1
The description of the economy
x = f (k, l) = k 1/2 + l1/2
y = f (k, l) = k 1/2 l1/2
uA (x, y) = xy
uB (x, y) = x + y
Person A owns firm x
Person B owns firm y and 10 units of k and of l
Normalize:
px = 1
2
Profit maximization
2.1
Profit maximization for firm producing y:
max py k1/2 l1/2 − pk k − pl l
Since the production function is constant returns to scale ("takal") we know
that in a competitive equilibrium, if the firm produces then it is indifferent about
the level of production. Specifically, if profits = py k 1/2 l1/2 − pk k − pl l > 0 for
some k ∗ , l∗ then by doubling to 2k ∗ and 2l∗ the firm will double its profits, and
hence at these prices the firm will produce an infinite amount and therefore this
cannot happen in a competitive equilibrium. So either the firm does not produce
and py k1/2 l1/2 − pk k − pl l ≤ 0, or it is indifferent about the level of production,
and then py k 1/2 l1/2 − pk k − pl l = 0. Let’s continue to solve assuming it does
produce. (We know that the firm will produce if person A has any income,
because if A has income then A will demand some y so in equilibrium the firm
must produce.)
So we have the equation
¡ ¢1/2 ¡ D ¢1/2
py kyD
− pk kyD − pl lyD = 0
ly
(1)
and of course the production equation
¡ D ¢1/2 ¡ D ¢1/2
ky
= yS
ly
(2)
Taking derivatives and set equal to zero gives: py M Pk = pk and py M Pl =
p l1/2
p k1/2
y
y
= pk , and 2l
= pl . Since the level of k and l are
pl . That is 2k
1/2
1/2
not determined by these conditions, because of the constant-returns-to-scale
technology discussed in the preceding paragraph, we know that these equations
only determine the relative use of k and l. So dividing one by the other we get
py l1/2 py k1/2
/ 2l1/2
2k1/2
= pk /pl , or
D
ly
kyD
=
pk
pl ,
so
lyD =
pk D
k
pl y
1
(3)
We can substitute equation 3 into equations 1 and 2 to obtain simpler relations.
¡ ¢1/2 ¡ D ¢1/2
¡ ¢1/2 ³ pk D ´1/2
−pk kyD −pl lyD = 0 so py kyD
−pk kyD −
ly
First, py kyD
pl ky
³
´
³ ´1/2
³ ´1/2
kyD − 2pk kyD = 0, so py ppkl
− 2pk = 0, so
pl ppkl kyD = 0, so py ppkl
³ ´1/2
, so
py = 2pk ppkl
1
(4)
(py )2 = pk pl
4
³ ´
¡ ¢1/2 ¡ D ¢1/2 ¡ D ¢1/2 ³ pk D ´1/2
D pk
Second y S = kyD
ly
= ky
k
=
k
y
pl y
pl . Thus
S
y =
2.2
kyD
µ
pk
pl
¶
(5)
Profit maximization for firm producing x:
¡
¢
max px k 1/2 + l1/2 − pk k = pl l
Solution (recall px = 1): 2k11/2 = pk and
kxD =
1
2l1/2
= pl , so
1
1
and lxD = 2
4p2k
4pl
(6)
Therefore
¡
¢
x = f kxD , lxD =
S
s
1
+
4p2k
s
1
1
1
=
+
2
4pl
2pk
2pl
(7)
and
π x = px x − pk k − pl l =
3
3.1
1
1
pk
pl
1
1
+
− 2 − 2 =
+
2pk
2pl
4pk
4pl
4pk
4pl
(8)
Consumer utility maximization
Consumer A
This person spends half his income on each good. His income is
his demand for x and y is
µ
¶
1
1
1
1
A
xD =
+
+
/2px =
4pk
4pl
8pk
8pl
¶
µ
1
1
1
1
A
yD =
/2py =
+
+
4pk
4pl
8py pk
8py pl
2
1
4pk
+
1
4pl ,
so
(9)
3.2
Consumer B
This person buys only y if py < 1, only x if py > 1 and is indifferent as to how
to spend her income if py = 1. Her income is 10pk + 10pl . We will look for an
interior solution where py = 1, and if this fails try an alternative. So for now
py
10pk + 10pl
= 1 and
D
= px xD
B + py yB
(10)
We will substitute this into all the equations above, and therefore will only
have pk and pl as variables to find. In particular, using equation 4 we have
pk pl = 1/4, so
1
(11)
pl =
4pk
4
4.1
Market clearing
Market for x
³
´
1
1
D
S
D
S
D
xD
+
x
=
x
,
so
using
equation
9
we
have
x
=
x
−
x
=
+
A
B
B
A
2pk
2pl −
³
´
1
1
so
8pk + 8pl
3
3
3
3pk
+
=
+
xD
(12)
B =
8pk
8pl
8pk
2
where the last equality uses equation 11. We continue to substitute equation
11 wherever possible below, and hence do not specify this each time.
D
Therefore by equation 10 we have yB
= 10pk + 10pl − 8p3k − 8p3 l = 10pk +
³ ´
17
10 4p1k − 8p3k − 31 = 17
2 pk + 8pk . Thus
8
4pk
D
=
yB
4.2
17
17
pk +
2
8pk
Market for y
³
D
D
yA
+ yB
= y S , so using equations 13 and 9 we have that y S = 17
2 pk +
³
´ ³
´ ³
´
1
1
17
17
1
1
9
8pk + 8pl =
2 pk + 8pk + 8pk + 8(1/(4pk )) = 9pk + 4pk
Therefore, using 5 we have
µ ¶
µ
¶
9
pk
pk
= kyD
= kyD
9pk +
= 4kyD p2k
4pk
pl
1/ (4pk )
4.3
(13)
17
8pk
´
+
(14)
Market for k
kxD + kyD = 10, so kyD = 10 − kxD , so using equation 6 we have
kyD = 10 −
3
1
2
4 (pk )
(15)
5
Putting the last two equations together and
obtaining all the values
From equations 14 and 15 we have
Ã
!
9
1
9pk +
= 4 10 −
(pk )2
2
4pk
4 (pk )
(16)
The only real solution is pk = 12 . So then using equation 11 we have
pl = 1/2.
Substituting this in all the other equations yields the following.
y
S
yS
kyD
lyD
D
yB
xD
B
xA
D
A
yD
xS
lxD
Check
µ
¶1/2 µ µ
¶¶1/2
1
pk
1
=
10 − 2
=9
10 − 2
4pk
pl
4pk
1
1
= 10pk + 10pl −
−
=8
2pk
2pl
1
= 10 − 2 = 9
4p
µ k
¶
pk
1
=
10 − 2 = 9
pl
4pk
3
3
= 10pk + 10pl −
−
= 8.5
8pk
8pl
3
3
=
+
= 1.5
8pk
8pl
1
1
=
+
= 0.5
8pk
8pl
1
1
=
+
= 0.5
8pk
8pl
1
1
=
+
=2
2pk
2pl
1
=
=1
4p2l
:
lxD + lyD = 1 + 9 = 10 = lS
4