Introduction

In Mathematics, maximum and minimum value of a function, known as extreme

which is the largest and smallest value that function that taken in a point whether from

its domain (extreme relative or vocal) or in the domain function as whole ( global

extreme). Pierre de Fermat is one of the first Mathematician that proposed general

technique ( adequality) for finding maximum and minimum. Finding extreme value is

the fundamental goal of optimization.

Part 1

a.

I. Mathematical optimization

An act, process, or methodology of making something (as a design, system, or

decision) as fully perfect, functional, or effective as possible using mathematical theory

or concept; specifically : the mathematical procedures (as finding the maximum of a

function) .

II.

Global Maximum :

A global maximum, also known as an absolute maximum, the largest overall value of

a set, function, etc., over its entire range.

Global Minimum

A global minimum, also known as an absolute minimum, is the smallest overall value

of a set, function, etc., over its entire range.

III.

Local Minimum :

It is the least value that locates within a set of points which may or may not be a

global minimum and it is not the lowest value in the entire set. It can also be termed as

"Relative Minimum"

Local Maximum :

It can also be expressed as "Relative Maximum". It is a greatest value in a set of

points but not highest when compared to all values in a set. The set of points can be

global maximum.

Example :

Graph 1

How to find

Maximum/Minimum

Quadratic equation

Method 1 : Calculus

If the quadratic is in the form y = ax2 + bx + c

The maximum or minimum value of a quadratic

function occurs at its vertex.

If a>0 is minimum., a<0 is maximum

For y = ax2 + bx + c, (c - b

2/4a) gives the y-value at its

vertex.

Method 2 : Completing square

If the quadratic is in the form y = a(x-h)2 + k

k gives us the maximum or minimum value of the

quadratic accordingly as a is negative or positive

respectively.

Method 3 : Differentiation

Using differentiation when the quadratic is in the form y = ax2+ bx + c

1. Differentiate y with respect to x. dy/dx = 2ax + b

2. Determine the differentiation point values in terms of dy/dx. It can be

found by setting these values equal to 0 and find the corresponding values.

dy/dx = 0. 2ax+b = 0, x = -b/2a

3. Substitute this value of x into y to get the minimum/maximum point

b) example

, therefore

Diagram 1

Diagram 2

Part 2

a) A P B

y

C Q D

x

Diagram 3

Let length AB=CD=X meter and width AC=PQ=BD=y meter

From excel, we obtain

length, width, Perimeter Area, x y (4y+2x) m^2

5 47.5 200 237.5

10 45 200 450

15 42.5 200 637.5

20 40 200 800

25 37.5 200 937.5

30 35 200 1050

35 32.5 200 1137.5

40 30 200 1200

45 27.5 200 1237.5

50 25 200 1250

55 22.5 200 1237.5

60 20 200 1200

65 17.5 200 1137.5

70 15 200 1050

75 12.5 200 937.5

80 10 200 800

85 7.5 200 637.5

90 5 200 450

95 2.5 200 237.5

100 0 200 0

Table 1

From this table, we can see dimension that can maximize the barn is

50 m × 25 m which gives total area of 1250

b)

h cm

(30-2h) cm

(30-2h) cm

Diagram 4

Volume = length × width × height

= (30-2h) × (30-2h) × h

=

= Volume , V =

؞

Part 3

I. Graph function of

t 0930 1030 1130 1230 1330 1430 1530 1630 1730 1830 1930 2030 2130 2230 2330 2430

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

P(t) 0 241.2 900 1800 2700 3359 3600 3358 2699 1799 899 240 0 242 901 1802

Graph 2

II. The peak hour reach at 1530/3.30pm, number of visitor on that hour are 3600

visitor

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

P(t) 0 241 900 1800 2700 3359 3600 3358 2699 1799 899 240 0 242 901 1802

0

500

1000

1500

2000

2500

3000

3500

4000

Vis

ito

r

P(t) Visitor versus time

III. At 7.30hour,

IV.

Further Exploration

b)

Cabinet type Price (RM)/Unit Area ( Volume (

100 0.6 0.8

200 0.8 1.2

Table 2

I. a)

≥

≥

≤

-----------(1)

≤ ] ÷ 100

≤ -------------(2)

[ ≤ 7.2] ÷ 0.2

3 ≤ 36 ------------(3)

b)

Graph 3

II.

We need to find the value of x and y that maximize the space of cabinet.

First method – Simultaneous equation

≤ 3 ≤ 36

= ---(1) 3 =36 ------(1)

= ---(3)

3 =36

=36 =

=6>>>> =3 -------(4) =

, V

Second method – Linear Programming

Graph 4

From the graph we get maximum value =3 , =

, V

III. From excel,

combination x y Area )

Volume

( )

cost

(RM)

A 4 5 6.4 9.2 1400

B 5 4 6.2 8.8 1300

C 6 4 6.8 9.6 1400

D 7 3 6.6 9.2 1300

E 8 3 7.2 10 1400

F 9 2 7 9.6 1300

Table 3

IV. If I were Aaron, I would choose combination E (x=8, y=3)

Because the cost is exactly RM1400 and it also maximize the are 7.2 and

volume 10 with high number of combination of cabinet 11.

Reflection

I’m hardworking

I’m explorer

I’m problem solver

I’m critical thinker

I’m info seeker

I’m genius

By Mathematics