1 Topic 7 Part I Partial Differentiation Part II Marginal Functions Part II Partial Elasticity Part III Total Differentiation Part IV Returns to scale Jacques (4th Edition):

2 Functions of Several Variables More realistic in economics to assume an economic variable is a function of a number of different factors: Y =f(X, Z) Demand may depend on the price of the good and the income level of the consumer Q d =f(P,Y) Output of a firm depends on inputs into the production process like capital and labour Q =f(K,L)

3 Graphically Sketching functions of two variables Y= f (X, Z) Sketch this function in 3-dimensional space or plot relationship between 2 variables for constant values of the third

4 For example I Consider a linear function form: y =a+bx+cz For different values of z we can represent the relationship between x and y

5 For example II Consider a non-linear function form: Y=X  Z  0<  < 1 & 0<  < 1 For different values of z we can represent the relationship between x and y

6 Part I: Partial Differentiation (Differentiating functions of several variables)

7 Consider our function of two variables: Y= f (x, z) = a + bx + cz

8 FOUR Second Order Partial Derivatives

9 Consider our function of two variables: Y= f (X, Z) = X  Z  First Partial Derivatives  Y/  X = f X =  X  -1 Z  > 0  Y/  Z = f Z =  X  Z  -1 > 0

10 Since  Y/  X = f X =  X  -1 Z   Y/  Z = f Z =  X  Z  -1 Second own partial  2 Y/  X 2 = f XX = (  -1)  X  -2 Z  < 0  2 Y/  Z 2 = f ZZ = (  -1)  X  Z  -2 < 0 Second cross partial  2 Y/  X  Z = f XZ =  X  -1 Z  -1 > 0  2 Y/  Z  X = f ZX =  X  -1 Z  -1 > 0

11 Example Jacques Y= f (X, Z) = X 2 +Z 3 f X = 2X > 0 Positive relation between x and y f XX = 2> 0 but at an increasing rate with x f XZ = 0, (= f zx ) and a constant rate with z impact of change in x on y is bigger at bigger values of x but the same for all values of z f Z = 3Z 2 > 0 Positive relation between z and y f ZZ = 6Z> 0 but at an increasing rate with z f ZX = 0 (= f xz ) and a constant rate with x impact of change in z on y is bigger at bigger values of z, but the same for all values of x

12 Example Jacques Y= f (X, Z) = X 2 Z f X = 2XZ>0 Positive relation between x and y f XX = 2Z>0 but at an increasing rate with x f XZ = 2X >0 and at an increasing rate with z impact of change in x on y is bigger at bigger values of x and bigger values of z f Z = X 2 >0 Positive relation between z and y f ZZ = 0 but at a constant rate with z f ZX = 2X>0 and an increasing rate with x impact of change in z on y is the same for all values of z but is bigger at higher values of x

13 Production function example Y= f(K L) = K 1/3 L 2/3 First partial derivatives of input gives Marginal product of input MPL=  Y/  L=Y L = ( 2 / 3 K 1/3 L 2/3 -1 ) = 2 / 3 Y/L >0 An increase in L holding other inputs constant will increase output Y MPK=  Y/  K = Y K =( 1 / 3 K 1/3-1 L 2/3 ) = 1 / 3 Y/K > 0 An increase in K holding other inputs constant will increase output Y

14 MPL =  Y/  L = ( 2 / 3 K 1/3 L -1/3 ) = 2 / 3 Y/L. MPK =  Y/  K = ( 1 / 3 K -2/3 L 2/3 ) = 1 / 3 Y/K. Second Own derivatives of input gives Marginal Returns of input (or the change in the marginal product of an input with the level of that input)   Y/  L 2 = - 2 / 9 K 1/3 L -4/3 < 0 Diminishing marginal returns to L (the change in MPL with L shows that the MPL decreases at higher values of L)   Y/  L 2 = - 2 / 9 K -5/3 L 2/3 < 0 Diminishing marginal returns to K (the change in the MPK with K shows that the MPK decreases at higher values of K)

