ETF2700/ETF5970
Mathematics for Business
Week 12: Unit Review
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Aims of the Unit
Mathematical representations (models) of
business and economics questions
Essential tools to solve mathematical problems
Interpretation of mathematical results
Real-world questions⇒ Mathematics⇒ Real world
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Topics
1 Mathematical preliminaries
2 Linear equations and matrix algebra
3 Calculus
Differentiation
Integration
Differential equation
4 Optimisation of a function of a single variable
5 Optimisation of a function of two/three
variables
6 Constrained optimisation: A function of two
variables
Linear programming
Lagrange method
7 Financial mathematics
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Preliminaries: Modeling
Real numbers
Most economic variables can be represented as real
numbers: in this class all numbers are real.
Integer: 0,1,−1,2,−2, . . .
Fraction: 12 ,−53 , 134513 , . . .
Decimal: 0.2,1.33,−22.15, . . .
Percentage: 20%,133%,−2215%, 53%, . . .
p% = p100 , NOT given in formula sheet
Irrational:
√
2, e, log(2), . . .
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Preliminaries: BEDMAS
HP 10bII+ Calculator
Default is in Chain Mode: NOT BEDMAS
Can change to Algebraic Mode: BEDMAS
Check the Calculator Manual on Moodle
Algebraic mode
7 + 7÷ 7 + 7× 7− 7 = 7 + 1 + 49− 7 = 50
42 × (5 + 6)− 10 = 16× 11− 10 = 176− 10 = 166
Chain mode
‘7 + 7÷ 7 + 7× 7− 7′ = ‘14÷ 7 + 7× 7− 7’
= ‘2 + 7× 7− 7′ = ‘9× 7− 7′ = ‘63− 7′ = 56
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Preliminaries: Percentage change
Not given in the formula sheet
%Change =
New Value−Old Value
|Old Value| × 100%
Percentage change 6= Change to percentage
The interest rate for a saving account was 2.50% in
2014 and 1.05% now.
−58% change in interest rate
1.05%− 2.50%
2.50%
× 100% = −58%
Change to interest rate is 1.05%− 2.50% = −1.45%
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Preliminaries: Miscellaneous
Interval
(a, b) excluding a and b open
[a, b ] including a and b closed
[a, b) including a , excluding b not open, not closed
(a, b ] excluding a , including b not open, not closed
Quadratic equation ax2 + bx + c = 0, a 6= 0
1 b2 − 4ac < 0: no real solutions
2 b2 − 4ac > 0: x1 = −b+
√
∆
2a , x2 =
−b−√∆
2a
3 b2 − 4ac = 0: one solution x = − b2a
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Linear equations and matrix algebra
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Solve a linear equation and
inequality
When a 6= 0
ax + b = 0 ⇔ ax = −b ⇔ x = −b
a
Inequality
a > 0: ax + b≥0 ⇔ ax≥− b ⇔ x≥− ba
No change of sign: Same for ≤, >, <.
a < 0: ax + b≥0 ⇔ ax≥− b ⇔ x≤− ba
Switch the sign direction: similar for ≤, >, <.
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System of two linear equations
Equation Form
a11x + a12y =b1
a21x + a22y =b2
Augmented matrix[
a11 a12 b1
a21 a22 b2
]
Matrix form
Ax = b, A =
[
a11 a12
a21 a22
]
x =
[
x
y
]
,b =
[
b1
b2
]
Using inverse matrix: x = A−1b, if A is invertible
Cramer’s Rule: not provided in the formula
sheet
Elimination method
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Matrix algebra
Provided in the formula sheet:
Inverse of a 2× 2 matrix
NOT provided in the formula sheet:
Determinant of a 2× 2 matrix
Arithmetic Rules for Matrices (sum, subtraction,
and multiplication)
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System of three equations
Equation Form
a11x + a12y =b1
a21x + a22y =b2
a31x + a32y =b3
Augmented matrix a11 a12 b1a21 a22 b2
a31 a32 b3

Forward elimination: Upper triangular form
No solution if the system is inconsistent, i.e. if
the upper triangular form has a row of the
form
[0 · · · 0 b ], where b 6= 0
If consistent: Backward substitution
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Calculus
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Single-variable Function and
Derivative
Linear function
f(x) = mx + c with slope m and intercept c
(Non-linear) functions f(x)
f ′(x) = lim
∆→0
f(x+∆)−f(x)
∆ = slope of the tangent line
Basic functions: Power, Exponential,. . .
Arithmetic rules (sum, subtraction, product,
quotation) and chain rule
Elasticity Elxf(x) =
xf ′(x)
f(x) : interpretation
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Formulae: Differentiation rules
Following formulae are provided in the exam:
Rule f(x) f ′(x)
Power Rule xp , p 6= 0 pxp−1
Constant K 0
Natural Exponential ex ex
Exponential ax ax ln(a)
Logarithm ln(x) 1/x
Product Rule u(x) · v(x) u ′(x) · v(x) + u(x) · v ′(x)
Quotient Rule u(x)v(x)
u′(x)·v(x)−u(x)·v′(x)
(v(x))2
Chain Rule u(v(x)) u ′(v(x)) · v ′(x)
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Indefinite integration
Given f ′(x) = g(x), what is f(x)?
