Solution Guide to Problem Set 1
Econ 121 Spring 2022
Instructor: Prof. Roland Strausz
Due: 02/10/2022
Problem 1
2
max U (x; a) = max e−(x−a)
x
x
Part 1.1
The first-order condition (FOC) is:
∂U (x; a)
=0
∂x
2
⇒e−(x−a) (−2(x − a)) = 0
2
⇒e−(x−a) (x − a) = 0
(1)
Part 1.2
Using (1) above,
2
e−(x−a) (x − a) = 0
2
⇒(x − a) = 0 [∵ e−(x−a) ̸= 0 for any real value of x]
⇒x∗ = a
(2)
1
Part 1.3
The second-order condition (SOC) is:
∂ 2 U (x; a)
∂x2
∂ ∂U (x; a)
(
)
∂x
∂x
∂ −(x−a)2
=
(e
(−2(x − a)))
∂x
=
2
2
= −2(e−(x−a) − 2(x − a)2 e−(x−a) )
2
= (4(x − a)2 − 2)e−(x−a)
(3)
Now, evaluating the expression in (3) at x∗ = a, we get
∂ 2 U (x; a)
|x∗ =a = −2 < 0
∂x2
(4)
Since the second derivate evaluated at the critical point (x∗ = a) is negative, the critical point is a local
maximum.
Part 1.4
dx∗
da
=
=1
da
da
Part 1.5
The maximum value of U (x; a) evaluated at the local maximum x∗ = a is given by:
2
U (x∗ ; a) = e−(a−a) = e0 = 1
Therefore,
∂U (x∗ ; a)
∂
=
1=0
∂a
∂a
Problem 2
max U (x1 , x2 ; a, b) = max ax21 + bx22 + (x1 + 1)x2
x1 ,x2
x1 ,x2
2
Part 2.1
The first order conditions with respect to x1 and x2 are:
∂U (x1 , x2 ; a, b)
= 2ax1 + x2 = 0
∂x1
∂U (x1 , x2 ; a, b)
= 2bx2 + x1 + 1 = 0
∂x2
(5)
(6)
Part 2.2
We can solve the system of two equations in (5) and (6) by using the method of substitution. To do this,
we first solve for x2 from (6) as x2 = − x12b+1 . Then we plug this value of x2 into (5) to get 2ax1 =
This gives us x∗1 =
1
4ab−1 .
x1 +1
2b .
2a
Then x∗2 = − 4ab−1
. These are the critical points.
Part 2.3
For the second order conditions, first note that the Hessian matrix is:

