Consumer choice

27 de octubre de 2011

3.1

Optimal choice (x1, x2) must meet the condition MRS(x1, x2) = −p1p2 .MRS is the ratio of the derivatives of the utility function, then we have:

−∂u(x1, x2)/∂x1∂u(x1, x2)/∂x2

= −p1p2⇒ ∂u(x1, x2)/∂x1

∂u(x1, x2)/∂x2=p1p2

If utility function is u(x1, x2) = x1x2 + x1, then :

x2 + 1

x1=p1p2

Substituting the conssumption bundle (6, 2) :

2 + 1

6=p1p2⇒ 1

2=p1p2

=⇒ p2 = 2p1

Substituting in budget constraint who must also meet the optimal choise:

6p1 + 2(2p1) = 100⇒ 10p1 = 100

{p1 = 10

p2 = 20

1

3.2

1.

Budget set: 2x1 + 3x2 5 10

Budget constraint: 2x1 + 3x2 = 10

2.

The consumer's utility is u(x1, x2) = x1 + x2 , good 1 gives thesame utility that good 2. That's because both goods are perfectsubstitutes (MRS is -1). However, the price of goods 1 and 2aren´t the same: good 1 costs 2 monetary units while the priceof good 2 is 3 monetary units. That explain why the optimalchoice will be (5, 0).

Conssumption bundle (5, 3) can't be optimal because is ouside thebudget set, the consumer doesn't have enough money to accessit.

Conssumption bundle (2, 2) is not optimal because the utility ofthis conssumption bundle (u(x1, x2) = 2 + 2 = 4) is less than theoptimal bundle (5, 0) where the value is 5.

2

3.

If prices where (5, 5), the slope of the budget constraint and theMRS would be the same (−1), then any point of the budgetconstraint would be an optimal solution.

3.3

1. With card system, our budget set will be: 80x1 + x2 ≤ 50000.

3

4

2. For the second tari�, our budget set will be: 40x1 + x2 = 48000.

5

3.

With the �rst option; 80x1+x2 ≤ 50000, for each unit of x1 we could buy80 units of x2. Then:

If α < 80⇒ x∗1 = 0 y x∗2 = 50000⇒ u(x∗1, x∗2) = 50000

If α = 80 , optimal conssumption will be any point of budget cons-traint ⇒ u(x∗1, x

∗2) = 50000

If α > 80⇒ x∗1 = 625 y x∗2 = 0⇒ u(x∗1, x∗2) = 625α

With the �rst option; 40x1+x2 ≤ 48000, for each unit of x1 we could buy40 units of x2. Then:

If α < 40⇒ x∗1 = 0 y x∗2 = 48000⇒ u(x∗1, x∗2) = 48000

If α = 40 , optimal conssumption will be any point of budget cons-traint ⇒ u(x∗1, x

∗2) = 48000

If α > 40⇒ x∗1 = 1200 y x∗2 = 0⇒ u(x∗1, x∗2) = 1200α

Therefore:

If α ≤ 40:

Choose 1º option: x∗1 = 0, x∗2 = 50000⇒ u(x∗1, x∗2) = 50000

If 40 < α ≤ 500001200 :

Choose 1º option: x∗1 = 0, x∗2 = 50000⇒ u(x∗1, x∗2) = 50000

If α < 500001200 :

Choose 2º option: x∗1 = 1200, x∗2 = 0⇒ u(x∗1, x∗2) = 1200α

3.4

1.Budget set of Plan A:{20x1 + 20x2 ≤ 8000, si x1 ≤ 200

(20 · 200) + 10(x1 − 200) + 20x2 ≤ 8000, si x1 > 200⇒

{20x1 + 20x2 ≤ 8000, si x1 ≤ 200

10x1 + 20x2 ≤ 6000, si x1 > 200

6

Budget set of Plan B: 20x2 ≤ 2000.

7

2. If his preferences were of the type u(x1, x2) = x1x2, the consumer chooseplan B since he has no limitation on the conssumption of good 1 and thereforecould get an in�nite utility.

