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1
2. a) Let M denote the meat and P denote potatoes.
Connie’s budget constraint is:
2P + 4M = 200
As shown in the …gure below, with M on the vertical axis, the vertical intercept is 50.
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b)
U (M; P )
=
2M + P
M UM
=
2
M UP
M UM
PM
M UP
PP
M UM
PM
=
1
2
4
1
2
M UP
1
=
PP
2
=
=
=
We conclude that any combination of meat and potatoes this the budget line
would provide her with maximum utility.
c)
d)
pp
=
4
pM
=
4
4P + 4M
M Up
PP
M UM
PM
M UM
PM
=
200
1
=
4
2
1
=
=
4
2
M Up
>
PP
Connie would spend all her money on meat, consuming 200/4=50 pounds of
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meat and no potatoes. This is a corner solution.
3. a) Suppose the wine is on the horizontal axis and cheese is on the vertical
axis.
p
=
y
=
!
q=1
(24 x)p
q
x + y = 24
b) The tax applies to sales beyond …ve bottles: the revenue from the …rst
…ve bottles is 5. While for another 19 bottles, due to 50% tax, he can only earn
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revenue 19/2=9.5. The maximum total income he can earn is 5 + 9:5 = 14:5; or
at most, he can obtain 14.5 units of cheese.
By selling 10 bottles of wine, it means he will possess 14 bottles of wine and
use the revenue from the sales of 10 bottles to buy cheese.
c) If a 25% of tax is applied to the sales of all units of wine, the new budget
constraint will become 0:75x + y = 18: The new budget line will cut through
the point (14; 7:5) which is the consumer’s choice in b)
MU
M Uy
Ux
Ux
= 1 y : With the tax policy, M
However, at (14:7:5); M0:5
0:75 <
1 ;
consuming more cheese would generate higher utility. John would sell more
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than 10 bottles wine in exchange for more cheese subject to a 25% tax.
Tax revenue:
Part b) total tax revenue = $5 * 0.5 = $2.5
part c) Suppose John sell a bottles of wine and a > 10:
Total tax revenue = a 0:25 > 10 0:25 = 2:5:
The government would collect more tax revenue in c).
4.
a) Nancy prefers (x1 ; x2 ) = (20; 70) as Professor Goodheart will pick only
the higher of the two scores as her score in the midterm, that means (x1 ; x2 )
= (20; 70) will give Nancy a total of 70 points in the midterms, but (x1 ; x2 ) =
(60; 60) will only give her 60 points. U (x1 ; x2 ) = maxfx1 ; x2 g
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b) U1 = 60; U2 = 70:
5. a) You are …nd a counterexample to prove it.
Or (x1 ; y1 )
(x2 ; y2 ) () x1 y1 x2 y2
1 & x2 y2
x1 y1 x2 y2 1
(x2 ; y2 ) (x3 ; y3 ) () 1 x2 y2 x3 y3 1
Then we have
2 x1 y1 x3 y3 2
x1 y1
For any (x1 y1 ) and (x3 ; y3 ) satisfying
1
x1 y1
x2 y2
2
x1 y1
x3 y3 <
we can not conclude (x1 ; y1 )
1; 1
x2 y2
1 or 1 < x1 y1
x3 y3
x3 y3
1; and
2;
(x3 ; y3 )
b)
)
)
(x1 ; y1 ) (x2 ; y2 ) () x1 y1
(x2 ; y2 ) (x3 ; y3 ) () x2 y2
x1 y1 x3 y3 > 2 > 1
(x1 ; y1 )
(x3 ; y3 )
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x2 y2 > 1
x3 y3 > 1
1 ()
1
6.
7. Given any value of x; max minfx; 3y + 2zg = max 3y + 2z:
As
M Uy
py
M Uz
pz
M Uy
py
=
=
>
3
9
2
8
M Uz
)z=0
pz
The optimization problem reduced to
max minfx; 3yg
s:t:x + 9y
=
8
)
x = 3y
x + 9y = 8
)
x=2
y = 32
8. a)
u1 (x; y)
= x+y
u2 (x; y)
=
M RS1
=
M RS2
=
(x + y)2
1
=1
1
2(x + y)
= 1 = M RS1 ; 8x; y > 0
2(x + y)
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Hence, u1 and u2 represent the same consumer preference.
b)
u1 (x; y)
=
x+y
u3 (x; y)
=
M RS3
=
x2 + y 2
x
y
There exist some consumption bundles (x; y) such that M RS3 6= 1; So u1
and u3 do not represent the same consumer preference.
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