comm5005 lecbture 4

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Quantitative Methods
for Business
COMM5005
Lecture 4
Anton Kolotilin

2. 1-1
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In this lecture we will look at how calculus allows us to examine
relationships found in business and the process of optimization.
In particular, we will consider
• The first derivative as a rate of change
• Marginal values
• Elasticity of demand
• Derivatives of multivariable functions
• Using the second derivative as an indicator of concavity
• Identifying maximum and minimum points
4-1
Lecture 4 topics

3. 1-2
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Readings
4-2
Readings for this lecture from Haeussler Paul and Wood:
Chapter Name Pages
11.1-11.2 Rules of Differentiation 482-498
11.3 The derivatives as a rate of change 499-509
12.1-12.2 Derivatives of Logarithmic and
Exponential Functions
532-542
12.3 Elasticity of Demand 543-548
13.1- 13.4,
13.6
Maxima and Minima 570-592,
603-614
17.1 – 17.2 Partial Derivatives 733-744

4. 1-3
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Your company is selling a new
generation smartphone. Previous models have been on the
market for some time and your staff are making estimates
about sales they will have in the (i) first month, (ii) second
month and (iii) twelfth month after the new phone is
introduced.
• Do you expect the sales volume to stay constant over
this period?
• Will sales be increasing or decreasing?
3-3
1. Rates of change

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In this function sales first increase more slowly, then
increase rapidly, reach a peak, flatten out and finally
decline. How do we measure the rate of change?
Time
Sales
3-5
Sales function

6. 1-6
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We could approximate the slope of
the dotted tangent line by calculating
the vertical distance ∆S over the
horizontal distance, ∆ t, but this will
not be very accurate for large
changes in t, i.e. large values of ∆ t.
∆ t
∆S
Time
Sales
3-6
Gradient of the tangent

7. 1-7
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• However if ∆ t is very small then we get a good
approximation for the slope of the tangent at t. The
instantaneous rate of change at t is the limit as the change
in t becomes very small, i.e.
𝑑𝑆
𝑑𝑡
= lim
!"→0
ΔS
Δ𝑡
• This is also called the first derivative of S with respect to t.
3-7
The derivative

8. 1-8
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• Note that the derivative has differing notation that may
be used. If y = f(x) then the first derivative can be
represented as either
$%
$&
or 𝑓' 𝑥 or 𝑦′.
• The sign of the derivative (positive +, negative -, zero)
shows whether the function is increasing, decreasing or
stationary at a point.
3-8
Notation and meaning

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3-9
Differentiation rules

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10
3-
1
0

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11
3-
1
1

12. 1-
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2
12
3-
1
2

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13
Let’s try some practice together to find the derivatives
of these functions
a) 𝑦 = 16𝑥( + 5𝑥
!
" + 8𝑥 − 40
b) 𝑝 = 2𝑞 + 8 )
c) 𝑟 =
!"#$
"!%&"%'(
3-
1
3
Example 1

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14
(𝑑) 𝑦 = 12𝑥𝑒!)"#'
𝐴.
*+
*)
= 72𝑥!𝑒!)"#'
𝐵.
*+
*)
= 12𝑒!)"#'
6𝑥&
+ 1
𝐶.
*+
*)
= 24𝑥,
− 12𝑥 𝑒!)"#!
𝐷. Something else
Find the derivative
3-
1
4

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(𝑒) 𝑟 = 2𝑠&
− 9 15 − ln 𝑠
𝑎𝑡 𝑠 = 4,
*-
*.
equals
𝐴. 1277.17
𝐵. 2249.58
𝐶. 1960.38
𝐷. Something else
Try this question too
3-
1
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• In economics the rates of change of certain functions
are commonly known as marginal values. They
represent the instantaneous rate of change at various
levels of output.
• For example: Marginal revenue, marginal cost,
marginal product
1. Marginal Values

