denoted by z = df = df x jx =x 0,y =y 0 dx jx =x 0 dx (x 0)

29 Consider function z = f (x, y). x y = y 0, vary only x: z = F (x) = f (x, y 0 ). The derivative of this function F (x) at x = x 0 is called the partial derivative of f with respect to x and denoted by z = df = df x jx =x 0,y =y 0 dx jx =x 0 dx (x 0) Notation: d = dee, δ = delta, = del

30 Consider function z = f (x, y). x y = y 0, vary only x: z = F (x) = f (x, y 0 ). The derivative of this function F (x) at x = x 0 is called the partial derivative of f with respect to x and denoted by z = df = df x jx =x 0,y =y 0 dx jx =x 0 dx (x 0) Notation: d = dee, δ = delta, = del It su ces to think of y and all expressions containing only y as exogenously xed constants. We can then use the familiar rules for di erentiating functions in one variable in order to obtain z x.

31 Consider function z = f (x, y). x y = y 0, vary only x: z = F (x) = f (x, y 0 ). The derivative of this function F (x) at x = x 0 is called the partial derivative of f with respect to x and denoted by z = df = df x jx =x 0,y =y 0 dx jx =x 0 dx (x 0) Notation: d = dee, δ = delta, = del It su ces to think of y and all expressions containing only y as exogenously xed constants. We can then use the familiar rules for di erentiating functions in one variable in order to obtain z x. Other common notations for partial derivatives are f x, f y or f x, f y orfx 0, fy 0.

32 Example: Let Then z x = 2y 3 x z (x, y) = y 3 x 2 z y = 3y 2 x 2

33 Example: Let z = x 3 + x 2 y 2 + y 4. Setting e.g. y = 1 we obtain z = x 3 + x and hence z x jy =1 z x jx =1,y =1 = 3x 2 + 2x = 5

34 For xed, but arbitrary, y we obtain z x = 3x 2 + 2xy 2 as follows: We can di erentiate the sum x 3 + x 2 y 2 + y 4 with respect to x term-by-term. Di erentiating x 3 yields 2x 2, di erentiating x 2 y 2 yields 2xy 2 because we think now of y 2 as a constant and d(ax 2 ) dx = 2ax holds for any constant a. Finally, the derivative of any constant term is zero, so the derivative of y 4 with respect to x is zero. Similarly considering x as xed and y variable we obtain z y = 2x 2 y + 4x 3

35 Example: The partial derivatives q q K and L of a production function q = f (K, L) are called the marginal product of capital and (respectively) labour. They describe approximately by how much output increases if the input of capital (respectively labour) is increased by a small unit. Fix K = 64, then q = K 1 6 L 1 2 = 2L 2 1 which has the graph 4 3 Q L

36 This graph is obtained from the graph of the function in two variables by intersecting the latter with a vertical plane parallel to L-q-axes K L The partial derivatives q q K and L describe geometrically the slope of the function in the K- and, respectively, the L- direction.

37 diminishing productivity of labour: The more labour is used, the less is the increase in output when one more unit of labour is employed. Algebraically: 2 q L 2 = L q L = 1 2 K L 2 = 2 6p K 2p L > 0, q = 1 L 4 K L 2 = 4 Exercise: Find the partial derivatives of 6p K 2p L 3 < 0, z = x 2 + 2x y 3 y x + 3y

38 Outline Objective 1 Objective 2 Isoquants 3

39 Objective For a function y = f (x) in one variable the derivative df dx jx =x 0 at x 0 is the slope of the tangent to the graph of f through the point (x 0, f (x 0 )) x The tangent of f (x) = x 2 at x = 1.

40 This tangent is the graph of the linear function y = l (x) = f (x 0 ) + df dx jx =x 0 (x x 0 ). This is so because l (x 0 ) = f (x 0 ), so the graph of l goes through the point (x 0, f (x 0 )). Moreover, y = l (x) = f (x 0 ) df x 0 + df x, dx 0 jx =x0 dx jx =x 0 so l (x) is indeed a linear function in x with intercept df f (x 0 ) dx 0 jx =x0 x 0 and the right slope df dx jx =x 0.