Indefinite integral:
∫
g(x)dx = f(x) + C ,
Outcome is a class of functions with the same
derivative function
Basic functions: power, exponential,
logarithm,. . .
Arithmetic rules: sum, subtraction, product with
constant
Advanced rules: integration by parts,
integration by substitution
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Definite integration
Think of f(x) = C ′(x) as a marginal cost function,
the variable cost for the a-th to b -th units∫ b
a
f(x)dx = C(b)− C(a)
Consumer Surplus
Market price P = P(Q) as a function of demand Q
CS =
∫ Q0
0
(P(Q)− P0)dQ =
∫ Q0
0
P(Q)dQ − P0 ·Q0
P0, Q0 are the current price and demand quantity
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Following formulae are provided in the exam:
Rule f(x)
∫
f(x)dx
Power Rule xp , p 6= −1 xp+1p+1 + C
One exception to power rule x−1 ln(x) + C
Integral of a constant K Kx + C
Natural Exponential ex ex + C
Integration by substitution∫ b
a
f(ϕ(t))ϕ′(t)dt =
∫ ϕ(b)
ϕ(a)
f(x)dx
Integration by parts∫
u(x) · v ′(x)dx = u(x)v(x)−
∫
v(x) · u ′(x)dx
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Differential equation
The unknown is a function rather than a real
variable
dy
dx
= f(x), y =
∫
f(x)dx
Formulae provided in the exam
dy
dx
= ky+c, y = Aekx−c
k
, A is an arbitrary constant
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Two-variable functions f(x, y)
Partial derivatives
fx(x, y): treat y as a known constant, and then
calculate the derive w.r.t. x
fy(x, y): treat x as a known constant, and then
calculate the derive w.r.t. y
Interpretation in view of approximation
Partial elasticity (how to interpret?)
Elxf(x, y) =
xfx(x,y)
f(x,y) , Elyf(x, y) =
yfy(x,y)
f(x,y)
No formulae of partial derivative/elasticity will
be provided.
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Optimisation: Single variable
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Optimisation over an open interval
A function f(x) defined on an open interval (a, b)
with derivative f ′(x).
1 We know there is a maximum/minimum point:
Solve f ′(c) = 0 to find all stationary points in (a, b)
Compare the f(c) and take the one(s) with
largest/smallest value.
2 We don’t know if there is an optimal point, but f
is convex/concave:
f is convex: f ′′(x) > 0 for all x ∈ (a, b)
f is concave: f ′′(x) < 0 for all x ∈ (a, b)
Concave: stationary point(s) is(are) maximum
point(s)
Convex: stationary point(s) is(are) minimum point(s)
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Optimisation over a closed interval
A function f(x) defined on a closed interval [a, b ]
with derivative f ′(x) (on (a, b)).
In this class we only work with continuous function,
so by extreme value theorem we know f has a
maximum and a minimum point.
Solve f ′(c) = 0 to find all stationary points in
(a, b): You may find nothing
Calculate the end-point values f(a) and f(b)
Compare the function values at stationary
points and end-points: the maximum/minimum
is the solution
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Optimisation: Two variables
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Optimisation on an open interval
Suppose you want to maximise/minimise a
two-variable function
f(x, y)
In this unit, we only work with cases that
the domain D is open, for example x, y > 0
the maximum/minimum point of f exists
Solve fx(x, y) = 0 and fy(x, y) = 0 to find the
stationary point(s)
The maximum/minimum value among the
stationary points is (are) the solution(s).
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Optimisation with a constraint
Suppose you want to maximise/minimise a
two-variable function
f(x, y)
subject to a (budget) constraint in form of
px + wy ≤ b
and some physical constraints on x and y .
In this unit: we assume the optimal point exists
f(x, y) = c1x + c2y is linear: Linear
programming
f(x, y) is not linear: Lagrange method
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Linear programming
Max/Min f(x, y) = c1x + c2y
subject to two (or more) linear constraints
a11x + a12y ≤ b1, a21x + a22y ≤ b2, . . .
and usually x ≥ 0, y ≥ 0
1 Extreme point theorem: there is an optimal
point among the corner points
2 The entire set of corner points can be found by
setting any two constraints (of the four) ‘active’.
3 Compare the value at all corner points: the
maximum/minimum one is an optimal solution
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Non-linear Objective Function
maximise/minimise f(x, y) subject to a (budget)
constraint in form of
px + wy ≤ b
with x > 0, y > 0 (or generally, (x, y) ∈ D a open set)
If
the (unconstrained) optimum point satisfies the
budget constraint
then it is also the solution for this constrained
problem.
Otherwise, you need to re-formulate the problem
and solve it by Lagrange Method.
