Optimisation

Optimisation



Problems and Solutions



with your fx-991ES or fx-


115ES Calculator












Dr Allen Brown








Cambridge 
Paperbacks

Cambridge Paperbacks

www.CambridgePaperbacks.com


First published by Cambridge Paperbacks 2019



© Allen Brown 2019

All rights reserved. No part of this publication may be reproduced or
transmitted in any form or by any means, electronic or mechanical, including
photocopy, recording, or any information storage and retrieval system without
permission in writing from the author.



Disclaimer

Although the author and publisher have made every effort to ensure that the
information in this book was correct during preparation and printing, the
author and publisher hereby disclaim any liability to any party for any errors
or omissions.

Read this First


Optimisation now features in some A Level Maths

syllabus and this flipbook offers you a brief
introduction to the subject. In particular it shows you

how you can use your fx-991ES or fx-115ES calculator
to further your understanding of the subject.


In optimisation problems you will be expected to find

the maximum or minimum value of a variable from
differentiating the objective function. This leads to the

critical value – the stationary value of the objective

function.

Once you have performed differentiation on the

objective function you can use the Y feature on

your calculator to determine whether it’s a minimum
or maximum value.


Since optimisation involves the extensive use of

numbers, you will most certainly find your fx-991ES or
fx-991ES calculator a key component in your skill set

development.

The flipbook is in two sections, the first section

considers general optimisation problems whereas the

second section looks at commercial examples in

particular.

The final two examples in the second section show

you how to arrive at a linear model for the sales of
tickets for an organised event.


Artificial Intelligence is receiving a lot attention these

days and you may be interested to know that
optimisation algorithms are widely used in AI systems.

If you are likely to pursue a career in AI in the future,

you will find this brief introduction useful; however it
is only one of the many mathematical devices used in

AI processes.




Dr Allen Brown

Cambridgeshire

Contents

1 Optimisation .............................................................................................. 2
Example 1 ..................................................................................................... 3

Example 2 ..................................................................................................... 4
Example 3 ..................................................................................................... 7

Example 4 ................................................................................................... 10
Example 5 ................................................................................................... 12

Example 6 ................................................................................................... 15
Example 7 ................................................................................................... 17
Example 8 ................................................................................................... 19

Example 9 ................................................................................................... 21
Example 10 ................................................................................................. 24

2 Optimising Commercial Performance ......................................................... 27
Example 1 ................................................................................................... 27

Example 2 ................................................................................................... 29
Example 3 ................................................................................................... 31

Example 4 ................................................................................................... 33
Example 5 ................................................................................................... 35
Example 6 ................................................................................................... 36

Example 7 ................................................................................................... 38
Example 8 ................................................................................................... 40

Example 9 ................................................................................................... 41
Example 10 ................................................................................................. 46




1

1 Optimisation

In many circumstances in life you will encounter
problems where you are expected to make the most

of the resources you have available. That is, you are

expected to optimise one or other resource. You will
also have constraints imposed which form part of the

model. Here is the general procedure for solving an

optimisation problem.

1. Draw a diagram if possible and attach the

information on the diagram.

2. Find the objective function which you need to
maximise or minimise.

3. Determine the constraints that are going to
affect your objective function.

4. Try and reduce the objective variable so that is

dependent on one variable only.
5. Differentiate the objective variable and equate

it to zero.

6. Find the critical values of the variables – these
are usually solutions of the differentiated

objective function.

7. Test the critical numbers by using the second
derivative leading to the correct solution.

2

From a mathematical point of view, you will need to

determine the objective function and then find its
maximum or minimum using differential calculus.


We shall start with a well known example of
maximising the area enclosed by a fence.



Example 1

Given a length of fence of length E (320 metres) what

is the maximum rectangular area that can be enclosed
by the fence?


