Economics 211
An example of an expenditure-minimizing consumer and a
utility-maximizing consumer
1. Consider a price-taking, expenditure-minimizing consumer in a two-good world. Denote goods prices by p1 , p2 , and suppose preferences are represented by the following utility
function
1/2
u (h1 , h2 ) = 2h1 + h2 .
(a) Use Lagrange’s method to derive the interior solution for the consumer’s Hicksian
demand functions, hi (p1 , p2 , u0 ), i = 1, 2, and the expenditure function e(p1 , p2 , u0 ). Then
deduce the corner solution from the interior solution.
(b) Show that Hicksian demands are homogeneous of degree zero in (p1 , p2 ) and e(p1 , p2 , u0 )
is homogeneous of degree one in p1 , p2 .
(c) Verify that Hicksian demands do not slope upwards in their own prices.
ANSWER
(a) Using the method of Lagrange multipliers:
1/2
L = p1 h1 + p2 h2 + µ u0 − 2h1 + h2
The first-order conditions for an interior solution are:
∂L
−1/2
= p1 − µh1
=0
∂h1
∂L
= p2 − µ = 0
∂h2
∂L
1/2
= u0 − 2h1 − h2 = 0
∂µ
Take the ratio of equations (1) and (2) and re-arrange:
−1/2
p1
µh1
=
p2
µ
h1 =
p22
.
p21
1
=
1
1/2
h1
or
(1)
(2)
(3)
Substitute this result into equation (3) and solve for h2 .
1/2
h2 = u0 − 2h1
= u0 − 2
p2
.
p1
Then
p22
e (p1 , p2 , u0 ) ≡ p1 h1 (p1 , p2 , u0 ) + p2 h2 (p1 , p2 , u0 ) = p2 u0 − .
p1
The results for the interior solution are the first row of the following table. Note that the
range follows directly from the restriction that demands cannot be negative. In this case
there can be no problem with h1 if prices are positive but h2 could be negative and so prices
and utility must be such that h2 ≥ 0. The corner solution follows from observing that if
utility and prices were such that h2 were negative then h2 must be set equal to zero. And
1/2
then from the utility function u0 = 2h1 or h1 = u20 /4 and then the expenditure function
follows from its definition immediately above.
Solution type
Interior
Corner
h1 (p1 , p2 , u0 )
p22 /p21
u20 /4
h2 (p1 , p2 , u0 )
u0 − 2p2 /p1
0
e (p1 , p2 , u0 )
p2 u0 − p22 /p1
p1 u20 /4
Range
u0 − 2p2 /p1 ≥ 0
u0 − 2p2 /p1 < 0
Note that the envelope theorem holds in this example. In the interior solution
p2
∂e (p1 , p2 , u0 )
= 22 = h1 (p1 , p2 , u0 )
∂p1
p1
∂e (p1 , p2 , u0 )
p2
= u0 − 2 = h2 (p1 , p2 , u0 )
∂p2
p1
And in the corner solution
u2
∂e (p1 , p2 , u0 )
= 0 = h1 (p1 , p2 , u0 )
∂p1
4
∂e (p1 , p2 , u0 )
= 0 = h2 (p1 , p2 , u0 )
∂p2
(b) Check that Hicksian demand functions are homogeneous of degree zero in prices. In
the interior solution
(αp2 )2
p22
=
= h1 (p1 , p2 , u0 )
p21
(αp1 )2
αp2
p2
h2 (αp1 , αp2 , u0 ) = u0 − 2
= u0 − 2 = h2 (p1 , p2 , u0 )
αp1
p1
h1 (αp1 , αp2 , u0 ) =
And in the corner solution
2
u20
= h1 (p1 , p2 , u0 )
4
h2 (αp1 , αp2 , u0 ) = 0 = h2 (p1 , p2 , u0 )
h1 (αp1 , αp2 , u0 ) =
And the expenditure function is homogeneous of degree 1 in prices. For the interior solution
p2
(αp2 )2
= αp2 u0 − α 2 = αe (p1 , p2 , u0 )
e (αp1 , αp2 , u0 ) = αp2 u0 −
αp1
p1
And for the corner solution
e (αp1 , αp2 , u0 ) = αp1
u20
= αe (p1 , p2 , u0 ) .
