ECON 490 - Y3

Homework 1 - Solution

Carlos HurtadoNumerical Methods: Economics

Due date: Sep 5th 2017 in class

1 [25pt] Utility Maximization

The constant elasticity of substitution (CES) utility function is de�ned as:

U(x, y) = (αxρ + (1− α) yρ)1ρ .

Denote by px and py the prices of goods x and y respectively. The consumer is endowed with income M . Hence, the constraintoptimization problem for the consumer is

maxx,y

U(x, y) subject to pxx+ pyy ≤M.

Derive the demand of good x and y in terms of the parameters α, ρ, the income M and the prices px and py.Solution: The utility maximization problem for the consumer is:

maxx,y

(αxρ + (1− α) yρ)1ρ s.t. pxx+ pyy ≤M.

The Lagrangian associated is

L(x, y, λ; px, py,M) = (αxρ + (1− α) yρ)1ρ + λ (M − pxx− pyy) .

The �rst-order (necessary) conditions are:

∂L∂x

=

(1

ρ

)(αxρ + (1− α) yρ)

1ρ−1ραxρ−1 − λpx = 0 (1)

∂L∂y

=

(1

ρ

)(αxρ + (1− α) yρ)

1ρ−1ρ (1− α) yρ−1 − λpy = 0 (2)

∂L∂λ

=M − pxx− pyy = 0 (3)

Dividing equation (1) by equation (2) we get:(1ρ

)(αxρ + (1− α) yρ)

1ρ−1ραxρ−1(

1ρ

)(αxρ + (1− α) yρ)

1ρ−1ρ (1− α) yρ−1

=λpxλpy

orαxρ−1

(1− α) yρ−1=pxpy

Then (x

y

)ρ−1

=(pxα

)( py1− α

)−1

or

y =(pxα

)− 1ρ−1

(py

1− α

) 1ρ−1

x.

Now substitute this expression for y into equation (3) (budget constraint):

M = pxx+ pyy

= pxx+ py(pxα

)− 1ρ−1

(py

1− α

) 1ρ−1

x

= x

(px + py

(pxα

)− 1ρ−1

(py

1− α

) 1ρ−1

)The demand function for good x is

x (px, py,M) =M

px + py(pxα

)− 1ρ−1

(py

1−α

) 1ρ−1

1

Simplifying we get

x (px, py,M) =M(pxα

) 1ρ−1

px(pxα

)− 1ρ−1 + py

(py

1−α

) 1ρ−1

By symmetry, the demand for good two is

y (px, py,M) =M(py

1−α

) 1ρ−1

px(pxα

)− 1ρ−1 + py

(py

1−α

) 1ρ−1

2 [25pt] A Toy Economy

Consider an economy with two goods. Let X denote leisure and Y denote food. The utility of the consumer is given by U(X,Y ) =

ln(X) + Y . The consumer has a total of 20 hours for leisure as initial endowment. The technology to produce food is Y = 2L12 ,

where L = 20−X, is the total number of hours that the consumer can work. The social planner problem is

maxx,y

U(X,Y ) s.t. Y = 2L12 & L = 20−X.

Find the optimal allocation of the social planer.Solution: The Social planner problem is

maxx,y

ln(X) + Y s.t. Y = 2L12 & L = 20−X

that is,

maxx,y

ln(X) + Y s.t. Y = 2 (20−X)12

ormaxx

ln(X) + 2 (20−X)12 .

The FOC is1

X− (20−X)−

12 = 0

or

1

X=

1

(20−X)12

1

X2=

1

20−X

equivalent to

X2 +X − 20 = 0

(X + 5) (X − 4) = 0

we discard the negative values, soX = 4; L = 16

3 [25pt] Three Equations in Three Unknowns

Use eliminations to reach upper triangular matrices. Solve by back substitution or explain why this is impossible. Exchangeequations when necessary. The only di�erence is the −x in the last equation

x+ y + z = 7

x+ y − z = 5

x− y + z = 3

x+ y + z = 7

x+ y − z = 5

−x− y + z = 3

Solution:

The extended system for the �rst set of equations is

2

1 1 1 71 1 -1 51 -1 1 3

So, multiply row 2 and 3 by −1, and add row 1 to row 2 and 3:

1 1 1 70 0 2 20 2 0 4

Exchange row 2 and row 3:

1 1 1 70 2 0 40 0 2 2

Then, back substitution gives z = 1, y = 2 and x = 4.The extended system for the second set of equations is

1 1 1 71 1 -1 5-1 -1 1 3

So, multiply row 2 by −1, and add row 1 to row 2 and 3:

1 1 1 70 0 2 20 0 2 10

By back substitution we get z = 5 and z = 1, so the system has a permanent failure (no solution).

4 [25pt] Inverse Matrix

Find the inverse matrix of the following matrices:

A =

0 0 0 20 0 3 00 4 0 05 0 0 0

B =

3 2 0 04 3 0 00 0 6 50 0 7 6

Solution: We get

A−1 =

0 0 0 1

2

0 0 14

00 1

30 0

12

0 0 0

B−1 =

3 −2 0 0−4 3 0 00 0 6 −50 0 −7 6

3