The area is shown on the left, from
the diagram we can see that,


= + + + = 2( + ) 1


The area is

= 2


The constraint is E = 320 metres. From Eq:1,


= − 3
2

The area becomes,


2
= ( − ) = − 4
2 2
3

To maximise the area, differentiate with respect to x,


= − 2 = 0 5
2

Therefore the critical value is,

320
= = = 80 6
4 4
From Eq:3,


320
= − = − 80 = 35 7
2 2

The maximum rectangular area
that can be enclosed by the

fence is therefore,

2
= 80 × 35 = 2,800 m




Example 2

The diagram on the left shown a rectangle of side
lengths 27 cm with the corners

removed (of length x). The idea

is to fold the material along the
dotted lines to form a tray.

Therefore select a value of x



4

which maximises the volume of the tray as you can

see in the diagram on the
left.


From this diagram, you can
see the volume of the tray

(the objective function) is,

2
= (27 − 2 ) 1

Expand this expression,


2
= (729 − 108 + 4 ) 2

Leaving,

3
2
= 4 − 108 + 729 3
To find the maximum volume of the tray, differentiate

Eq:3 with respect to x,

2
= 12 − 216 + 729 4


This is equal to zero leaving a quadratic equation. Use
your fx-991ES to find the solutions. Enter the

keystrokes (MODE)53,







5

Now enter the coefficient values,

12=z216=729=


=n =n







There are two solutions, the first solution will be

ignored leaving the critical value of x to maximise the

tray volume of 4.5 cm. You can test this by
determining the value of the second derivative. If it’s

negative, we have a maximum. You can use the Y

function on your fx-991ES to calculate the second
derivative. Given,


2
= 12 − 216 + 729 4


use the following keystrokes, (MODE)1

qY12Q(X)dp216J(X)

+729$4.5











6

=










Showing that 4.5 is a maximum value.




Example 3

A manufacturer of soft drinks wants to make a

3
cylindrical tin to hold 0.2 litre (200 cm ) of fizzy drink.
Determine the height, diameter and area of the tin so

the amount of material is minimised.


A diagram of the tin is shown on
the left of diameter D and height F.

The volume of the tin is,


2
= ( ) 1
2

3
The constraint is B = 200 cm . Need
to minimise the surface area of the tin which leads to

a minimum amount of material needed in its

manufacture. From Eq:1, make F the subject,




7

2 2
= ( ) 2


The surface area of the tin comprises the end pieces

and a rectangular area folded round to make a
cylinder. This is expressed as,


2
( ) = + 2 ( ) 3
2
From the diagram on the left

you can see where the

terms come from in the last
equation. Use Eq:2 to

replace F in Eq:3 to give,


2
2 2 2
( ) = ( ) + 4
4

This becomes,

4 2
( ) = + 5
2

Differentiate this objective function with respect to D,

4
= − + = 0 6
2

Rearranging to give the critical value,



8

3 4
= √ 7


The height of the tin is from Eq:2 is,

4
= 8
2


The minimum surface area of the tin from Eq:5 is,

4 2
= + 9
2

When using your calculator to determine Amin you

could perform a three stage calculation,


1. Calculate Dmin
2. Calculate the height F (Eq:8)

3. Calculate the area A (Eq:9)


The keystrokes are,

CQ(D)Q(=)qS4Q(B)aqL

$$Q(:)

Q(F)Q(=)4J(B)aqLJ(D)d$

Q(:)


4J(B)aJ(D)$+aqLJ(D)d

R2r


9

B? 200= the volume constraint

= = =





Diameter cm Height cm Surface area cm 2

You will note the diameter is the same as the height
for a minimum amount of metal.


As you can see in the sample tin
shown on the left, the choice of

having the height the same as the

diameter is not one chosen by
manufacturers of soft drinks.






Example 4
A cone is produced by cutting a

segment out of a circular piece

of paper as shown in the
diagram on the left. The

diameter is D and its radius is

D/2. The paper is folded around
to from a cone as seen in the second diagram on the

10

left. The height of the cone is F.

If the value of D is 12.5 cm,
what is the maximum volume

which can be contained within

the cone?