3.

u(x1, x2) = min

{x1,

1

2x2

}At the optimum we know that:

x1 =1

2x2 ⇒ x∗2 = 2x∗1

With the plan A:

1rst stretch: 20x1 + 20x2 = 8000

x∗1 =400

3, x∗2 =

800

3

u(400

3,800

3) = min

{400

3,800

3

}=

400

3

2ndstretch: 10x1 + 20x2 = 6000

x1 = 120 x2 = 240⇒ pero 120 < 240!︸ ︷︷ ︸contradiction!

With the plan B:

6000 + 20x2 = 8000⇒ x∗2 = 100

x2 = 2x1 ⇒ x∗1 = 50

u(50, 100) = min {50, 100} = 50

Then:

Like400

3> 50, will choose the planA!

8

3.5

1.Maxx1,x2

x1/41 x

3/42

s.a. 100x1 + 2500x2 ≤ 40000

L(x1, x2, λ) = x1/41 x

3/42 − λ(100x1 + 2500x2 − 40000)

∂L∂x1

= 0⇒ 14x

−3/41 x

3/42 − 100λ = 0

∂L∂x2

= 0⇒ 34x

1/41 x

−1/42 − 2500λ = 0

∂L∂λ = 0⇒ 100x1 + 2500x2 − 40000 = 0

We divide the �rst ecuation by the second:

1/4x−3/41 x

3/42

3/4x1/41 x

−1/42

=100λ

2500λ

x23x1

=100

2500

x2x1

=300

2500

x1 =25

3x2

Substituing x1 in the budget constraint and found x2:

100(25

3x2) + 2500x2 − 40000 = 0

100(25

3x2) +

7500

3x2 −

120000

3= 0

2500x2 + 7500x2 = 120000

x2 = 12

9

Optimal conssumption will be:{xopt1 = 100

xopt2 = 12

Graphical representation:

2.Maxx1,x2

x1/41 x

3/42

s.a. 100x1 + 2250x2 ≤ 39900

L(x1, x2, λ) = x1/41 x

3/42 − λ(100x1 + 2250x2 − 39900)

10

∂L∂x1

= 0⇒ 14x

−3/41 x

3/42 − 100λ = 0

∂L∂x2

= 0⇒ 34x

1/41 x

−1/42 − 2250λ = 0

∂L∂λ = 0⇒ 100x1 + 2250x2 − 39900 = 0

We divide the �rst ecuation by the second:

1/4x−3/41 x

3/42

3/4x1/41 x

−1/42

=100λ

2250λ

x23x1

=100

2250

x2x1

=300

2250=

2

15

x1 =15

2x2

Substituing x1 in the budget constraint and found x2:

100(15

2x2) + 2250x2 − 39900 = 0

750x2 + 2250x2 − 39900 = 0

x2 =39900

3000

x2 = 13,3

Optimal conssumption will be:{xopt

′

1 = 99,75

xopt′

2 = 13,3

11

Graphical representation:

3. We need to �nd were the optimal conssumption produces more utility:

u(xopt1 , xopt2 ) = 1001/4 · 123/4 ' 20,38

u(xopt′

1 , xopt′

2 ) = 99,751/4 · 13,33/4 ' 22

u(xopt′

1 , xopt′

2 ) > u(xopt1 , xopt2 )

Yes, the consumer will become a member.

12

3.6

1. Budget constraint is: 2x1 + 5x2 = 40. In this case MRS is − 83 while the

slope of the budget constraint is − 25 . Optimal choice will be (20, 0) and is on

the budget constraint: 2(20) + 5(0) = 40.

2.MRS(x1, x2) = −

p1p2

−12x1x

−1/22 · x2

12x1x

−1/22 · x1

= −2

5

2x1 = 5x2

x1 =5

2x2

13

Substituing in budget constraint:

2(5

2x2) + 5x2 = 40

x2 = 4

Optimal choice will is (10, 4) and is on budget constraint:

2(10) + 5(4) = 40

3.Optimal conditions are: {

x1 = 3x2

2x1 + 5x2 = 40{x1 = 3x2

2(3x2) + 5x2 = 40{x1 = 120/11

x2 = 40/11

Optimal choice is (120/11, 40/11) and is on the budget constraint:

2(120/11) + 5(40/11) = 40

14

4.

5.

15