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Re-formulating the problem
When a budget is insufficient, the original problem is
then equivalent to
maximise/minimise f(x, y)
subject to a (budget) constraint in form of
px + wy=b
with x > 0, y > 0 (or generally, (x, y) ∈ D a open
set).
the budget has to be exhausted
In this unit, you are directly given this
formulation. Read the question carefully
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Lagrange Method
Lagrangian function
L(x, y, λ) = f(x, y) + λ(b − px − wy)
where λ is called Lagrange multiplier
Find the stationary points of L(x, y, λ):
Lx(x, y, λ) = 0, Ly(x, y, λ) = 0, Lλ(x, y, λ) = 0
Compare the values at the stationary point(s):
the maximal/minimal one gives your optimal
solution x∗, y∗, and the optimal value f(x∗, y∗)
As a byproduct, you can obtain λ∗, and you
should be able to interpret it.
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Interpreting the Lagrange Multiplier
If the budget increases from b by 1 unit, the
maximal/minimal value will change
approximately by λ∗ units.
If a question asks you: “If the budget is
increased (or decreased) by . . . , compute the
resulting change in the maximal (or minimal)
value of . . . using the Lagrange multiplier
method”, usually both of the following answers
are accepted:
an approximate change using λ∗
an exact change by solving the problem with both
new and old budgets
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Financial Mathematics
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Financial Maths in Exam
Do NOT expect: you just need to plug in the
(correct) formula all the time
A question starts from a real-life problem, and
you need to ‘translate’ it into a mathematical
problem first
Read the questions carefully, and find out the
information you need
If you believe some information is missing, you
need to think of what assumption you
should/can make, state the assumption clearly
and answer the question. It is part of the
assessment requirement.
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Sequence and series
Arithmetic sequence: (a,a + d ,a + 2d , . . .):
sum of the first n(n ≥ 1) terms is
Sn =
n∑
i=1
Ti =
n
2
[2a + (n − 1)d ]
Geometric sequence: (a,ar,ar2, . . .):
sum of the first n(n ≥ 1) terms is
Sn =
n∑
i=1
Ti =
a(1− rn)
1− r
Do not mix up with the interest rate r : here r is the common ratio.
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Interest rate and discounting
Report annual rate rather than sub-period rate
Nominal rate often given in % form: r = 2%
Effective rate
reff =
(
1 +
r
m
)m
− 1,
where m is number of payment periods in a year
Calculator: by default m = 12
When m = 1 (annual payments):
P0 =
PT
(1 + r)T
(in Formulae Sheet)
General: P0 = PT(1+r/m)T , with T number of months
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Present value and IRR
A cash flow stream (x0, x1, . . . , xT)
Cash flows are received at m equal periods per
year
Present value
Given annual (nominal) interest rate r ,
PV = x0 +
x1
1 + r/m
+
x2
(1 + r/m)2
+ . . .+
xT
(1 + r/m)T
Internal rate of return (Annualised)
The interest rate r that solves PV= 0.
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PV and IRR: Explanations
Little (even no) marks on calculator instructions
Write down the formula you used: including
plugging the value of the cash flow stream,
interest rate, number of payment periods, etc.
Often the answer shall be reported in
percentage: please round off to the correct
number of decimal places as required in the
question
By default, the displayed result by a calculator
is already in percentage, rounded to 2 decimal
places
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Suppose the principal L is paid over t years at
interest rate r , with number of payments m per year.
Debt repayment (Formula will be provided)
A0 = L · r/m
1− (1 + rm)−n
where n = m × t is the total number of payments
Advice: give the formula, state all the values of
the input variables, and then provide the result
rounded to the correct decimal places (if
required)
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Depreciation and inflation
Please read the question carefully: straight-line
or reducing balance method?
If the depreciate rate i is not explicitly given:
you need to figure it out.
If no information about inflation: you should not
worry about it, i.e. take ri = 0.
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Difference equation
Such a question may involve solving a difference
equation
Yt+1 = aYt + b
with initial condition of Y0
Often, this equation will NOT be explicitly given:
you will need to read the questions and
formulate it.
You should solve it, step by step, as taught in
our lecture.
No formula will be provided in the exam.
Be careful whether a = 1 or a 6= 1.
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Exam
There are 7 questions, most of which have sub
questions.
Mark allocation is provided.
Exam represents 60% of overall assessment
There is a pass hurdle (see the exam front page)
10 min reading, 2 hours answering, and 30 min
scanning and uploading
It won’t be abnormal if you cannot finish all
questions. Attempt as many as you can.
All unit materials are examinable.
Questions are similar to those in lectures,
tutorial, and assignments.
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Exam tips
Scan or take photos of your handwritten pages,
and then upload them to the submission folder
— similar to the submission way of your second
assignment
Decide the order of questions that you will
answer
Go for the easy marks first — build up your
confidence
Explain your answers whenever necessary:
emphasise main points and write adequately
Read each question carefully for what is
required, and answer it
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Exam preparation
Summarise lectures and tutorials: make some
quick notes for yourself in several A4 pages
DO Exercises (questions)
Review assignments and solutions
Consultation schedule during the 3 weeks of
exam: A schedule to be available on Moodle
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Conclusion words
Thank you so much for taking this unit,
and wish you success in the exam and in the future.
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