Mathematics for Economics
Homework 3
Do the following problems in the Textbook.
Section 7.2 #2 and #4 in p. 213: For d2 Y /dI 2 in #4 (b), you need to use chain rule (since
Y is a function of I) based on the answer (a).
Section 12.2 #7 and #9 in p. 411
Se
Mathematics for Economics - ECON 205
Final Exam
Duration: 2 hours
Assistant Professor DO Quoc-Anh, Singapore Management University
November, 2009
Introduction
This nal examination evaluates your capacity to do simple mathematic calculations and proofs in
Functions of One Variable
[SHS] Chapter 4
1
Intervals
A set is a collection of objects. Each object in the set is called
an element of the set.
An interval is a set of (real) numbers between, and possibly including,
two numbers.
The interval from a to b i
Here are selected questions from examinations from the past few years (some have been modified
slightly). More past exam questions have appeared in your recent assignments.
AT
Question 1 (differentiation)
(a)
Write down the definition of the derivative f
Lecture 8
1. Su cient Condition for Local Optimum: Many Variables
2. Su cient Condition for Global Optimum: One Variable
3. Su cient Condition for Global Optimum: Many Variables
Lecture 8
(Singapore Management University)
Optimization II
2010
1 / 25
Revie
Lecture 10
1. Su cient Condition: Global Optimum
2. Optimization with n Variables and m Constraints
3. Envelope Theorem without constraint
4. Envelope Theorem with constraint
Lecture 10
(Singapore Management University)
Optimization IV
2010
1 / 41
Roadmap
Lecture 11
1. Optimization with Inequality Constraint: Kuhn-Tucker Conditions
2. Su ciency of Kuhn-Tucker Conditions
Lecture 11
(Singapore Management University)
Optimization V
2010
1 / 36
Review of Envelope Theorem
Suppose we maximize
f (x, r ) =
x 2 + 4
Intermediate Mathematics for Economics
Final Exam
Instructions
1. This is a closed book exam.
2. No calculators or mobile phones are permitted to use during the exam.
3. You are bound by the SMU Examination Regulation.
4. The exam is designed to take 2 ho
Lecture 7
1. What is Optimization?
2. Existence of Optimum: The Extreme-Value Theorem
3. Necessary Condition: One variable
4. Necessary Condition: Many Variables
5. Su cient Condition: One Variable
Lecture 7
(Singapore Management University)
Optimization
Lecture 6
1. Integration
2. Interest Rate and Present Discount Value
3. Quadratic Forms
Lecture 6
(Singapore Management University)
PDV and Quadratic Form
2010
1 / 32
1. Intergrals (Ch 9)
What integral (integration)? Consider two functions f (x ) and
s
F
Lecture 4
1. Introduction to Matrices
2. Cramer Rule
s
3. Inverse Matrices
Lecture 4
(Singapore Management University)
Matrices
2010
1 / 39
Solutions of Systems of Equations: Objective
Consider a system of two equations in two variables x and y :
ax + by
Lecture 1
Dierentiation (Chapter 6)
Dierentiability
Rules for Dierentiation
Relationship between Shape of Function and Derivatives
Continuity
Intermediate Value Theorem
Level Curve
1
1. Derivative and Dierentiability
How can we measure the steepnes
1. (10 points) If a function f is continuous on a compact set, then the function f attains
both a maximum and a minimum on the set. For example, consider
max f (x) = x s.t. x [1, 1].
x
The theorem provides only the sucient condition, not necessary conditi
Intermediate Mathematics for Economics
Solution to Sample Questions, 2010
1. A rm maximizes its prot, given xed prices p and w:
max = pf (L, a) wL,
L
where f (L, a) = La and a > 0 is constant.
a. Provide the sucient condition for the solution(s) to exist
Mathematics for Economics
Solution to Homework 1
Do the following problems in the textbook.
Chapter 6.10. #2 and #6 in page 206
Chapter 6.11. #2 and #8 in page 212.
Answers for #9 are in the textbook Appendix.
Chapter 7.4. #6 in page 230
Chapter 7.5
Students Manual
Essential Mathematics for
Economic Analysis
rd
3 edition
Knut Sydster
Arne Strm
Peter Hammond
For further supporting resources please visit:
www.pearsoned.co.uk/sydsaeter
Preface
This students solutions manual accompanies Essential Mathema
Mathematics for Economics
Solution 2
Textbook questions:
Section 7.2 #2 and #4 in page 222-223:
Section 12.2 #6 and #8 in Page 421
Section 12.3 #4 in page 426
1
1. Use the chain rule:
dZ
Q L Q K
=
+
dx
L x K x
= L1 K 1 x1 y 1 + (1 )L K (x1 y 1 )
1
y 1
Mathematics for Economics
Solution to Homework 11
Section 14.8:
#2. Form the Lagrangian:
L = x2 + 2y 2 x (x2 + y 2 1)
a. FOC are
L01 = 2x 1 2x = 0
L02 = 4y 2y = 0
0, x2 + y 2 1, (x2 + y 2 1) = 0
b. Consider two cases: (i) = 0. Then, x = 1 and y = 0. (ii)
Solution to Homework 12
Mathematics for Economics
1. The general solution of x + 1 x =
2
1
4
is
t
1
x = Ce 2 + .
2
The equilibrium state of the equation is
1=C+
y
1
2
and it is stable. When x(0) = 1,
1
1 t 1
1
C = x = e 2 +
2
2
2
2
1.0
0.8
0.6
0.4
0
1
2
Mathematics for Economics
Solution to Homework 10
Section 14.7 #1. Assume 0 a <
m
p
in a consumers problem:
max x + a ln y s.t. px + qy = m
x,y
1. a. Find the solution (x , y ).
L = x + a ln y (px + qy m).
FOC are
L01 = 1 p = 0
a
L02 =
q = 0
y
px + qy =
Mathematics for Economics
Solution 8
13.2 #6
#6. The prot function of a rm is (x, y) = px + qy x2 y 2 where p and q are prices
and x2 + y 2 are costs when the rm produces quantities, x and y.
a. First-order conditions are
01 = p 2x = 0
02 = q 2y = 0.
Thus
Mathematics for Economics
Solution to homework 9
Section 14.1 #4
#4 Find x and y in the following problems:
a. min x2 + y 2 s.t. x + 2y = 4 :
L = x2 + y 2 (x + 2y 4)
L01 = 2x = 0
L02 = 2y 2 = 0
x + 2y = 4
From the rst two equations, = 2x and 2y = 2, we nd
Mathematics for Ecoomics
Solution to Homework 5
Section 12.11. #3
Section 12.11. #4 and #6
Section 15.8. #2, #6 and #8 in page 584-585
1
1. In the equilibrium,
q1 = f (q2 , c1 )
q2 = g(q1 , c2 )
a. The dierentials of the pair of equations are
0
0
dq1 = f1
Mathematics for Economics
Solution to Homework 4
Section 15.2. #4 in page 540
Section 15.3. #2 and #4 in page 544-545
Section 15.4. #5 and #6 in page 551
1
Section 15.5. #2 in page 553
Section 16.1. #4 and #6 in page 576
2
Section 16.2 #6 in page 580: (Fo
Mathematics for Economics
Solution to homework 7
Section 8.4. #2
Section 8.4. # 8
Section 8.5. #6 (Q) is stationary at
Q =
From the second-order condition,
P
ab
1
b1
.
00 (Q) = ab(b 1)Qb2 < 0 for all Q > 0.
Thus, this is a maximum point.
1. The objectiv