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For a total revenue function, R = f(q), the rate of
change of R with respect to q is called the marginal
revenue
𝑀𝑅 = lim
∆+→,
∆-
∆+
=
$-
$+
We usually interpret
marginal revenue as the
approximate revenue of
one additional unit of output.
R
q
∆𝑅
∆𝑞
R = f(q)
Marginal revenue

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𝑀𝐶 = lim
∆+→,
∆𝐶
∆𝑞
=
𝑑𝐶
𝑑𝑞
We usually interpret marginal
cost as the approximate cost
of one additional unit of
output.
C
q
For a total-cost function, C = f(q), the rate of change
of C with respect to q is called the marginal cost.
∆𝐶
∆𝑞
C = f(q)
Marginal cost

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A publishing company’s revenue for a new travel
guidebook is given by
𝑅 = 2.8𝑞
(
. + 0.5𝑞
where R is total revenue and q is the number of copies
printed. What is its marginal revenue?
MR =
𝑑𝑅
𝑑𝑞
= 2.8
3
4
𝑞/
0
. + 0.5 = 2.1𝑞/
0
. + 0.5
Example 1

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Example 2
A firm’s revenue is given by the function
For what values of q is revenue increasing?
𝑅 = −1.8𝑞#
+ 900𝑞

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The demand for rooms in the Waterview Hotel
is given by the equation 𝑝 = −0.8𝑞 + 200.
a) What is the marginal revenue at q = 40?
A. 136 B. -0.8
C. 168 D. Something else
b) For what values of q is revenue increasing (decreasing)?
Example 3

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A company’s costs are given by the formula
𝐶 =
3𝑞!
0.4𝑞 − 2
+ 6000
a) What is its marginal cost?
b) What is MC at q = 50?
c) How do we interpret the answer?
d) How much different is the marginal cost at q =100?
Example 4

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Average cost is
𝐴𝐶 = 2𝑞!
− 28𝑞 + 200 −
100
𝑞
a) What is marginal cost when q= 80?
b) At what rate is average cost changing when q = 80?
Example 5

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The spread of a computer virus
can be approximated by the formula
𝑁 = 4.25𝑒1.3)"
where N is the number of devices affected and t is time in days.
What is the rate of increase five days after it was released (i.e.
after time zero)?
Example 6

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Example 7
After t years, the value of a principal of $565,768 invested at 5.5% compounded
continuously is given by the formula
(a) Find and interpret it.
(b) What is the rate of change after 4 years?
𝑆 = 𝑃𝑒$%
𝑑𝑆
𝑑𝑡

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• Price elasticity of demand measures the responsiveness of
quantity demanded to changes in price.
• We could measure responsiveness by finding the ratio of the
percentage change in quantity to the percentage change in
price.
𝐸 =
Δq
q ∗ 100%
Δp
p
∗ 100%
=
𝑝
𝑞
Δ𝑞
Δ𝑝
• As ∆p becomes very small we get the point elasticity of demand.
𝜂 = lim
&'→)
Δ𝑞
Δ𝑝
𝑝
𝑞
=
d𝑞
d𝑝
𝑝
𝑞
2. Elasticity of demand

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𝜂 is the Greek letter eta and denotes elasticity.
To determine elasticity we look at the absolute value of η.
If 𝜂 > 1 or 𝜂 < −1, demand is elastic.
If 𝜂 < 1, or −1 < 𝜂 < 0, demand is inelastic.
If 𝜂 = 1, or 𝜂 = −1, there is unit elasticity.
𝜂 =
d𝑞
d𝑝
𝑝
𝑞
Interpreting the elasticity of demand

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Marginal revenue is
MR =
𝑑𝑅
𝑑𝑞
=
𝑑 𝑝𝑞
𝑑𝑞
Use the product rule:
MR =
𝑑𝑝
𝑑𝑞
𝑞 + 𝑝 = 𝑝
𝑑𝑝
𝑑𝑞
𝑞
𝑝
+ 1
so
MR = 𝑝
1
𝜂
+ 1
Marginal revenue and elasticity

