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