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Question1
a)
= 3xy;
=



We set the tangency condition:

=


and the budget line:
+
=
Therefore;
x*(
,
,I) =


,
x*(
,
,I) =

V(
,
,I)=

*


b)
(
,
,I) =

(
(
,
,I) =

(
,
,I) =

(
(
,
,I) =

E(
,
,U) =

( V(
,
,U) =

󰇛

*

)
c)
= 3xy;
=



We set the tangency condition:

=


and the budget line:
+
=
We solve to find:
X =


, Y=


=
, so Tom spends
of his budget on ketchup and
on mustard.
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For I=$18,
=$2, and
=$2, we find
=6,
=3.
d)
Replacing
=$6 instead of
=$2 above, we find
=2,
=3
e)
=
=
=72
We are looking for
and
so that:
= 72

=
󰇛󰇜
=3
=
=

=4.160
f)
Suppose
=$3;
We find
and
=3
but
=3
3
+3
= 72
3(
+
) = 72
+
= 24
=24-3= 21
Tom’s condiments budget is $18; this is how much he spends on bundle A at initial prices, and
bundle C at final prices. To buy basket B at final prices, he would need
= $6 4.160+ $3
4.160 = $37.44.The compensating variation is the difference between I and
:
= 37.44-18=
$19.44
Question 2
a)
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The budget constraint is represented by a straight downward sloping line, of |slope|=
= 2
b)
For the portion where 20 or fewer pretzels are bought, the budget line stays the same as before.
However, the 21
st
pretzel now only costs $2.50*40% = $1 =
of a frozen yogurt. The budget line
on this portion has |slope|=
=5.
When she buys 20 pretzels it costs her 20*2.5=$50, which leaves $10 to spend on yogurt:
x=

=2.
For the price of the 2 remaining cups, Nadia can get 2*
= 10 pretzels in return. So the
intersection with the y axis is at y=20+10=30
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