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MR = 𝑝
1
𝜂
+ 1
• When demand is elastic, 𝜂 < −1, marginal revenue is
positive, i.e., revenue is increasing in quantity. Thus, a
fall in price and increase in quantity gives an increase in
revenue.
• When demand is inelastic, −1 < 𝜂 < 0, marginal
revenue is negative, i.e., revenue is decreasing in
quantity. Thus, a fall in price and increase in quantity
gives a decrease in revenue.
Prices fall and we know there is unit elasticity of demand.
What happens to revenue?
Marginal revenue and elasticity

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If the demand function for tickets to an all day music event is
𝑞 = 25,000 − 50𝑝
Find the values of p for which demand is
a) Elastic/Inelastic
b) If the current ticket price is $200, should the organizer
increase the price to raise more revenue? What if the
ticket price is $300?
Example 8

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Example 8 Answer
Find where there is unit elasticity (h = –1)
By testing prices above and below 250 we see
For p > 250 h < –1 ( > 1, demand is elastic)
For p < 250 0 >h > –1 ( < 1, demand is inelastic)
𝑑𝑞
𝑑𝑝
= −50 𝜂 =
𝑝
𝑞
×
𝑑𝑞
𝑑𝑝
=
𝑝
25,000 − 50𝑝
×−50
𝜂 =
−50𝑝
25,000 − 50𝑝
−50𝑝 = −25,000 + 50𝑝
100𝑝 = 25,000
𝑝 = 250
𝜂
𝜂

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If the demand function for Mountain bikes is
𝑞 = 30,000 − 2𝑝1
Find the elasticity of demand at p = $100. Interpret the
results.
Example 9

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Now we want to know the rates of change of multivariable
functions.
Example
The production function of a firm is given by Q= 2𝐿
#
$ + 𝐾
#
$,
a) What is the change to production output when the firm
employs one more worker?
b) What is the change to production output when the firm
employs one more unit of capital?
3. Partial derivatives

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If 𝑧 = 𝑓 𝑥, 𝑦 , the partial derivative of 𝑓 with respect to 𝑥 is given by
𝜕
𝜕𝑥
𝑓 𝑥, 𝑦 = lim
∆&→,
𝑓 𝑥 + ∆𝑥, 𝑦 − 𝑓(𝑥, 𝑦)
∆𝑥
the partial derivative of 𝑓 with respect to 𝑦 is given by
𝜕
𝜕𝑦
𝑓 𝑥, 𝑦 = lim
∆%→,
𝑓 𝑥, 𝑦 + ∆𝑦 − 𝑓(𝑥, 𝑦)
∆𝑦
Partial derivatives

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• Essentially, to find
4
4&
𝑓 𝑥, 𝑦 , we treat 𝑦 as a constant
and differentiate 𝑓 with respect to 𝑥 in the usual way.
• To find
4
4%
𝑓 𝑥, 𝑦 , we treat 𝑥 as a constant and
differentiate 𝑓 with respect to 𝑦 in the usual way.
Example
The function 𝑧 = 𝑎𝑥 + 𝑏𝑦, where 𝑎 and 𝑏 are constants,
would have two first partial derivatives:
𝜕𝑧
𝜕𝑥
= 𝑎 and
𝜕𝑧
𝜕𝑦
= 𝑏.
Partial derivatives

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Here is the plot of
𝑧 = 6𝑥1
− 27𝑦 + 30
When we find partial derivatives, the
first step is like making a cut in one
direction. If we keep 𝑥 constant and
differentiate with respect to 𝑦, the
cut surface is a line.
If we keep 𝑦 constant and differentiate with respect to 𝑥,
the cut surface is a parabola.
Second step is to find the gradient of the tangent to the
line or parabola.
Visual representation

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Find the first order partial derivatives of the functions
a) 𝑝 = 3𝑞0
#
$
− 4𝑞1
%
&
+ 60
b) C = 46𝑞0
1
− 15𝑞0 + 52𝑞1
%
$
− 20𝑞1
In some multivariable functions we have to be more careful in
determining the derivatives where there are products of two
variables involved.
Example 10