41 The graph of a function in one variable is a curve and its tangent is a line. By analogy, the graph of a function z = f (x, y) in two variables is a surface and its tangent at a point (x 0, y 0 ) is a plane

42 Non-vertical planes are the graphs of linear functions z = ax + by + c. The tangent plane of the function z = f (x, y) in (x 0, y 0 ) is, correspondingly, the graph of the linear function z = l (x, y) = f (x 0, y 0 ) + f x jx =x 0,y =y 0 (x since the graph of this linear function contains the point (x 0, y 0, f (x 0, y 0 )) and has the right slopes in the x- and y-directions. x 0 ) + f y jx =x0,y =y 0 (y y 0 ) (1)

43 Exercise: This is the example shown in the above graph. The tangent plane of z = f (x, y) = p x + p y at the point (1, 1) is obtained as follows: We have z x = 1 1 p 2 x and z y = 1 1 p 2 y and hence z x jx =1,y =1 = 1 2 and z y jx =1,y =1 = 1 2. Since f (1, 1) = 2 the tangent plane is the graph of the linear function z = l (x, y) = (x 1) (y 1) = x y

44 Outline Objective 1 Objective 2 Isoquants 3

45 Using the abbreviations dx = x x 0, dy = y y 0 and dz = z z 0 the formula for the tangent can be neatly rewritten as dz = z x z dx + dy (2) y an expression which is called the total di erential.

46 Outline Objective 1 Objective 2 Isoquants 3

47 The marginal rate of substitution xy = c x

48 Suppose (x 0, y 0 ) is a point on the level curve f (x, y) = c. Then the slope of tangent to the level curve in this point in the x-y-plane is dy dx jx =x 0,y =y 0 = z z (3) x jx =x 0,y =y 0 y jx =x0,y =y 0 provided z y 6= 0. This is known as the rule for implicit jx =x 0,y =y 0 di erentiation: If the function y = y (x) is implicitly de ned by the equation f (x, y (x)) = c, then its derivative dy dx is given by the above formula.

49 Remark: A quick way to memorize the formula (3) is to use the total di erential as follows: In order to stay on the level curve one must have dz = z z 0 = 0, so 0 = dz = z z dx + x y dy hence or dy = dy dx = z z dx x y z z x y.

50 Example: A linear function z = ax + by + c has the partial derivatives z z x = a and y = b. According to the above formula its level curves have the slope z z x y = a b. Indeed, its level curves are the solutions to the equations l = ax + by + c with some constant l. Solving the equation for y we obtain y = a b x + l c b which is a linear function with slope a b.

51 Example: An isoquant of the production function Q = K 6 1 L 1 2 has according to the above formula the slope dk Q Q 1 = dl L K = 2 K L 2 6 K L 2 6 = 2 K K 6 K 6 L 2 L 2 = 3 L. in the point (K, L). Indeed, an isoquant is the set of solutions to the equation q = K 1 6 L 1 2 for xed q. Solving for K we see that the isoquant is the graph of the function K = ql = q 6 L 3.

52 Di erentiation yields dk dl = 3 q6 L 4 and since (K, L) is on the isoquant dk dl = 3 K 1 6 L L 4 = 3 K L.

53 dk dl Economically, is the marginal rate of substitution of labour for capital: If labour is reduced by one unit, how much more capital is (approximately) needed to produce the same output? Just reducing labour by one unit reduces output by the marginal product of labour q q L. Each additional unit of capital produces K the marginal product of capital more units. Therefore, if now capital input is increased by q L by q L +. q K units, output changes overall q q q L K K = 0.

54 Remark: The implicit function theorem states that if z y jx =x 0,y =y 0 6= 0 for some point (x 0, y 0 ) on a level curve f (x, y) = c, then one can nd a function y = g (x) de ned near x = x 0 such that the level curve is locally the graph of this function, i.e. one has f (x, g (x)) = c for x near x 0. Exercise: For the production function Q = p KL calculate the marginal rate of substitution for K = 3 and L = 4.

55 Outline Objective 1 Objective 2 Isoquants 3

56 The Chain rule Objective Suppose that z = f (x, y) is a function in two variables x, y. Suppose that x depends on the variable t via x = g (t) and that y depends on the variable t via y = h (t). Then we can de ne the composite function in one variable z = F (t) = f (x (t), y (t)) which has t as the independent and z as the dependent variable. The derivative of this function, if it exists, is calculated as dz dt = z dx x dt + z dy y dt. Notice that this is just the total di erential divided by dt.

57 Example: Suppose z = xy, x = t 3 + 2t, y = t 2 according to the chain rule, t. Then, dz dt = z dx x dt + z dy y dt = y 3t x (2t 1) = t 2 t 3t (2t 1) t 3 + 2t Since z = xy = t 3 + 2t t 2 the product rule: t we get the same result using dz dt = 3t2 + 2 t 2 t + t 3 + 2t (2t 1).