ACKNOWLEDGEMENT
Overall this math project was not only tough but challenging as well. I
had to acquire assistance from a few people throughout this project in
order to complete this project.
Firstly, I would like to thank my group of friends, Yung Wei Xian,
Ong Jin Liang, Chan Jun Cheng, Low Tze Wei, Jason Lem, Abishek
Suresh Kumar, Ngu Bing Han and Leong Wai Hyn who gave me the
most help in completing this project. Despite the arguments we had, we
managed through them and completed the project by working together as
a team.
I would also like to thank my teacher, Pn Teng Loy Ying who
showed us and guided us by briefing us about the procedures and
advised us in carrying out this project.
Lastly, I would like to thank my mother and father as well for they
were the ones who had gave me moral support in finishing this project.
1
AIMS AND OBJECTIVES
Nationwide in Malaysia, every Form 5 student taking Additional
Mathematics is required to work on the Additional Mathematics Project
Work 2015. The project is to be done with aims to create a future with
improvements:
i. To apply and adapt a variety of problem-solving strategies to solve
problems.
ii. To improve thinking skills.
iii. To promote effective mathematical communication.
iv. To develop mathematical knowledge through problem solving in a
way that increases students’ interest and confidence.
v. To use the language of mathematics to express mathematical ideas
precisely.
vi. To provide learning environment that stimulates and enhances
effective learning.
vii. To develop positive attitude towards mathematics.
2
INTRODUCTION
In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x2. The output of a function fcorresponding to an input x is denoted by f(x) (read "f of x").Functions of various kinds are "the central objects of investigation" in most fields of modern mathematics. There are many ways to describe or represent a function. Some functions may be defined by a formula or algorithm that tells how to compute the output for a given input. Others are given by a picture, called the graph of the function. In science, functions are sometimes defined by a table that gives the outputs for selected inputs. A function could be described implicitly, for example as the inverse to another function or as a solution of a differential equation.
The red curve is the graph of a function f in the Cartesian plane, consisting of all points with coordinates of the form (x,f(x)). The property of having one output for each input is represented geometrically by the fact that each vertical line (such as the yellow line through the origin) has exactly one crossing point with the curve.
3
Pierre de Fermat developed the technique of adequality (adaequalitas in Latin) to calculate maxima and minima of functions, tangents to curves, area, center of mass, least action, and other problems in mathematical analysis. According to André Weil, Fermat "introduces the technical term adaequalitas, adaequare, etc., which he says he has borrowed from Diophantus. As Diophantus V.11 shows, it means an approximate equality, and this is indeed how Fermat explains the word in one of his later writings." (Weil 1973). Diophantus coined the word παρισὀτης (parisotēs) to refer to an approximate equality.Claude Gaspard Bachet de Méziriac translated Diophantus's Greek word into Latin as adaequalitas.[citation needed] Paul Tannery's French translation of Fermat’s Latin treatises on maxima and minima used the words adéquation and adégaler.
Fermat used adequality first to find maxima of functions, and then adapted it to find tangent lines to curves. To find the maximum of a term p(x ), Fermat equated (or more precisely adequated) p(x ) and p(x+e ) and after doing algebra he could cancel out a factor of e and then discard any remaining terms involving e. To illustrate the method by Fermat's own example,
consider the problem of finding the maximum of p ( x )=bx−x2. Fermat adequated bx−x2 with
b ( x+e )−( x+e )2=bx−x2+be−2 ex−e2. That is (using the notation to denote
adequality, introduced by Paul Tannery): bx−x2 bx−x2+be−2 ex−e2 Canceling terms and dividing by Fermat arrived at b 2 x+e Removing the terms that contained Fermat arrived at the desired result that the maximum occurred when x=b /2.
4
PART 1
a1)Mathematical optimization
Mathematical optimization,alternatively, optimization or mathematical programming is the selection of a best element with regard to some criteria from some set of available alternatives.