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A production function is given by the formula 𝑞 =
1
(
𝑙
#
$𝑘
#
&
where 𝑙 is labour and 𝑘 is capital.
a) What is the marginal product of labour?
b) What is the marginal product of capital?
𝐴.
4+
45
=
0
)
𝑙
#
$𝑘
#
& 𝐵.
4+
45
=
0
01
𝑙/
#
$𝑘/
%
&
𝐶.
4+
45
=
0
)
𝑙
#
$𝑘/
%
& 𝐷.
4+
45
=
0
(
𝑙/
#
$𝑘
#
&
Example 11

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A production function is given by the formula
Q =
𝑙𝑘
𝑙 + 2𝑘
Find both marginal productivities, i.e. marginal
product of labour and marginal product of capital.
Example 12

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The demand functions for two products A and B are
given by
𝑞6 = 𝑓 𝑝6, 𝑝7 and 𝑞7 = 𝑔 𝑝6, 𝑝7 .
If
4+'
48(
> 0 and
4+(
48'
> 0, then A and B are substitutes.
If
4+'
48(
< 0 and
4+(
48'
< 0, then A and B are complements.
Complements and substitutes

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The demand functions for two products A and B are given by
𝑞6 = 20 − 5𝑝6 +
(
.
𝑝7 and 𝑞7 = 50 + 2 𝑝6 − 15𝑝7.
Determine if A and B are complements or substitutes?
Example 13

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What happens if there are more variables?
Consider the function 𝑊 = 3𝑥𝑦 + 5𝑦𝑧1
− 6𝑥1
𝑧
Now we obtain three partial derivatives, allowing one
variable at a time to vary and keeping the others constant.
Find
49
4&
,
49
4%
and
49
4:
.
Example 14

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Next, we look at optimisation using second order
derivatives and linear programming.
Notices

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• So far we saw how the first derivative could be used to
show rates of change.
y
x
𝑓! 𝑥 > 0
𝑓!
𝑥 < 0
𝑓! 𝑥 = 0
𝑓! 𝑥 = 0
𝑓! 𝑥 > 0
4. The second derivative and
the shape of a curve

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• If 𝑓*
𝑥 > 0 , which of the following graphs describe the true
shape of 𝑓 𝑥 ?
y
x
𝑓! 𝑥 > 0
𝑓!
′ 𝑥 > 0
y
x
𝑓! 𝑥 > 0
𝑓!′ 𝑥 < 0
y
x
𝑓! 𝑥 > 0
• The second derivative gives different information. In
conjunction with the first derivative it gives a more complete
picture of a function.
𝑓!
′ 𝑥 > 0
then 𝑓!′ 𝑥 < 0
The second derivative and
the shape of a curve

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• The second derivative shows the concavity of the
function, i.e. whether it is upward or downward facing
in a particular section.
• To calculate the second derivative we simply
differentiate the first derivative.
• The notation used is
$$%
$&$ or 𝑦″ or 𝑓″(𝑥)
The second derivative and
the shape of a curve

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CONVEX (CONCAVE UP)
𝑓 is said to be concave up on the interval 𝑎, 𝑏 if 𝑓' 𝑥 is
increasing on 𝑎, 𝑏 , i.e., 𝒇'' 𝒙 > 𝟎 for all 𝑥 in 𝑎, 𝑏 .
y
x
Slope increasing
y
x
y
x
Slope increasing
Slope increasing
𝑓!! 𝑥 > 0
𝑓!! 𝑥 > 0
𝑓!! 𝑥 > 0
Convexity

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48
CONCAVE (CONCAVE DOWN)
𝑓 is said to be concave down on the interval 𝑎, 𝑏 if 𝑓' 𝑥
is decreasing on 𝑎, 𝑏 , i.e., 𝒇'' 𝒙 < 𝟎 for all 𝑥 in 𝑎, 𝑏 .
y
x
Slope decreasing
y
x
y
x
Slope decreasing Slope decreasing
𝑓!!
𝑥 < 0
𝑓!!
𝑥 < 0 𝑓!! 𝑥 < 0
Concavity

