asmt2_fall2010_sol

# asmt2_fall2010_sol - University of Victoria October 22,...

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University of Victoria October 22, 2010 Economics 203 Assignment #2 Due: Monday, November 1 , 2010 , @ 12 noon. (In the boxes marked A ECON 203" near the Economics Department Office) Show all your work. Put your LAB section on your assignment . Questions from Chapter 9: 1) Problem 4, page 271. (3 Marks) (V3. page 333) 2) Problems 7, page 271 (5 Marks) (V3. page 333) Questions from Appendix Chapter 9: 3) Problem 2, page 648 (1 Mark) (V3. page 344) 4) Problem 4, page 648 (2 Marks) (V3. page 344) 5) Problem 7, p 649 (6 Marks) (V3. page 345) 6) Problem 8, page 649 (2 Marks) (V3. page 345) Questions from Chapter 10: 7) Problem 4, page 301 (3 Marks) (V3. page 373) 8) Problem 8, page 302 (3 Marks) (V3. page 374) 9) Problem 12, page 302 (3 Marks) (V3. page 374) Questions from Appendix Chapter 10: 10) Problem 1, page 659 (2 Marks) (V3. page 385) 11) Problem 3, page 659 (3 Marks) (V3. page 385) 12) Problem 4, page 659 (2 Marks) (V3. page 385) (Total: 35 Marks)

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Questions from Chapter 9: 1) Problem 4, page 271. (3 Marks) The missing numbers for the Total product column are 180, 320, and 480. The missing numbers for the Average product column are [– (not defined)], 160, and 140. The missing numbers for the Marginal product column are 180, 100, and 60. 2) Problem 7, page 271 (5 marks) 7. a) (1 mark) b) Maximum production (Q = 9) occurs at L = 6. [Calculus trained students: To find the maximum, take the first derivative and set it equal to zero and solve for L: dQ/dL = 3 – (1/2)L = 0, which yields L = 6, Q = 9.] (2 Marks) c) For 0 < L < 2, MP L [= (5/2)L] is increasing. For 2 < L < 7, MP L [=3 – ½ L] is decreasing. (1 mark) d) MP L = 0 for L = 6, and MP L < 0 for L > 6. Past L = 6, adding L lowers output Q. (1 mark) Questions from Appendix Chapter 9: 3) Problem 2, page 648 (1 mark) For K = 27, we have Q = 6L 1/3 , and so MP L = dQ/dL = 2L –2/3 . 4) Problem 4, page 648 (2 marks) We are given that Q = F(K, L) = min (2K, 3L) with K = 6 and L = 5. Then Q = min (12, 15) = 12. MP K = F(7, 5) – F(6, 5) = 14 – 12 = 2. MP L =F(6, 6) – F(6, 5) = 12 – 12 = 0. Q 9 26 7 L
5) Problem 7, p 649 (6 marks) (a) The graph for Q = K .5 L .5 appears in Figure 9A-6 (page 645 of the text). For Q =KL (see 7.d below), the isoquant for Q = 1 is identical to the Q=1 isoquant in Figure 9A-6, and the Q=2 isoquant is intermediate between the isoquants labelled Q=1 and Q=2, and passes through the point [L = (2) .5 , K= (2) .5 ]. (2 marks) (b) MP L = .5K .5 L –.5 ; MP K = .5L .5 K –.5 , and MRTS = MP L / MP K = K/L. (1 mark) (c) Substituting into Q = K .5 L .5 , we have 4 = K .5 (9) .5 , which solves for the optimal K = 16/9 machine-days per period. The MRTS at (L = 9, K = 16/9) = K/L = 16/81. [You can satisfy yourself, by plugging the values for K and L into the appropriate formulas, that MP L = .5(16/9) .5 (9) –.5 = 2/9; AP L = Q/L = 4/9 (= 2 MP L ), MP K = .5(9) .5 (16/9) –.5 = 9/8, and AP K = Q/K = 4/(16/9) = 9/4 (= 2 MP K )]. (1 mark) (d) For the diagram, see 7.a above. In this case, MP L = K; MP K = L, and MRTS = MP L /MP K = K/L. Substituting into Q = KL, we have 4 = K(9), which solves for the optimal K = 4/9 machine-days per period. The MRTS at (L = 9, K = 4/9) = K/L = 4/81. [You can satisfy yourself, by plugging the values for K and L into the appropriate formulas, that MP L = K = 4/9; AP L = Q/K = 4/9 (= MP L ), MP K = L = 9, and AP K = Q/K = 4/(4/9) = 9(= MP K ). It is useful to compare your results for this case and the initial case, possibly even using graphs of MP L and AP L as functions of L and of MP K and AP K

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## This note was uploaded on 08/03/2011 for the course ECON 203 taught by Professor Okhan during the Spring '11 term at University of Victoria.

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asmt2_fall2010_sol - University of Victoria October 22,...

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