f What is the consumers short run indirect utility function 19 This question is

# F what is the consumers short run indirect utility

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(f) What is the consumer’s (short-run) indirect utility function? 19. This question is a minor variation of Question 19 in Shannon (1995, p.508). Consider a firm that that is a price-taker in input markets. Suppose that this firm has a one output and two input production technology that can be represented by a production function of the form Q = f ( L, K ) = AK α L β . This is known as a generalised Cobb-Douglas production function. The input price for labour is w and the input price for capital is r . Suppose that this firm wants to minimise the cost of producing a particular output level, Q . Its cost minimisation problem is min wL + rK : AK α L β = Q . 6
(a) Find the Lagrangian for this constrained minimisation problem. (Use λ to denote the Lagrange multiplier.) (b) Obtain the values of K , L and λ that satisfy the first-order con- ditions for this problem. (c) What are the firm’s (minimum) cost function and output-conditional input demand functions? (d) What is the partial derivative of the firm’s (minimum) cost func- tion with respect to the chosen output level, Q ? (e) Compare your answer to part (d) of this question with the value for λ that you found in part (b) of this question. What is the interpretation of the value for λ that you found in part (b) of this question? 1 Partial Derivatives Answers See separate document (which is an extract from Shannon and McDonald 1995). 2 Unconstrained Optimisation Answers 2.1 Question 11 2.1.1 The objective function π ( q 1 , q 2 ) = - 90 + 20 q 2 1 + 5 q 2 2 - 8 q 1 - 5 q 1 q 2 . 2.1.2 The first-order partial derivatives ∂f ∂q 1 = 40 q 1 - 8 - 5 q 2 . ∂f ∂q 2 = 10 q 2 - 5 q 1 . 2.1.3 The critical points The first-order conditions for an unconstrained optimisation problem are: ∂f ∂q 1 = 40 q 1 - 8 - 5 q 2 = 0, 7
and ∂f ∂q 2 = 10 q 2 - 5 q 1 = 0. These FOCs can be rearranged to obtain the following system of two equations in two unknown endogenous variables: 40 q 1 - 5 q 2 = 8 - 5 q 1 + 10 q 2 = 0 . (In this example, both equations are linear. This will not always be the case.) Upon solving this linear equation system, we obtain the following unique critical point for the objective function: ( q * 1 , q * 2 ) = 16 75 , 8 75 . 2.1.4 The second-order partial derivatives 2 f ∂q 2 1 = 40. 2 f ∂q 2 2 = 10. 2 f ∂q 2 ∂q 1 = 2 f ∂q 1 ∂q 2 = - 5. 2.1.5 The Hessian matrix The Hessian matrix for the objective function is H = 40 - 5 - 5 10 . 2.1.6 The second-order conditions Note that det( H 1 ) = 40 > 0 and det( H 2 = det( H ) = (40(10) - ( - 5)( - 5) = 400 - 25 = 375 > 0 . Thus we can conclude that H is a positive definite matrix for all values of ( q 1 , q 2 ). This means that the critical point will yield a minimum value of the objective function. 8
2.1.7 The optimal value of the objective function The minimum value of the objective function is π ( q * 1 , q * 2 ) = π 16 75 , 8 75 = - 90 + 20( q * 1 ) 2 + 5( q * 2 ) 2 - 8 q * 1 - 5 q * 1 q * 2 = - 90 + 20 16 75 2 + 5 8 75 2 - 8 16 75 - 5 16 75 8 75 = 506 , 250 + 5 , 120 + 320 - 9 , 600 - 640 5 , 625 = 511 , 690 - 10 , 240 5 , 625 = 501 , 450 5 , 625 = 102 , 338 1 , 125 . 2.2 Question 12 2.2.1 The objective function π ( L, K ) = 500 L 0 . 3 K 0 . 7 - 2 L - 5 K .

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