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Find the second derivative of the function
𝑅 = −12𝑞! + 80𝑞 + 24
Answer
𝑑𝑅
𝑑𝑞
= −24𝑞 + 80
𝑑!
𝑅
𝑑𝑞! = −24 < 0 for all 𝑞.
Example

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Find the second derivative of the function
𝑦 = −4𝑥( + 15𝑥1 + 48
and identify values of 𝑥 for which it is concave down.
A. −12𝑥1 + 30𝑥, 𝑥 > 2.5
B. −24𝑥 + 30, 𝑥 > 1.25
C. −24𝑥 + 30, 𝑥 < 1.25
D. −12𝑥1 + 30𝑥, 𝑥 < 2.5
Example 1

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Determine where the function is concave up and where
it is concave down. Identify its inflection point.
𝑦 = 𝑥( − 12𝑥1 + 1.
Inflection point: A function 𝑓 has an inflection point at
𝑎 if 𝑓 is continuous and changes concavity at 𝑎.
Example 2: Inflection point

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52
An aspect of business performance that is often under
scrutiny is how to find where peaks and troughs occur.
To find and identify turning points we must look at two
characteristics:
1. The first derivative is zero. Such points are stationary
points, where the function is neither increasing nor
decreasing.
2. The second derivative is then used to help us establish
whether a stationary point is a local (or relative)
maximum or minimum.
5. Identifying maxima and minima

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y
x
𝑓!
𝑥 > 0
𝑓! 𝑥 < 0
𝑓! 𝑥 = 0
𝑓!! 𝑥 < 0
𝑓!
𝑥 = 0
𝑓!! 𝑥 > 0
𝑓! 𝑥 > 0
Identifying maxima and minima

54. 1-
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54
First-derivative test
• Identify all critical values such that 𝑓*
𝑎 = 0.
• For each critical value at which 𝑓 is continuous, there is a local
maximum if 𝑓*
𝑥 changes from + to -, and a local minimum if
𝑓*
𝑥 changes from - to +.
Second-derivative test
• If 𝑓**
𝑎 < 0, i.e., 𝑓 is concave down at
𝑎, then 𝑓 has a local maximum at 𝑎.
• If 𝑓**
𝑎 > 0, i.e., 𝑓 is concave up at 𝑎,
then 𝑓 has a local minimum at 𝑎.
local maximum
local
minimum
Conditions for maximum and minimum

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55
Here are the local maximum and minimum points of a function.
Does this function have an absolute maximum? an absolute
minimum?
In order to see if the turning point is the absolute
minimum/maximum we need to consider the whole function.
Relative extrema and Absolute extrema

56. 1-
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56
Identify all the stationary points of the function
𝑦 = −12𝑥3
+ 6𝑥1
+ 40
a) Are they local maxima or minima?
b) Is there an absolute maximum or minimum at either
of these points?
c) Find the absolute extrema of this function on the
closed interval [0, 2].
Example 3

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57
For the function 𝑦 = 6𝑥3 − 5𝑥.
a) Identify any stationary points on this function.
b) Identify whether these are turning points. If so, are they
local maxima or minima?
c) Is there an absolute maximum or minimum at either of
these points?
Example 4

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59
Suppose that the demand equation for a monopolist’s
product is 𝑝 = 260 − 1.8𝑞 and the average-cost function is
𝑐 = 1.4𝑞 + 4 +
3,
+
, where q is number of units, and both p
and c are expressed in dollars per unit.
a. Determine the level of output and price at which profit
is maximized.
b. Determine the maximum profit.
Example 5

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• Moodle test will be held during the Thursday lecture slot in
Week 5 (45 minutes with an access window 18:00-19:30).
• Topics examined in this test will be Lectures 1-4.
• Open-book mid-term, so textbook and notes are allowed.
• You will need a calculator to solve problems.
1-
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0
Notices