The volume of a cone is

⅓base_area  height which is,

2 2 3
2
= [( ) − ] = ( ) − 1
3 2 3 2 3

Differentiate V with respect to F,

2
2
= ( ) − = 0 2
3 2

Rearranging Eq:2 to give the critical value,


= 3
2√3

2
2
You can see from Eq:2, the second derivative d V/dF
is negative for all positive values of F which indicates

a maximum value. Substituting Eq:3 into Eq:1, the
volume is expressed as,


3
= 4
24√3



11

The keystrokes for performing this calculation on your

fx-991ES calculator are,

CqLQ(D)q(x )a24s3r
3

D? 12.5=









The maximum volume cone you can make with a

circular piece of paper of radius of 12.5 cm is 147.6
2
cm .




Example 5
Bricks manufactured by the

Littleport Brick Co. have a

2
surface area 720 cm . Derive
an object function for the

brick’s volume and determine the value of x which

maximises its volume.

The surface area of the brick is given by,


= 2(5 × ) + 2( × ) + 2(5 × )


12

or

2
= 12 + 10 1

The volume of the brick is given by,


2
= 5 2

From Eq:1 the value of F is given by,

−10 2
= 3
12

Replacing F in Eq:2 gives,


2
= 5 ( −10 2 ) = 5 − 25 3 4
12 12 6

Eq:4 is the objective function; differentiate this to give
the maximum,


5 75 2
= − = 0 5
12 6

This gives a critical value of,



= √ 6
30


2
We know that A = 720 cm , leaving = 4.9 cm. To
show this is a maximum (the second derivative should





13

be negative) use Eq:5; the keystrokes for your fx-

991ES calculator are,

CqYa5O720R12$
pa75Q(X)dR6$$4.9







=






Since this number is negative, = 4.9 gives the

maximum values which is,
5×4.9×720 25×4.9 3
= − 7
12 6

The keystrokes for this calculation are,

Ca5O4.9O720R12

$pa25O4.9q(x )R6=
3







3
The brick volume is therefore 980 cm .




14

Example 6

The diagram on the left shows
a sector of circle whose radius

is x. The perimeter length of the

sector is F. The area A of the
2
sector is 52 cm . Determine the
minimum length of the perimeter of the whole sector.


You will know from your knowledge of geometry, the
area of the sector is given by,


1 2
= 1
2

The arc length is , so the perimeter of the sector is,

= + 2 2


From Eq:1 we can see that,

2
= 3


From Eq:2 the perimeter of the sector becomes,

2
( ) = + 2 4


This is the objective function which we need to

minimise. Differentiating Eq:4,



15

2
= − + 2 = 0 5
2

This leads to a critical value,


= √ = √52 = 7.211 cm 6

To confirm this is a minimum, the second derivative of

Eq:4 should be positive, using Eq:5, the keystrokes for
your fx calculator are,


CqYz2O52aQ(X)d$
+2$7.211









=









Which verifies that = 7.211 is the minimum value.

The minimum perimeter length is, using Eq:4 is,

52
= 2 ( + ) = 2 ( 7.211 + 7.211) 7


Leaving = 28.8 cm.


16

Example 7

A compound shape is made
from a rectangle and a

semicircle as shown in the

diagram on the left. The
perimeter of the shape E is 86

m. Determine the maximum area of the shape.


The shape consists of a rectangle and a semicircle, the
area A is given by,


2
3
= 3 + ( ) 1
2 2
The perimeter of the shape is,


1
86 = 2 + 3 + × 3 2
2
Making y the subject of this equation, which becomes,


3
= 43 − (1 + ) 3
2 2

Substituting Eq:3 into Eq:1,

9 2 5
= 129 − ( + ) 4
4 2

This is the objective function. Differentiating this,



17

9 5
= 129 − ( + ) = 0 5
2 2
Leaving a critical value of,
86
= = 5.08 m 6
5
3( + )
2
Substituting Eq:6 into Eq:4,
9×5.08 2 5
= 129 × 5.08 − ( + ) 7
4 2
The keystrokes for this calculation are,

C129O5.08pa9O5
.08dR4$(qL+2.5

)=








2
The maximum area of the shape is 327.7 m . To show
the calculated value of x is a maximum, enter the

following keystrokes from Eq:5,
CqY129pa9QR2$

(qL+2.5)$5.08=









Since this is negative, x is a maximum value.