An optimization problem consists of maximizing or minimizing a real function by systematically choosing input values from within an allowed set and computing the value of the function. The generalization of optimization theory and techniques to other formulations comprises a large area of applied mathematics. More generally, optimization includes finding "best available" values of some objective function given a defined domain (or a set of constraints), including a variety of different types of objective functions and different types of domains.
a2) Global maximum and minimum
A function f has a global maximum or minimum at c if f(c) ≥ f(x) for all x in D, where D is the domain of f. The number f(c) is called the maximum value of f on D. Similarly, f has a global minimum or maximum at c if f(c) ≤ f(x) for all x in D and the number f(c) is called the minimum value of f on D. The maximum and minimum values of f are called the extreme values of f.
a3) Local maximum and minimum
A function f has a local maximum (or relative maximum) at c if f(c) ≥ f(x) when x is near c. This means that f(c) ≥ f(x) for all x in some open interval containing c. Similarly, f has a local minimum or maximum at c if f(c) ≤ f(x) when x is near c.
5
b) Ways to find maximum and minimum value of a quadratic function
Completing the square
y=a(x+ b2 a
)2
+ 4 ac−b2
4
Example:
y = 2x2 + 4x + 5 y = -2x2 + 4x + 5a = 2 a = -2a¿0 a ¿0
y minimum = 4 (2 ) (5 )−42
4 (2)
= 40−16
8
y maximum = 4 (2 ) (5 )−42
4 (−2)
= −40−16
−8
6
DifferentiationCompleting the square
Ways to find maximum and
minimum value
Value of a quadratic function
a<0a>0
y minimum =
4 ac – b2
4 a
y minimum =
4 ac – b2
4 a
= 3 = 7
Differentiation
Example:
y = -2x2 + 4x + 5 y = 2x2 + 4x + 5dydx
=−4 x+4dydx
=4 x+4
−4 x+4=0 4 x+4=0x=1 x=−1d2 yd x2 = -4
d2 yd x2 = 4
d2 yd x2 <0
d2 yd x2 >0
7
= 2a
2a
2ax + b = 0
x = -
= 2a
2a
Maximum Minimum
y maximum=−2 (1 )2+4 (1 )+5¿7
y minimum=2(−1)2+4 (−1 )+5 ¿3
PART 2
a) Encik Shah has been involved in a sheep farming business for several years and he supplies meats and milk products to the communities. He wished to build a rectangular sheep pen with two parallel partitions using 200 meters fence. Find the dimensions of the rectangle that will maximize the total area of the pen. Hence state the maximum area of the pen.
Calculation
The rectangular pen should look like this, with 2 parallel partitions:
y y y
x x x x
Let 3y = length of rectangular pen and x = width of rectangular pen.
As the total fence is 200m long, thus an equation can be formed:4x+6y = 200 6y = 200-4x
y = 200−4 x
6 ------ (1)
Assume A is the area of the rectangular pen:A = 3y × x
8
A = 3 xy
A = 3 x¿ )
A = 100 x−2 x2-----(2)
Method 1
To find the maximum area, we must first differentiate equation (2), that is to find dAdx
.
A = 100 x−2 x2
dAdx
= 100−4 x
At maximum point, dAdx
= 0
0 = 100−4 x
4 x = 100
x = 25
Sub x = 25 into equation (1):
y = 200−4 x
6
y = 200−4(25)
6
y = 200−100
6
y = 100
6
9
Length = 3 y
= 3(100
6 )
= 50 m
Width = x
= 25m
∴ The dimension of the rectangle that will maximise the total area is 50 m x 25 m.
Sub x = 25 into equation (2):
A = 100 x−2 x2
A = 100(25)−2(25)2
A = 2500−2¿625)
A = 2500−1250
A = 1250 m2
∴ The maximum area of the pen is 1250 m2.
Method 2
Perform completing the square to find the maximum point.
From equation (2):
A = 100 x−2 x2
A = −2 x2+100 x
A = −2(x2−50 x )
A = −2(x2−50 x+252−252)
A = −2 [(x−25)2−625 ]
A = −2 ( x−25 )2+125010
∴ The point is (25, 1250), with x value=25 and y value=1250
Sub x = 25 into equation (1):
y = 200−4 x
6
y = 200−4(25)
6
y = 200−100
6
y = 100
6
Length = 3 y
= 3(100
6 )
= 50 m
Width = x
= 25m
∴ The dimension of the rectangle that will maximise the total area is 50 m x 25 m.
Sub x = 25 into equation (2):
A = 100 x−2 x2
A = 100(25)−2(25)2
A = 2500−2¿625)
A = 2500−1250
A = 1250 m2
∴ The maximum area of the pen is 1250 m2.
