Lectures in Microeconomics-Charles W. Upton
Extending the Problem
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc
Extending the Problem
A Golden Oldie
• Demand for the park:VC =180,000 - 20,000pc
VA = 120,000 - 10,000pa
Extending the Problem
A Golden Oldie
• Demand for the park:VC =180,000 - 20,000pc
VA = 120,000 - 10,000pa
• Capacity = 200,000
Extending the Problem
A Golden Oldie
• Demand for the park:VC =180,000 - 20,000pc
VA = 120,000 - 10,000pa
• Capacity = 200,000• Adding Capacity $3 a visit.
Extending the Problem
A Golden Oldie
• If demand is to be restricted to 200,000 what single fee would you recommend?
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?
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?
Extending the Problem
Charging a Single Fee
• Demand for the park:VC =180,000 - 20,000pc
VA = 120,000 - 10,000pa
Extending the Problem
Charging a Single Fee
• Demand for the park:VC =180,000 - 20,000pc
VA = 120,000 - 10,000pa
• When p = 0, demand is 300,000
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000pa
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000paac ppTrip 1000020000
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000papTrip
ppTrip ac
30000
1000020000
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000papTrip
ppTrip ac
30000
1000020000
To eliminate 100,000 trips set p = $3.33
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000papTrip
ppTrip ac
30000
1000020000
To eliminate 100,000 trips set p = $3.33
But is there a better way of doing
it, so as to minimize DWL?
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000pa
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000pa
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000pa
Dead Weight Loss from Children is (1/2) (20,000pc )(pc )=
10,000pc2
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000pa
Dead Weight Loss from Adults is (1/2) (10,000pa )(pa )=
5,000pa2
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000pa
DWL= 10,000pc2 +5,000pa
2
Extending the Problem
Charging a Single Fee
Children Adults
V=180,000-20,000pc
V=120,000-10,000pa
20,000pc 10,000pa
DWL= 10,000pc2 +5,000pa
2
ac ppTrip 1000020000
Extending the Problem
Charging Separate Fees
• Many different adult and child’s fees will cut demand
Extending the Problem
Charging Separate Fees
• Many different adult and child’s fees will cut demand
• But
20,000pc + 10,000pa=100,000
Extending the Problem
Charging Separate Fees
• Many different adult and child’s fees will cut demand
• But
20,000pc + 10,000pa=100,000
pa = 10– 2pc
Extending the Problem
Charging Separate Fees
• Deadweight loss is
10,000pc2 + 5,000pa
2
• The two prices are
pa = 10– 2pc
• Deadweight loss is
10,000pc2 + 5,000(10-2pc)2
Extending the Problem
Charging Separate Fees
10,000pc2 + 5,000(10-2pc)2
Extending the Problem
Charging Separate Fees
10,000pc2 + 5,000(10-2pc)2
10,000pc2 + 5,000(100-40pc+4pc
2)
Extending the Problem
Charging Separate Fees
10,000pc2 + 5,000(10-2pc)2
10,000pc2 + 5,000(100-40pc+4pc
2)
10,000pc2 + 500,000-
200,000pc+20,000pc2
Extending the Problem
Charging Separate Fees
10,000pc2 + 5,000(10-2pc)2
10,000pc2 + 5,000(100-40pc+4pc
2)
10,000pc2 + 500,000-
200,000pc+20,000pc2
DWL =500,000-200,000pc+30,000pc2
Extending the Problem
Charging Separate Fees
500,000-200,000pc+30,000pc2
Extending the Problem
Charging Separate Fees
500,000-200,000pc+30,000pc2
• In sum, we can achieve our objective with many different values of pc (and pa).
Extending the Problem
Charging Separate Fees
500,000-200,000pc+30,000pc2
• In sum, we can achieve our objective with many different values of pc (and pa).
• Lets find the one that minimizes DWL.
Extending the Problem
Charging Separate Fees
500,000-200,000pc+30,000pc2
• In sum, we can achieve our objective with many different values of pc (and pa).
• Lets find the one that minimizes DWL.
• To do that….
Extending the Problem
Charging Separate Fees
500,000-200,000pc+30,000pc2
• In sum, we can achieve our objective with many different values of pc (and pa).
• Lets find the one that minimizes DWL.
• To do that….– Take the derivative– Set it equal to zero
Extending the Problem
Charging Separate Fees
500,000-200,000pc+30,000pc2
• Differentiating
-200,000+60,000pc
Extending the Problem
Charging Separate Fees
500,000-200,000pc+30,000pc2
• Differentiating
-200,000+60,000pc
33.3$cp
Extending the Problem
Charging Separate Fees
500,000-200,000pc+30,000pc2
• Differentiating
-200,000+60,000pc
33.3$cp
pa = 10– 2pc
Extending the Problem
Charging Separate Fees
500,000-200,000pc+30,000pc2
• Differentiating
-200,000+60,000pc
33.3$cp
pa = 10– 2pc
pa=$3.33
Extending the Problem
The Dead Weight Loss
DWL = 500,000-200,000pc+30,000pc2
Extending the Problem
The Dead Weight Loss
DWL = 500,000-200,000pc+30,000pc2
DWL = 500,000-200,000($3.33)+30,000($3.33)2
Extending the Problem
The Dead Weight Loss
DWL = 500,000-200,000pc+30,000pc2
DWL = 500,000-200,000($3.33)+30,000($3.33)2
DWL = $55,556
Extending the Problem
The Dead Weight Loss
DWL = 500,000-200,000pc+30,000pc2
DWL = 500,000-200,000($3.33)+30,000($3.33)2
DWL = $55,556 ?DWL
Cost
Extending the Problem
End
©2004 Charles W. Upton