15 Part II: Partial Elasticity

16 e.g. Y= f(K L) = K 1/3 L 2/3

17 Q = f(P,P S,Y) = 100-2P+P S +0.1Y P = 10, P S = 12, Y= 1000  Q=192 Partial Own-Price Elasticity of Demand  QP =  Q/  P. P/Q = -2 * (10/192) = Partial Cross-Price Elasticity of Demand  QPS =  Q/  P S. P S /Q = +1 * (12/192) = 0.06 Partial Income Elasticity of Demand  QI =  Q/  Y. Y/Q = +0.1 * (1000/192) = 0.52 e.g. demand function Q = f( P, P S, Y)

18 Part III: Total Differential Total Differential: Y= f (X)  Y= dY/dX.  X If  X =10 and dY/dX = 2,   Y = = 20 Total Differential: Y= f (X, Z)  Y=  Y/  X.  X +  Y/  Z.  Z or dY=  Y/  X. dX +  Y/  Z. dZ

19 Example: Y= f (K, L) = Y = K 1/3 L 2/3 dY=  Y/  K. dK +  Y/  L. dL or dY= ( 1 / 3 K -2/3 L 2/3 ).dK + ( 2 / 3 K 1/3 L -1/3 ).dL Rewriting: dY= ( 1 / 3 K 1/3 K -1 L 2/3 ).dK + ( 2 / 3 K 1/3 L 2/3 L -1 ).dL or dY= 1 / 3. Y / K. dK + 2 / 3. Y / L. dL To find proportionate change in Y dY / Y = 1 / 3. dK / K + 2 / 3. dL / L

20 Y= A f (K, L) = Y = A K  L  dY=  Y/  K.dK +  Y/  L.dL +  Y/  A.dA or dY= .AK  -1 L  dK + .AK  L  -1.dL + K  L . dA dY= .Y/K. dK + .Y/L. dL + Y/A. dA Or for proportionate change in Y: dY/Y= .dK/K + .dL/L + dA/A Or for proportionate change in A: dA/A = dY/Y–( .dK/K+ .dL/L) Total Differential: Y= f (X, Z)

21 Part IV: Returns to Scale Returns to scale shows the change in Y due to a proportionate change in ALL factors of production. So if Y= f(K, L) Constant Returns to Scale f( K, L ) = f(K, L) = Y Increasing Returns to Scale f( K, L ) > f(K, L) > Y Decreasing Returns to Scale f( K, L ) < f(K, L) < Y

22 Cobb-Douglas Production Function: Y = AK  L  Quick way to check returns to scale in Cobb- Douglas production function Y = AK  L  then if  +  = 1 : CRS if  +  > 1 : IRS if  +  < 1 : DRS

23 Y * = f( K, L) = A ( K)  ( L)  Y * = A  K   L  =  +  AK  L  =  +  Y  +  = 1 Constant Returns to Scale  +  > 1 Increasing Returns to Scale  +  < 1 Decreasing Returns to Scale Example: Y= f(K, L) = A K  L 

24 Homogeneous of Degree r if: f( X, Z ) = r f(X, Z) = r Y Homogenous function if by scaling all variables by, can write Y in terms of r Note superscripts! Note then, for cobb-douglas Y = K  L , the function is homogenous of degree  + 

25 X f X + Z f Z = r f(X, Z) = rY E.g. r =1, Constant Returns to Scale If Y= f(K, L) = A K  L (1-  ) Does K f K + L f L = rY = Y? K (  Y/K) + L((1-  )Y/L) = ?  Y + (1-  )Y =  Y + Y -  Y = Y Thus, Eulers theorem shows (MPk * K ) + (MPL * L) = Y in the case of homogenous production functions of degree 1 Eulers Theorem

26 then Y( K, L) = ( K) ½ ( L) ½ = ½ K ½ ½ L ½ = 1 K ½ L ½ = Y homogenous degree constant returns to scale Eulers Theorem: show that K.  Y /  K + L.  Y /  L = r.Y = Y (r=1 as homog. degree 1) = K. (½.K ½ -1.L ½ ) + L.(½.K ½.L ½ -1 ) = K(½. Y/K) + L(½.Y/L) = ½.Y + ½.Y = Y Example.. If Y = K ½ L ½

27 Summary: Function of Two Variables Partial Differentiation - Production Functions– first derivatives (marginal product of K or L) and second derivatives (returns to K or L) Partial Elasticity – Demand with respect to own price, price of another good, or income Total Differentials Returns to Scale Plenty of Self-Assessment Problems and Tutorial Questions on these things….