18

Example 8

A hollow half cylinder is shown on
the left with a length F and a

width of D (the diameter of the

semicircle). The amount of sheet
metal used in its construction if

2
1,720 cm . Find the values of D and F to give a
maximum capacity.

By considering the area, we know that,


2
1720 = + 1
4 2

This is the sum of the end pieces and the curved
section. Now make F the subject of Eq:1,


3440
= − 2
2

The volume of the half cylinder is given by,

2
1
= ( ) 3
2 2

Substitute Eq:2 into Eq:3,

3
= 430 − 4
16

This is the objective function to maximise.

19

3 2
= 430 − = 0 5
16

Rearranging Eq:5 to give a critical value of,


6880
= √ 6
3


Leaving = 27 cm. The length of the half cylinder
is, using Eq:2,


3440 27
= − = 27 cm 7
×27 2
The volume is, using Eq:4,


×27 3
= 430 × 27 − 8
16

The keystrokes for this calculation are,

C430O27paqLO27

q(x )R16=
3








3
The volume of the half cylinder is 7,747.2 cm .







20

Example 9

A lamp is suspended above a
round table of diameter D. The

illumination from the lamp follows

a Lambertian Cosine Distribution.
This means the intensity of the

light at the edge of the table is,

cos( )
= 1

2
What is the distance the bulb should be above the

table to give maximum illumination at the perimeter

of the table?

As you can see from the diagram,


2

2
2
= ( ) + 2
2
and

cos( ) = = 3
2
√ ( ) + 2
2


= 2 4
2


2
√ ( ) + 2 [( ) + ]
2 2
Leaving,
21


= 3 5
2
2
2
(( ) + )
2
Need to determine the derivative,


3 3
2 2 2 2 2 2
= {(( ) + ) − (( ) + ) } 6
2 3 2 2
2
(( ) + )
2
3 1
2 2
2 2 2 2 2
= 3 {(( ) + ) − 3 (( ) + ) } 7
2 2 2
2
(( ) + )
2
For maximum illumination, this derivative is zero,
therefore,


3 1
2 2
2 2
2
2
2
(( ) + ) − 3 (( ) + ) = 0 8
2 2
Leaving,
3 1
2 2
2 2 2 2 2
(( ) + ) = 3 (( ) + ) 9
2 2

This reduces to,

2
2 2
( ) + = 3 10
2





22

2
Subtracting x to both sides and taking the square root
to give a critical value of,


= 11
2√2

Maximum illumination occurs when x has this value.
Substituting this back into Eq:5 to give,


1.5
4 2
= ( ) 12
2
√2 3
The keystrokes for calculating this are,

C4as2$$(2a3$)^
1.5=









Leaving the maximum illumination is,


= 1.539
2

This occurs when the lamp is at a distance of 1.414

above the table.






23

Example 10

A right angled prism is shown on
the left which has a volume of

3
1,440 cm . Its height is 8x and its
base length is 6x. Derive an
expression for its surface area

and determine the value x which

give a minimum value to its surface area.