11
From both of the methods used, we can conclude that:
(i) The dimension of the rectangle that will maximise the total area is 50 m x 25 m.(ii) The maximum area of the pen is 1250 m2.
b) Reza is helping En Shah to make a box without the top. The box is made by cutting away four squares from the corners of a 30cm square piece of cardboard as shown in Figure 1 and bending up the resulting cardboard to form walls of the box.
Find the largest possible volume of the box.
Solution:
Volume ¿ (30−2h ) (30−2h ) (h )
¿4 h3−120 h2+900 h
12
To find maximum volume, we must differentiate the equation, that is to find dVdh
.
V=4 h3−120 h2+900 h
dVdh
=12 h2−240 h+900
At maximum volume, dVdh
=0 .
12 h2−240 h+900=0
h2−20 h+75=0
(h−15 ) (h−5 )=0
h=15 ,5
dVdh
=12 h2−240 h+900
d2Vd h2 =24 h−240
When h=15,
d2Vd h2 =24 (15)−240
¿120>0
∴ Volume is minimum if h=15
When h=5,
d2Vdh2 =24 (5)−240
¿−120<0
13
∴ Volume is maximum if h=5
Largest possible value of the box ¿4 (5)3−120 (5 )2+900 (5)
¿500−3000+4500
¿2000 cm2
PART 3
A market research company finds that traffic in a local mall over the couse of a day could be
estimated by the function P(t) = -1800cos[π6
( t )]+1800 where P, is the number of people going to
the mall, and t is the time, in hours, after the mall opens. The mall opens at 9.30 am.
(i) Sketch the graph of function P(t).
When cos ,x should be an angle. An angle has two units, that is degree and radian. To solve this equation, we must first confirm if the angle is in degree or radian.
A clock is divided into 12 parts, each part representing an hour.
14
For example, if it is 1 hour, the angle will be1
12 of the clock.
If it is 2 hours, the angle will be 2
12of the clock.
If it is 3 hours, the angle will be 3
12of the clock.
Thus, as a conclusion, we can say that if it is t hours, the angle will bet
12of the clock.
For angle in unit degree, the calculation to find the angle will be:
Angle = t
12×360 °
= t (30 °¿
As for angle in unit radian, the calculation to find angle will be:
Angle = t
12×2 π
= π6
( t )
15
Based on both equation above, it is proved that the angle is in radian mode, as the angle is as same as in equation P(t).
(We take π as 3.142)
Time Hours / t Number of people / P(t)9:30 a.m. 0
P(0) = -1800cos[π6
(0 )]+1800
= -1800cos (0) +1800 = -1800(1) +1800 = -1800+1800 = 0
11:30 a.m. 2P(2) = -1800cos[
π6
(2 )]+1800
16
= -1800cos (1.0473) +1800 = -1800(0.49991) +1800 = -899.84+1800 = 900.16 ≈ 900(since number of people cannot be a decimal number, thus the nearest integer is taken)
3:30 p.m. 6P(6) = -1800cos[
π6
(6 )]+1800
= -1800cos (3.142) +1800 = -1800(-1.00000) +1800 = 1800+1800 = 3600
5:30 p.m. 8P(8) = -1800cos[
π6
(8 )]+1800
= -1800cos (4.1893) +1800 = -1800(-0.49956) +1800 = 899.21+1800 = 2699.2 ≈ 2699
9:30 p.m. 12P(0) = -1800cos[
π6
(12 )]+1800
= -1800cos (6.284) +1800 = -1800(1.0000) +1800 = -1800+1800 = 0
17
Number of people / P(t)
A: 9:30 a.m. C: 3:30 p.m.B: 11:30 a.m. D: 5:30 p.m.E: 9:30 p.m.
18
Hour / t
(ii) When does the mall reach its peak hours and state the number of people
From the graph, we can see that during the 6th hour, that is by 3:30 p.m. , the graph reaches its maximum point, which is 3600.
∴ The mall reaches its peak hour by 3:30 p.m. and has 3600 people
(iii) Estimate the number of people in the mall at 7:30 p.m.
From the table drawn, at 7:30 p.m., there is estimated 899 people in the mall.
(iv) Determine the time when the number of people in the mall reaches 2570.