4
(c) Check that Hicksian demands do not slope upwards in their own prices. For the
interior solution
∂ p22 p−2
∂h1 (p1 , p2 , u0 )
1
=
= −2p22 p−3
1 < 0
∂p1
∂p1
∂h2 (p1 , p2 , u0 )
1
= −2 < 0.
∂p2
p1
And in the corner solution
∂h1 (p1 , p2 , u0 )
∂h2 (p1 , p2 , u0 )
=0=
.
∂p1
∂p2
2. Consider a price-taking consumer in a two-good world. Denote goods prices by p1 , p2
and wealth by w. Write the budget constraint as
w − p1 x1 − p2 x2 ≥ 0.
Suppose preferences are represented by the following utility function
1/2
u (x1 , x2 ) = 2x1 + x2 .
(a) Use Lagrange’s method to derive the consumer’s ordinary demand functions, xi (p1 , p2 , w), i =
1, 2, and the indirect utility function V (p1 , p2 , w) for the interior solution. Then deduce the
corner solution from the interior solution.
(b) Show that the ordinary demands are homogeneous of degree zero in (p1 , p2 , w)
3
ANSWER
(a) Using the method of Lagrange multipliers:
1/2
L = 2x1 + x2 + λ (w − p1 x1 − p2 x2 )
The first-order conditions for an interior solution are:
∂L
−1/2
= x1
− λp1 = 0
∂x1
∂L
= 1 − λp2 = 0
∂x2
∂L
= w − p 1 x1 − p 2 x2 = 0
∂λ
(1)
(2)
(3)
Take the ratio of equations (1) and (2) and re-arrange:
−1/2
x1
1
p1
λp1
=
λp2
p2
p2
=
p1
p22
= 2
p1
=
1/2
x1
x1
Substitute this result into equation (3) and solve for x2 .
w − p1
p22
p21
− p 2 x2 = 0
p22
p1
w
p2
=
−
p2 p 1
p 2 x2 = w −
x2
Then
V (p1 , p2 , w) ≡ u (x1 (p1 , p2 , w) , x2 (p1 , p2 , w))
2 1/2 p2
w
p2
= 2 2
+
−
p1
p2 p1
w
p2
=
+
p2 p1
The results for the interior solution are the first row of the following table. As with question
1, the range follows directly from the restriction that demands cannot be negative, and the
corner solution follows from x2 being zero in such a case.
4
Solution type
Interior
Corner
x1 (p1 , p2 , w) x2 (p1 , p2 , w)
p22 /p21
w/p2 − p2 /p1
w/p1
0
V (p1 , p2 , w)
w/p2 + p2 /p1
2 (w/p1 )1/2
Range
w/p2 − p2 /p1 ≥ 0
w/p2 − p2 /p1 < 0
(b) Checking homogeneity for the interior solution
(αp2 )2
p22
=
= x1 (p1 , p2 , w)
p21
(αp1 )2
αw
αp2
w
p2
x2 (αp1 , αp2 , αw) =
−
=
−
= x2 (p1 , p2 , w)
αp2 αp1
p2 p1
x1 (αp1 , αp2 , αw) =
And in the corner solution
αw
w
=
= x1 (p1 , p2 , w)
αp1
p1
x2 (αp1 , αp2 , αw) = 0 = x2 (p1 , p2 , w)
x1 (αp1 , αp2 , αw) =
Note that by substituting e(p1 , p2 , u0 ) for w, and V (p1 , p2 , w) for u0 , the ordinary demands
become the Hicksian demands and vice-versa.
For the interior solution,
p22
= h1 (p1 , p2 , u0 )
p21
(p2 u0 − p22 /p1 ) p2
p2
x2 (p1 , p2 , e(p1 , p2 , u0 )) =
−
= u0 − 2 = h2 (p1 , p2 , u0 )
p2
p1
p1
x1 (p1 , p2 , e(p1 , p2 , u0 )) =
p22
= x1 (p1 , p2 , w)
p21
w
p2
p2
w
p2
h2 (p1 , p2 , V (p1 , p2 , w)) =
+
−2 =
−
= x2 (p1 , p2 , w)
p2 p1
p1
p2 p1
h1 (p1 , p2 , V (p1 , p2 , w)) =
For the corner solution,
p22
= h1 (p1 , p2 , u0 )
p21
2
1/2
2 pw1
w
=
= x1 (p1 , p2 , w)
h1 (p1 , p2 , V (p1 , p2 , w)) =
4
p1
x1 (p1 , p2 , e(p1 , p2 , u0 )) =
5