The area of the prism is made up of the two end pieces

and the three long pieces. Therefore the area A is,

1
2
2
= 2 × × 6 × 8 + 6 + 8 + √(6 ) + (8 )
2
or


2
2
2
= 48 + [ 14 + √36 + 64 ]
Which becomes,

2
= 48 + 28 1

The volume is given by,


1 2
= × 6 × 8 × = 24 2
2
3
We know that = 1440 cm , therefore,


24

1440
= 3
24 2

Substituting Eq:3 into Eq:1,

2
= 48 + 28 ( 1440 )
24 2

which becomes,


2
= 48 + 1620 4

This is the objective function, differentiating Eq:4


1620
= 56 − = 0 5
2
Rearranging to give a critical value of,



3 1620
= √ = 3.07 6
56

The keystrokes on your fx-991ES calculator to

calculate the second derivative at this value are Eq:5,

CqY56Q(X)p1620a

J(X)d$$3.07=











25

Since this is a positive number, we have a minimum

value. The surface area of the prism is given by, using
Eq:4


2
= 48 × 3.07 + 1620 7
3.07
The keystrokes for this calculation are,


C48O3.07d+1620
a3.07=










2
The minimum surface area is 654.35 cm .


























26

2 Optimising Commercial Performance

When considering the performance of a typical

commercial enterprise the normal parameters are

costs and number of units produced and sold. In the
first example we shall consider the making of green

widgets.


Example 1
Splax Ltd is a manufacturer of green

widgets. The volume production cost

uses the linear model,

( ) = 10,000 − 1


where x is the number of units made. The income to

the company B(x) in £ during any month is given by,

( ) = ( ) 2


Substituting Eq:1 into Eq:2


( ) = (10,000 − ) 3

The model for monthly expense C(x) is,


( ) = (1000 + 2 ) 4





27

Once we know the monthly income and expenditure

we are able to determine the profit D(x) per month
which is given by,


( ) = ( ) − ( ) 5


Substitute Eq:2 and Eq:4 in Eq:5 to give,

( ) = (10,000 − ) − (1000 + 2 )


Leaving

2
( ) = 9000 − 3 6

This is the objective function which we need to
maximise. The derivative of Eq:6 is,



= 9000 − 6 = 0 7


We arrive at a critical value of = 1,500 units. To test

this is maximum, use the Y feature on your fx-
991ES calculator with the following keystrokes,


CqY9000p6Q(X)$2
500










28

=









Since the second derivative is negative, it shows we

have a maximum. The profit every month Eq:6,

2
(1500) = 9000 × 1500 − 3 × 1500 = £6,750,000



Example 2

7-Down manufactures really expensive

yellow-orange widgets for €2,000 each
and they sell 1,000 of them a month.

However it seems that by reducing the

cost by €5 they are able to sell 50 widgets more. What
price should 7-Down charge to maximise it profits?


The unit price becomes 2000 − 50 . The number

sold every month is 1000 + 50 . The revenue each
month is therefore,


( ) = (2000 − 50 )(1000 + 50 ) 1


Expanding Eq:1 you get,



29

4
3 2
6
( ) = −2.5 × 10 + 5 × 10 + 2 × 10 2
This is the object function. Differentiating,
3 4
= −5 × 10 + 5 × 10 = 0 3

Leaving a critical value of,

5 × 10 4
= = 10
5 × 10 3
To show this is a maximum, enter the following

keystrokes on your fx-991ES calculator,
CqYz5K3Q(X)+5K4

$10








=







Since this is a negative number it shows that = 10

is a maximum. Substitute this value into Eq:1,

6
= (2000 − 500)(1000 + 500) = 2.25 × 10

The income to 7-Down is therefore €2,250,000 a

month.

30

Example 3

Potwells have discovered there is a market
demand for ornate metal vases. Looking at

the manufacturing costs, each will cost £9 to

make with an overhead of £500. Their
marketing graduate has suggested the

selling price should be,


( ) = 30 − 0.2√ 1

If R(x) is the revenue, C(x) the cost function, the profit

function will be,


( ) = ( ) − ( ) 2

The revenue will be,


( ) = ( ) 3


Substituting Eq:1 into Eq:3,

3
( ) = (30 − 0.2√ ) = 30 − 0.2 2 4


The cost function is,


( ) = 500 + 9 5

The profit function is, substituting Eq:4 and Eq:5 into

Eq:2,

31

3
( ) = 30 − 0.2 2 − (500 + 9 ) 6


which is the same as,

3
( ) = 21 − 0.2 2 − 500 7


Differentiating this objective function,

0.6√
= 21 − = 0 8
2
Rearranging the give the critical value if x,


2
2×21
= ( ) = 4900 9
0.6
Test to see if this is a maximum, enter the following

keystrokes into your calculator (Eq:8),


CqY21pa0.6sQ(X)
R2$$4900=








The negative value (just) indicates it a maximum
value. Using Eq:7, the profit from the sale of a critical

