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A2(solutions)

# A2(solutions) - University of Victoria Economics 203...

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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 .
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A2(solutions) - University of Victoria Economics 203...

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