19
-1800cos [π6
(t ) ] +1800 = 2570
-1800cos [π6
(t ) ] = 2570 – 1800
-1800cos [π6
(t ) ] = 770
cos [π6
(t ) ] = 770
−1800
cos [π6
(t ) ] = -0.42778
π6
(t ) = cos−1−0.42778
3.1426
( t ) = 2.0128
t = 2.0128 ×6
3.142
t = 3.8437
t = 3 hours + (0.8437x60) minutes
t = 3 hours 51 minutes
∴ The time the number of people in the mall reaches 2570 is 9:30 a.m. + 3 hours 51 minutes = 1:21 p.m.
20
FURTHER EXPLORATION
a) Conduct a research on Linear Programming (LP) and then write notes of your findings. You may include the historical aspects and its importance to real life situations. You must also give at least two examples on how the LP is being applied in various field of study.
Introduction
Linear programming (LP; also called linear optimization) is a method to achieve the best outcome (such as maximum profit or lowest cost) in a mathematical model whose requirements are represented by linear relationships. Linear programming is a special case of mathematical programming (mathematical optimization).
More formally, linear programming is a technique for the optimization of a linear objective
function, subject to linear equality and linear inequality constraints. Its feasible region is
a convex polytope, which is a set defined as the intersection of finitely many half spaces, each of
which is defined by a linear inequality. Its objective function is a real-valued affine
function defined on this polyhedron. A linear programming algorithm finds a point in the
polyhedron where this function has the smallest (or largest) value if such a point exists.
Linear programs are problems that can be expressed in canonical form:
where x represents the vector of variables (to be determined), c and b are vectors of (known)
coefficients, A is a (known) matrix of coefficients, and is the matrix transpose. The
expression to be maximized or minimized is called the objective function (cTx in this case). The
inequalities Ax ≤ b and x ≥ 0 are the constraints which specify a convex polytope over which the
objective function is to be optimized. In this context, two vectors are comparable when they have
the same dimensions. If every entry in the first is less-than or equal-to the corresponding entry in
the second then we can say the first vector is less-than or equal-to the second vector.
Linear programming can be applied to various fields of study. It is used in business
and economics, but can also be utilized for some engineering problems. Industries that use linear
programming models include transportation, energy, telecommunications, and manufacturing. It
21
has proved useful in modeling diverse types of problems in
planning, routing, scheduling, assignment, and design.
History
The problem of solving a system of linear inequalities dates back at least as far as Fourier, who in
1827 published a method for solving them, and after whom the method of Fourier–Motzkin
elimination is named.
The first linear programming formulation of a problem that is equivalent to the general linear
programming problem was given by Leonid Kantorovich in 1939, who also proposed a method
for solving it. He developed it during World War II as a way to plan expenditures and returns so
as to reduce costs to the army and increase losses incurred by the enemy. About the same time as
Kantorovich, the Dutch-American economist T. C. Koopmans formulated classical economic
problems as linear programs. Kantorovich and Koopmans later shared the 1975 Nobel Prize in
economics. In 1941, Frank Lauren Hitchcock also formulated transportation problems as linear
programs and gave a solution very similar to the later Simplex method; Hitchcock had died in
1957 and the Nobel Prize is not awarded posthumously.
During 1946-1947, George B. Dantzig independently developed general linear programming
formulation to use for planning problems in US Air Force. In 1947, Dantzig also invented
the simplex method that for the first time efficiently tackled the linear programming problem in
most cases. When Dantzig arranged meeting with John von Neumann to discuss his Simplex
method, Neumann immediately conjectured the theory of duality by realizing that the problem he
had been working in game theory was equivalent. Dantzig provided formal proof in an
unpublished report "A Theorem on Linear Inequalities" on January 5, 1948. Postwar, many
industries found its use in their daily planning.
Dantzig's original example was to find the best assignment of 70 people to 70 jobs. The
computing power required to test all the permutations to select the best assignment is vast; the
number of possible configurations exceeds the number of particles in the observable universe.
However, it takes only a moment to find the optimum solution by posing the problem as a linear
program and applying the simplex algorithm. The theory behind linear programming drastically
reduces the number of possible solutions that must be checked.
The linear-programming problem was first shown to be solvable in polynomial time by Leonid
Khachiyan in 1979, but a larger theoretical and practical breakthrough in the field came in 1984
22
when Narendra Karmarkar introduced a new interior-point method for solving linear-
programming problems.