number of stools is,

3
(4900) = 21 × 4900 − 0.2 × 4900 2 − 500 10

32

The keystrokes for this calculation are,

C21O4900p0.2O4

900^1.5$p500=








The profit from selling 4,900 would be £33,800. The

unit price, using Eq:1,


(4900) = 30 − 0.2√4900 = £16

Potwells will have to sell an awful of them to reach the

optimum number. The original model Eq:1 is

therefore questionable.



Example 4

The van hire company HIRE has a fleet of vans which
they hire out. They use the model,


( ) = 750 − 5 1

where x is measured in €.

How much should they

charge their customers to maximise their return E(x)?



33

The amount the earn each day (revenue) is given by,

( ) = ( ) 2


Substituting Eq1 into Eq:2, gives,


( ) = (750 − 5 ) 3


Expanding Eq:3,

2
( ) = 750 − 5 4

This is the objective function which we need to
maximise. Differentiating Eq:4,


= 750 − 10 = 0 5

Leaving
10 = 750

The critical value of x is therefore €75, the amount

they should charge each day for a van. To show this is
a maximum the second derivative of Eq:4 should be

negative. The keystrokes for your fx-991ES are (using
Eq:5),

CqY750p10Q(X)$1
0








34

=







Since this is a negative number, a maximum has been

determined. The daily revenue is therefore,
2
(750) = 750 × 75 − 5 × 75 = €28,125
Example 5

Lumus-12 has started to make blue

widgets for €28 and is able to sell them

for €70. They are selling 50 of these
every day. After a pilot scheme it has

been found that if they reduce the cost

by €2.4 they sell an extra 20 of them.
What is the best selling price to maximise

the daily income?


The revenue E per day is given by,

( ) = (50 + 20 )(70 − 2.4 − 28)


This becomes,

( ) = (50 + 20 )(42 − 2.4 ) 1


This is the object function to maximise; differentiate,



35


= (50 + 20 ) (42 − 2.4 ) + (42 − 2.4 ) (50 + 20 ) = 0


Leaving,

(50 + 20 )(−2.4) + (42 − 2.4 )(20) = 0 2


Leaving a critical value of,

42×20−50×2.4
= = 7.5 3
40×2.4

Profit before reduction,


= 50(70 − 28) = €2,100

Profit after reduction,


= (50 + 7.5 × 20)(70 − 2.4 × 7.5 − 28) = €4,800



Example 6

Kinetic-9 Ltd makes orange widgets

where the price of each widget is
given by,


( ) = 270 − 0.02 1


is the cost in £ per item. To manufacture x widgets it
costs,


( ) = 900 + 32 2

36

What is the value of x to maximise profit?

If the revenue is R(x), we know that the profit is,


( ) = ( ) − ( ) 3


Since,


( ) = ( ) 4

Substituting Eq:1 into Eq:4 we get,

2
( ) = (270 − 0.02 ) = 270 − 0.02 5

Now substitute Eq:5 and Eq:2 into Eq:3 to give,


2
( ) = 270 − 0.02 − 900 − 32

Which becomes,

2
( ) = 238 − 0.02 − 900 6

This is the objective function. The derivative is given

by,

( )
= 238 − 0.04 = 0 7


Rearranging Eq:7,

238
= = 5,950 8
0.04




37

To confirm this is a maximum value, use the following

keystrokes on your fx-991ES calculator (Eq:7),

CqY238p0.04Q(X)

$5950=









As you can see the result is negative indicating a
maximum value. The sale price becomes, using Eq:1,


( ) = 270 − 0.02 × 5950 = £151 9

This means the profit for each widget is £151 provided

they manufacture 5,950 widgets.