 
∂2U
U11 U12   ∂x21

= 2
∂ U
U21 U22
∂x1 ∂x2



2a 1 
=

1 2b
∂2U
∂x2 ∂x1 
∂2U
∂x22
Then the sufficient second-order conditions for the critical point to be a local maximum are: U11 < 0 and
1
2a
(U11 U22 − U12 U21 ) > 0. Therefore, if we want ((x∗1 , x∗2 ) = ( 4ab−1
, − 4ab−1
)) to be local maximum, we need
2a < 0 ⇒ a < 0. And, from the second condition, we also need 4ab − 1 > 0 ⇒ ab >
Thus, the joint conditions on a and b are that a < 0 and b <
1
4
⇒b<
1
4a
[∵ a < 0].
1
4a .
Part 2.4
∂x∗2
∂
−2a
(4ab − 1)2 − 8ab
2
=
(
)=−
=
>0
∂a
∂a 4ab − 1
(4ab − 1)2
(4ab − 1)2
Therefore x∗2 increases as a increases.
Part 2.5
2
1
4a
We know U (x∗1 , x∗2 ; a, b) = a (4ab−1)
2 + b (4ab−1)2 − (1 +
Therefore,
dU (x∗1 ,x∗2 ;a,b)
da
=
1−4ab−a(−4b)
(1−4ab)2
=
1
(1−4ab)2
1
2a
4ab−1 ) 4ab−1
=
a
1−4ab .
> 0. So the maximum value increases as a increases.
3
Problem 3
1
3
max U (x1 , x2 ) = x14 x24
x1 ,x2
subject to p1 x1 + p2 x2 = I
Part 3.1
To simplify our calculations, we will consider the following increasing transformation of the utility function: V (x1 , x2 ) = (U (x1 , x2 ))4 = x1 x32 . As we saw in lectures, because an increasing transformation
preserves the underlying preference denoted by the utility function, this will yield the same demand.
The Lagrangian for the consumer’s utility maximization problem is then
L(x1 , x2 , λ; p1 , p2 , I) = x1 x32 + λ(p1 x1 + p2 x2 − I)
(7)
Part 3.2
The FOCs are:
Using (8) and (9), we get
3x1
x2
=
p2
p1 .
∂L
= x32 + λp1 = 0
∂x1
(8)
∂L
= 3x1 x22 + λp2 = 0
∂x2
(9)
∂L
= p 1 x1 + p 2 x2 − I = 0
∂λ
(10)
Using this condition together with (10) gives us (x∗1 =
I
∗
4p1 , x2
=
3I
4p2 ).
This is the critical point.
Part 3.3
To ensure that the critical point is indeed the solution to the consumer’s maximization problem, we need
to check the second-order conditions.
4
Approach 1: Check that the determinant of the bordered Hessian matrix is positive at (x∗1 , x∗2 ). The
bordered Hessian matrix is:
 

L
L
L
1λ
2λ 
 λλ
 0
  2

  ∂ L
L
 λ1 L11 L12  =  ∂λ∂x1
 

∂2L
Lλ2 L21 L22
∂λ∂x2
∂L2
∂x1 ∂λ
∂2L
∂x21
∂2L
∂x2 ∂x1
 

0
p
p
0
p
p
1
2 
1
2
 

 