Uses
Linear programming is a considerable field of optimization for several reasons. Many practical problems in operations research can be expressed as linear programming problems. Certain special cases of linear programming, such as network flow problems and multicommodity flow problems are considered important enough to have generated much research on specialized algorithms for their solution. A number of algorithms for other types of optimization problems work by solving LP problems as sub-problems. Historically, ideas from linear programming have inspired many of the central concepts of optimization theory, such as duality, decomposition, and the importance of convexity and its generalizations. Likewise, linear programming is heavily used in microeconomics and company management, such as planning, production, transportation, technology and other issues. Although the modern management issues are ever-changing, most companies would like to maximize profits or minimize costs with limited resources. Therefore, many issues can be characterized as linear programming problems.
23
Example 1:
Suppose that a farmer has a piece of farm land, say L km2, to be planted with either wheat or
barley or some combination of the two. The farmer has a limited amount of fertilizer,
F kilograms, and insecticide, P kilograms. Every square kilometer of wheat
requires F1 kilograms of fertilizer and P1 kilograms of insecticide, while every square kilometer
of barley requires F2 kilograms of fertilizer and P2 kilograms of insecticide. Let S1 be the selling
price of wheat per square kilometer, and S2 be the selling price of barley. If we denote the area of
land planted with wheat and barley by x1 and x2 respectively, then profit can be maximized by
choosing optimal values for x1 and x2. This problem can be expressed with the following linear
programming problem in the standard form:
Maximize: (maximize the revenue—revenue is the "objective
function")
Subject to: (limit on total area)
(limit on fertilizer)
(limit on insecticide)
(cannot plant a negative area).
Which in matrix form becomes:
Maximize
subject to
24
Example 2:
You need to buy some filing cabinets. You know that Cabinet X costs $10 per unit, requires six square feet of floor space, and holds eight cubic feet of files. Cabinet Y costs $20 per unit, requires eight square feet of floor space, and holds twelve cubic feet of files. You have been given $140 for this purchase, though you don't have to spend that much. The office has room for no more than 72 square feet of cabinets. How many of which model should you buy, in order to maximize storage volume?
The question ask for the number of cabinets I need to buy, so my variables will stand for that:
x: number of model X cabinets purchasedy: number of model Y cabinets purchased
Naturally, x > 0 and y > 0. I have to consider costs and floor space (the "footprint" of each unit), while maximizing the storage volume, so costs and floor space will be my constraints, while volume will be my optimization equation.
cost: 10x + 20y < 140, or y < – ( 12
) x + 7
space: 6x + 8y < 72, or y < – ( 34
) x + 9
volume: V = 8x + 12y
This system (along with the first two constraints) graphs as:
25
When you test the corner points at (8, 3), (0, 7), and (12, 0), you should obtain a maximal volume of 100 cubic feet by buying eight of model X and three of model Y.
b) Aaron owns a shipping company. He plans to move into his new office which is near to the city centre. He needs some filing cabinets to organize his files. Cabinet x which cost RM 100 per unit, requires 0.6 square meters of the floor space and can hold 0.8 cubic meters of files. Cabinet y which cost RM 200 per unit, requires 0.8 square meters of the floor space and can hold 1.2 cubic meters of files. The ratio of the number of cabinet x to the number of cabinet y is not less than 2:3. Aaron has an allocation of RM 1400 for the cabinets and the office has room for no more than 7.2 square meters.
i) Using the given information,
(a) write the inequalities which satisfy all the above constraints.
Cabinet x Cabinet y
Cost RM 100 RM 200Floor space 0.6 square meters 0.8 square metersStorage volume 0.8 cubic meters 1.2 cubic meters
I)xy
≥23
3 x ≥ 2 y
II) 100 x+200 y≤ 1400x+2 y≤ 14
III) 0.6 x+0.8 y ≤7.23 x+4 y ≤ 36
(b) construct and shade the region that satisfies all the above constraints.
26
ii) Using two different methods, find the maximum storage volume.
Method 1
From the storage volume of both cabinets, we can form an equation:
Volume = 0.8 x+1.2 y
As volume can be any constant, we’ll take the constant by multiplying 0.8 and 1.2 together.
0.8 x+1.2 y = 0.8 ×1.20.8 x+1.2 y = 0.962 x+3 y = 2.4
The line is drawn on the graph.A ruler is used to find the maximum point by moving the ruler parallel to the line and the maximum point is the point where the ruler last touched in the shaded region.
27
3 x=2 y
x+2 y=14
3 x+4 y=36
The maximum point is (8, 3).Thus, this means that the maximum storage volume obtained is by having 8 cabinet x and 3 cabinet y.
