1. Lectures in Microeconomics-Charles W. Upton Extending the Problem

2. A Golden Oldie Demand for the park: V C =180,000 - 20,000p c V A = 120,000 - 10,000p a

3. Extending the Problem A Golden Oldie Demand for the park: V C =180,000 - 20,000p c V A = 120,000 - 10,000p a Capacity = 200,000

4. Extending the Problem A Golden Oldie Demand for the park: V C =180,000 - 20,000p c V A = 120,000 - 10,000p a Capacity = 200,000 Adding Capacity $3 a visit.

5. Extending the Problem A Golden Oldie If demand is to be restricted to 200,000 what single fee would you recommend?

6. Extending the Problem A Golden Oldie If demand is to be restricted to 200,000 what single fee would you recommend? If children and adults can be charged a separate fee, what fees would you recommend?

7. Extending the Problem A Golden Oldie If demand is to be restricted to 200,000 what single fee would you recommend? If children and adults can be charged a separate fee, what fees would you recommend? Should the park expand capacity in that case?

8. Extending the Problem Charging a Single Fee Demand for the park: V C =180,000 - 20,000p c V A = 120,000 - 10,000p a

9. Extending the Problem Charging a Single Fee Demand for the park: V C =180,000 - 20,000p c V A = 120,000 - 10,000p a When p = 0, demand is 300,000

10. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a

11. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c

12. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a

13. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a

14. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a

15. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a To eliminate 100,000 trips set p = $3.33

16. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a To eliminate 100,000 trips set p = $3.33 But is there a better way of doing it, so as to minimize DWL?

17. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a

18. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a

19. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a

20. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a Dead Weight Loss from Children is (1/2) (20,000p c )(p c )= 10,000p c 2

21. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a Dead Weight Loss from Adults is (1/2) (10,000p a )(p a )= 5,000p a 2

22. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a DWL= 10,000p c 2 + 5,000p a 2

23. Extending the Problem Charging a Single Fee Children Adults V=180,000- 20,000p c V=120,000- 10,000p a 20,000p c 10,000p a DWL= 10,000p c 2 + 5,000p a 2

24. Extending the Problem Charging Separate Fees Many different adult and child’s fees will cut demand

25. Extending the Problem Charging Separate Fees Many different adult and child’s fees will cut demand But 20,000p c + 10,000p a =100,000

26. Extending the Problem Charging Separate Fees Many different adult and child’s fees will cut demand But 20,000p c + 10,000p a =100,000 p a = 10– 2p c

27. Extending the Problem Charging Separate Fees Deadweight loss is 10,000p c 2 + 5,000p a 2 The two prices are p a = 10– 2p c Deadweight loss is 10,000p c 2 + 5,000(10-2p c ) 2

28. Extending the Problem Charging Separate Fees 10,000p c 2 + 5,000(10-2p c ) 2

29. Extending the Problem Charging Separate Fees 10,000p c 2 + 5,000(10-2p c ) 2 10,000p c 2 + 5,000(100-40p c +4p c 2 )

30. Extending the Problem Charging Separate Fees 10,000p c 2 + 5,000(10-2p c ) 2 10,000p c 2 + 5,000(100-40p c +4p c 2 ) 10,000p c 2 + 500,000- 200,000p c +20,000p c 2

31. Extending the Problem Charging Separate Fees 10,000p c 2 + 5,000(10-2p c ) 2 10,000p c 2 + 5,000(100-40p c +4p c 2 ) 10,000p c 2 + 500,000- 200,000p c +20,000p c 2 DWL =500,000- 200,000p c +30,000p c 2

32. Extending the Problem Charging Separate Fees 500,000-200,000p c +30,000p c 2

33. Extending the Problem Charging Separate Fees 500,000-200,000p c +30,000p c 2 In sum, we can achieve our objective with many different values of p c (and p a ).

34. Extending the Problem Charging Separate Fees 500,000-200,000p c +30,000p c 2 In sum, we can achieve our objective with many different values of p c (and p a ). Lets find the one that minimizes DWL.

35. Extending the Problem Charging Separate Fees 500,000-200,000p c +30,000p c 2 In sum, we can achieve our objective with many different values of p c (and p a ). Lets find the one that minimizes DWL. To do that….

36. Extending the Problem Charging Separate Fees 500,000-200,000p c +30,000p c 2 In sum, we can achieve our objective with many different values of p c (and p a ). Lets find the one that minimizes DWL. To do that…. –Take the derivative –Set it equal to zero

37. Extending the Problem Charging Separate Fees 500,000-200,000p c +30,000p c 2 Differentiating -200,000+60,000p c

38. Extending the Problem Charging Separate Fees 500,000-200,000p c +30,000p c 2 Differentiating -200,000+60,000p c

39. Extending the Problem Charging Separate Fees 500,000-200,000p c +30,000p c 2 Differentiating -200,000+60,000p c p a = 10– 2p c

40. Extending the Problem Charging Separate Fees 500,000-200,000p c +30,000p c 2 Differentiating -200,000+60,000p c p a = 10– 2p c p a =$3.33

41. Extending the Problem The Dead Weight Loss DWL = 500,000-200,000p c +30,000p c 2

42. Extending the Problem The Dead Weight Loss DWL = 500,000-200,000p c +30,000p c 2 DWL = 500,000- 200,000($3.33)+30,000($3.33) 2

43. Extending the Problem The Dead Weight Loss DWL = 500,000-200,000p c +30,000p c 2 DWL = 500,000- 200,000($3.33)+30,000($3.33) 2 DWL = $55,556

44. Extending the Problem The Dead Weight Loss DWL = 500,000-200,000p c +30,000p c 2 DWL = 500,000- 200,000($3.33)+30,000($3.33) 2 DWL = $55,556

45. Extending the Problem End ©2004 Charles W. Upton