 
2
∂ L =
 = p
2 
0
3x
p
V
V
1
1
11
12




2 
∂x1 ∂x2
 

 
∂2L
2 6x x
p
V
V
p
3x
2
21
22
2
1 2
2
∂x2
∂L2
∂x2 ∂λ


(11)
2
The determinant of this is given by:
0 − p1 (p1 6x1 x2 − 3x22 p2 ) + p2 (p1 3x22 ) = 6p1 x2 (p2 x2 − p1 x1 )
Evaluating this at (x∗1 =
I
∗
4p1 , x2
=
3I
4p2 ),
6p1 x∗2
(12)
we get
I
3I
−
4
4
= 3p1 x∗2 I > 0,
(13)
as required.
Approach 2: Equivalently, since the constraint takes the form p1 x1 + p2 x2 − I = 0, we know from
lectures that the sufficient condition simplifies to the requirement that at (x∗1 , x∗2 ), we have
p21 V22 − 2p1 p2 V12 + p22 V11 < 0.
Thus, at (x∗1 , x∗2 ), we want
p21 6x1 x2 − 2p1 p2 3x22 = −6p1 x2 (p2 x2 − p1 x1 ) < 0.
This is equivalent to what we verified in (13) above.
Part 3.4
From the optimal solutions we can write p1 x∗1 =
on good 1, and the remaining
3
4
1
4 I,
p2 x∗2 =
3
4 I.
Therefore,
1
4
of the income is spent
is spent on good 2. This is a consequence of the Cobb-Douglas utility
function, where the exponents on x1 , x2 specify the fraction of total income that would be spent on each
good.
5
Part 3.5
∂x∗1
∂
I
=
(
)=0
∂p2
∂p2 4p1
∂x∗2
∂ 3I
3I
=
(
)=− 2 <0
∂p2
∂p2 4p2
4p2
Therefore, if p2 increases then the consumption of good 1 does not change, but the consumption of good
2 decreases.
Part 3.6
Suppose income doubles, and the new income I ′ = 2I. Then the optimal consumption values can be found
2I 3∗2I
I
3I
∗∗
by replacing I by I ′ . The new optimal consumption values will be (x∗∗
1 , x2 ) = ( 4p1 , 4p2 ) = ( 2p1 , 2p2 ).
The proportions of income spent on each good does not change. This can be seen by replacing I with I ′
in the consumer’s budget constraint. This replaces all instances of I in the FOCs, SOCs with I ′ , and the
final solutions will turn out to be (x∗∗
1 =
I′
∗∗
4p1 , x2
=
3I ′
4p2 ).
Part 3.7
To solve using the method of substitution, we first find the value of x2 in terms of x1 and other parameters
of the problem using the consumer’s budget constraint. This gives us:
x2 =
p1
I
− x1
p2 p2
(14)
Now we can write the consumer’s utility maximization problem as:1
1
max U (x1 ) = x14 (
x1
3
p1
I
− x1 ) 4
p2 p2
1
Again, we could perform an increasing transformation, for example Ũ (x1 ) = x1 ( pI2 −
below.
6
(15)
p1
x )3 ,
p2 1
to simplify calculations
The FOC is:
3
1
p1
p1
1 − 43 I
3 1 I
p1
x1 ( − x1 ) 4 + x14 ( − x1 )− 4 (− ) = 0
4
p2 p2
4
p2 p2
p2
3
1
3
1
1 − I
I
p1
p1
3
p1
⇒ x1 4 ( − x1 ) 4 = x14 ( − x1 )− 4
4
p2 p2
4
p2 p2
p2
p1
I
⇒ p2 ( − x1 ) = 3p1 x1
p2 p2
(16)
⇒ I − p1 x1 = 3p1 x1
⇒ 4p1 x1 = I
I
4p1
I
p1 I
3I
⇒ x∗2 =
−
=
using (15)
p2 p2 4p1
4p2
⇒ x∗1 =
These are exactly the same answers we found in part 2.2
Problem 4
max U (x1 , x2 ) = 10x31 x42
x1 x2
subject to p1 x1 + p2 x2 = I
Part 4.1
The Lagrangian function is
L(x1 , x2 , λ; I, p1 , p2 ) = 10x31 x42 + λ(p1 x1 + p2 x2 − I)
The three FOCs are:
∂L
= 30x21 x42 + λp1 = 0
∂x1
∂L
= 40x31 x32 + λp2 = 0
∂x2
∂L
= p 1 x1 + p 2 x2 − I = 0
∂λ
2
(17)
(18)
(19)
To verify that the second order conditions also hold, you can take the second derivate of the modified utility function
with respect to x1 and then evaluate the second derivative at the point x1 = 4pI 1 . The sign of this should be negative.
7
Analogous to Problem 3, using the first two FOCs we get
x1
x2
=
3p2
4p1 .
Plugging this into (19), we get
3I
(x∗1 , x∗2 ) = ( 7p
, 4I ). Substituting the values of I, p1 , p2 we get (x∗1 , x∗2 ) = (2, 2). The resulting level of
1 7p2
utility is U (2, 2) = 10(2)3 24 = 1280.
Part 4.2
For the second consumer we can write the Lagrangian function as:
L = 3log(x1 ) + 4log(x2 ) + λ(p1 x1 + p2 x2 − I)
The three FOCs are:
∂L
3
=
+ λp1 = 0
∂x1
x1
4
∂L
=
+ λp2 = 0
∂x2
x2
∂L
= p1 x1 + p2 x2 − I = 0
∂λ
Using the first two FOCs we get
x1
x2
=
3p2
4p1 .
(20)
(21)
(22)
Then using this and the third FOC we get the optimal
3I
4I
∗∗
solutions (x∗∗
1 , x2 ) = ( 7p1 , 7p2 ) = (2, 2).
The optimal consumption bundle of the second consumer is exactly equal to the optimal consumption
bundle of the first consumer when they face the same prices and have the same income. This is because
the underlying preferences of the two consumers are the same. This can be proved by showing that the
