14.01 F all 2010: Final Exam Solution S et

[Pages:12]

ABAB

$1000 40% 35% $200

(21+40%))40%$4020($1000404%02%)0(%20%2)47%$(55$662000/(0$1420000)

 $200 $100

PX/PY=$100/$200=1/2MRSYforX=3

log(y - 60)

y < 160

U = u(y) =

1 80

y

y 160

1 2

1 2

1

1

1

E[UBill, B]

=

log 54 + log 55 = log 2970

2

2

2

1

1

1

E[UBill, A]

=

log 40 + log 70 = log 2800

2

2

2

2970-50

q

q = f (K, L) = 4K + 2L

r = 4 w = 4

K

Expension path

Isoquants L

Image by MIT OpenCourseWare.

-1/2

q = 4K + 2L qL

K= - 4 2

q

L(q) = 0

q q = 4K K(q) =

4

q

q T C(q) = rK(q) + wL(q) = 4 + 4 ? 0 = q

4

MC(q) = 1

q p = 7 - q

max

qA

qA(p

-

M

C

(qA

))

max

qA

qA(7

-

qA

-

qB

-

1)

qA

6 - 2qA - qB = 0

qA

=

6

- qB 2

qB

=

6

- qA 2

qA

qA

=

3

-

1 2 qB

=

3

-

3 2

+

qA 4

3

3

4 qA = 2

qA = 2

qA = 2 qB = 2 Q=4

p = 7-4=3 A = 2(3 - 1) = 4 B = 4

max q(7 - q - 1)

q

6 - 2q = 0 q = 3

M=3(4-1)=9

t = 0, 1, 2 t = 1

1 1

U1(N1, C1) = N12 C1

2

C1 + w1N1 = 24w1

w1 = 10 t = 2

1 2

U2(N2, C2) = N23 C2

3

w2 = 20

C2 + w2N2 = 24w2

t = 1, 2

MN2R=St=8N=11Hw=2M2=1R21S6H=C1w2C=1=213=2202Cw12CN=12=12w01N1t=2w2=w1+2.5(H1-8)=20

t = 1

 t = 0 1/(1.1) 1/(1.1)2

P V = 120/(1 + i) + 320/(1 + i)2 = 120 0.9 + 320 0.8 = 108 + 256 = 364

t = 0 t = 1 w1 = 20 t = 2 w2 = 30

C2=480H1=12H2=1C61=120C2=C1160=240

N P V = -200 + 120/(1.1) + 160/(1.1)2 = -200 + 108 + 128 = 36

QUS = 120 - 2pUS QEU = 60 - pEU MC = 10

pUS pEU

? QUS = 120 - 2PUS

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