Volume = 0.8 x+1.2 y
= 0.8 (8 )+1.2(3) = 10.0 m3
Method 2
To find the volume, the formula Volume = 0.8 x+1.2 y is used.
Since every points in the shaded region is a possible answer, thus every point is being calculated to find their storage volume.
Number of cabinet x Number of cabinet y Storage volume (m3)0 0 0.0
28
3 x=2 y3 x+4 y=36
x+2 y=14
2 x+3 y = 2.4
1 0 0.81 1 2.02 0 1.62 1 2.82 2 4.02 3 5.23 0 2.43 1 3.63 2 4.83 3 6.03 4 7.24 0 3.24 1 4.44 2 5.64 3 6.84 4 8.04 5 9.25 0 4.05 1 5.25 2 6.45 3 7.65 4 8.86 0 4.86 1 6.06 2 7.26 3 8.46 4 9.67 0 5.67 1 6.87 2 8.07 3 9.28 0 6.48 1 7.68 2 8.88 3 10.09 0 7.29 1 8.49 2 9.610 0 8.010 1 9.211 0 8.812 0 9.6
29
As we can see from the table above, the highest storage volume is by having 8 cabinet x and 3 cabinet y. The storage volume achieved is 10.0 m3.
iii) Aaron plans to buy cabinet x in a range of 4 to 9 units. Tabulate all the possible combinations of the cabinets that he can purchase. Calculate the cost of each combination.
The cost can be calculated by the formula Cost = 100 x+200 y.
Number of cabinet x Number of cabinet y Cost (RM)4 0 4004 1 6004 2 8004 3 10004 4 12004 5 14005 0 5005 1 7005 2 9005 3 11005 4 13006 0 600
30
6 1 8006 2 10006 3 12006 4 14007 0 7007 1 9007 2 11007 3 13008 0 8008 1 10008 2 12008 3 14009 0 9009 1 11009 2 1300
iv)If you were Aaron which combination would you choose? Justify your answer and give your reasons.
I would choose the combination of 8 cabinet x and 3 cabinet y. This is because:-
(a) It has the maximum storage volume compared to other combination, thus I can store more files.
(b) Although its price is RM1400, which is the highest price among all combinations, but it is still in the range of allocation of money, which is the maximum of RM1400. Thus it can be accepted.
(c) It occupies 7.2 square meters, which still satisfies the maximum floor space given.
31
CONCLUSION
Prior to the end of this project, it was such an extensive temporal length
to go through to finish the task. Much time was spent but it was worth it.
Time management had become an important habit to adopt along the
entire time I worked to complete the project.
32
One of the benefits of doing this project is the improvement of the
mathematical skill which may be helpful in solving mathematical
problems. As it turns out, the concept of differentiation, trigonometric
function and linear programming become clearer in our mind. We
understand them well and learned their application in real life.
Now, for us additional mathematics is not just a subject for exam
only but we are keen to apply what we had learned to real life. All the
work of many mathematicians should certainly be appreciated as many
of them have contributed and shaped the world as it is today. We found
that out when researching for the history of linear programming.
REFLECTIONPOEM
Many years ago I say, Math was just so fun,Fun to do, easy to learn, and always useful in life.Now after years of learning it, after PMR was done,Add Maths came into my life and "cut" me like a knife.
It's interesting oh yes it is, this cannot be denied,
33
But it's just too hard and complicated and annoys that brain of mine.The answers and working are just too long, though yes, it is its pride, Sometimes, no choice, I have to give up, and tell myself I've tried.
Add Maths, I have a question to ask, should I love you or should I hate you?Because when I believe I understand I realise I'm still a jerk.But sometimes no matter how much I try, I never get a clue,Therefore I choose to close my book, look at it and then just smirk.
And then again, it pops in my mind, there's SPM ahead,No choice, no choice, not a choice at all, I can't laze around anymore.Okay friends and everyone, now that I have said,I'll just do my Add Maths homework now, and see what my brain will store.
34
REFERENCES
https://www.google.com.my
https://en.wikipedia.org/wiki/Linear_programming
https://en.wikipedia.org/wiki/Maxima_and_minima
http://www.ithink.org.my/Home/Index
https://en.wikipedia.org/wiki/Trigonometric_functions
https://en.wikipedia.org/wiki/Pierre_de_Fermat
35
APPENDIX
36