utility function of the second consumer is a strictly increasing transformation of the utility function of
the first consumer. More specifically,
V (x1 , x2 ) = 3log(x1 ) + 4log(x2 )
= log(x31 x42 ) + log(10) − log(10)
= log(10x31 x42 ) − log(10)
= log(
U (x1 , x2 )
)
10
u
Here the transformation is v = log( 10
). It can be show that this is a strictly increasing transformation
because
∂v
∂u
=
1
u
> 0 for all u > 0.
8
Part 4.3
Denote the two utility functions as U (x1 , x2 ) = 10x31 x42 , and V (x1 , x2 ) = 3log(x1 ) + 4log(x2 ). Then the
marginal utility of good 1 for the two utility functions evaluated at x2 = 1 are:
∂U (x1 , x2 )
∂x1
∂V (x1 , x2 )
∂x1
x2 =1
x2 =1
= 30x21 x42
=
3
x1
x2 =1
x2 =1
=
= 30x21
(23)
3
x1
(24)
To see when the marginal utility of good 1 increases or decreases in the consumption of good 1, we take
the derivative of the marginal utilities with respect to x1 .
∂2U
= 60x1 > 0 for all x1 > 0
∂x21
∂2V
3
= − 2 < 0 for all x1
2
∂x1
x1
Therefore, for the second utility function marginal utility is always decreasing, and for the first utility
function marginal utillity is always increasing (assuming that consumption values can never be negative).
Part 4.4
Indifference curves for the two utility functions can be represented as
10x31 x42 = c̄
3log(x1 ) + 4log(x2 ) = ĉ,
where c̄, ĉ are arbitrary constants.
1
−3
c̄ 4
) x1 4 . Then the slope of the first
To find the slope of the first indifference curve we write x2 = ( 10
indifference curve is:
dx2
dx1
1
−7
c̄ 4
= − 34 ( 10
) x1 4 . The first thing to note is that the slope is negative, which
means that the indifference curve is downward sloping. To show how this slope changes with the value
of x1 , we take another derivative with respect to x1 as follows:
d2 x2
dx21
slope of the first indifference curve increases with the value of x1 .
9
=
11
21 c̄ 14 − 4
16 ( 10 ) x1
> 0. Therefore, the
To compute the slope of the second indifference curve we do the following:
3log(x1 ) + 4log(x2 ) = ĉ,
⇒ log(x31 x42 ) = ĉ
⇒ x31 x42 = eĉ
ĉ
− 34
⇒ x2 = e 4 x1
⇒
3 ĉ − 7
dx2
= − e 4 x1 4 < 0
dx1
4
So, the slope is again negative for the second indifference curve. To find how the slope changes with x1 ,
we take a second derivative with x1 .
d2 x2
21 ĉ − 11
= e 4 x1 4 > 0
2
16
dx1
So, the slope again increases with x1 .
The fact that the indifference curves are downward sloping and that their slope is increasing in x1 implies
that the upper contour sets are (strictly) convex (see lecture 5 for more details). Therefore, the two utility
functions are (strictly) quasiconcave. This property provides an alternative formalization of the notion
of “diminishing marginal utility,” and unlike the requirement that
∂2U
∂x21
< 0, this property is preserved
under increasing transformations.
An indifference curve that violates this property can be drawn as follows. Here the slope of the curve
decreases as the value of x1 increases.
10
In the figure below, we show two indifference curves for the same utility function. Under monotonicity,
the curve in orange provides higer utility than the one in black. The line in blue is the budget constraint.
If we solve the first order conditions using this utility function, then we will get point A as the criticial
point, where the black indifference curve is tangent to the budget constraint. However, as shown both
points B and C on the budget constraint provide higher utility than point A. In fact, among all indifference
curves that can be drawn here, the one in black provides the least utility subject to the budget constraint.
Therefore, if we just select the critical point (solution to FOCs) as the solution we will end up minimizing
and not maximizing utility. However, if we check the SOCs then we can show that point A is not the
maximum point.
The take-away from this exercise is that unless we have indifference curves whose slopes increase
(equivalently, utility functions that are quasiconcave), we cannot directly use the critical point as the
utility maximizing bundle. If second order conditions for a maximum are to hold then the upper contour
sets have to be convex, which is exactly what this property ensures.
11