Example 7

Cronix-15 have just started to make

red widgets and they are selling

boxes of 100 for £12. It seems that
when they increase the price of a

box by £1 they sell 5 less boxes a
day. What price should Cronix-15 sell the boxes to

maximise their profits?

38

The revenue is given by,

( ) = (12 + )(100 − 5 ) 1


Expanding this expression,


2
( ) = 1200 + 36 − 5 2

This is the objective function; differentiate,


= 36 − 10 = 0 3


The critical value for x is therefore,

36
= = 3.6 4
10

To show this is a maximum, use the following

keystrokes on your fx calculator (Eq:3),

CqY36p10Q(X)$3.

6=








The fact it’s negative shows it’s a maximum. The new
price should therefore be,


= 12 + 3.6 = £15.6 per box



39

The revenue is given by Eq:1, leaving

( ) = (12 + 3.6)(100 − 5 × 3.6) = £1,279


Before the adjustment, the revenue was £1,200.



Example 8

Luxis Ltd has been making yellow green

widgets and needs to change it strategy

to maximise its profits. The company has
been selling them for £7 each with a sale

certainty of 4,000 a month. After a brief

discount period, they discovered that if they decrease
the price by £1.40 they sell 950 more in a month. With

this information, how can they maximise their profits?


The revenue is given by,

( ) = (7 − 1.4 )(4000 + 950 ) 1


Expanding Eq:1,

2
( ) = 28000 + 1050 − 1330 2

This is the objective function, differentiate,


= 1050 − 2660 = 0 3


40

Leaving the critical value,

105
= = 0.394 4
266

Using Eq:1, the maximum revenue is,


= (7 − 1.4 × 0.394)(4000 + 950 × 0.394)

= £28,207


The revenue before adjustment was,


7  4000= £28,000

Not a huge increase in profit, their original price was

very close to the maximum price they should charge.



Example 9

The Littleport Symphony Orchestra is
holding another concert. The average

number of people attending has been
600 when the price was £10. In the

previous concert the ticket price was

lowered to £8.50 and the attendance
rose to 812. Determine the linear demand function

and find the optimum ticket price.




41

Assuming the demand function ( ) is a linear

function, it will pass through the coordinates (10, 600)
and (8.50, 812). The gradient is given by,


600−812
( ) = ( − 10) + 600 1
10−8.5
Leaving,


( ) = 2013 − 141.3 2


The revenue = price  demand, which is

2
( ) = 2013 − 141.3 3

This is the objective function; now differentiate,


= 2013 − 282.6 = 0 4

The critical value is therefore,


2013
= = 7.12 5
282.6

This suggests the optimum ticket price sale should be
£7.12. The revenue will be from Eq:3,


2
(7.12) = 2013 × 7.12 − 141.3 × 7.12
The keystrokes are,






42

C2013O7.12p141

.3O7.12d=









which is £7,169.44. The number of tickets that are

expected to sell is Eq:2,

(7.12) = 2013 − 141.3 × 7.12 = 1007 tickets


Here is the graph showing the objective function and

the coordinates of the two original ticket prices.


















From the two coordinates we were able to create the

objective function. This calculation can be generalised
and performed on your fx-991ES calculator. Let A be

the average number of people attending at ticket




43

price B and let E be the trial price where F tickets are

sold. The demand function becomes,

−
( ) = ( − ) + 6
−

Let C become,

−
= 7
−

or


( ) = − + = ( − ) + 8

The revenue is ( )


2
( ) = − ( − ) 9
This objective function becomes,



= 2 − + = 0 10


The critical value for X is,

−
= 11
2

which is new price for the tickets. The demand will be

Eq:8 and the revenue will by Eq:9. Performing these
calculations on your fx-991ES, use the multi-stage

feature.


44