ADDITIONAL MATHEMATIC PROJECT WORK 2015

APPRECIATION

Firstly, I,would like to give a big thanks to my parent for providing everything, such as money, to

buy anything that are related to this project work,their advice and support. Then, I want to

thank my teacher, for teaching me Additional Mathematics form 5 and guiding me throughout

this project. Last but not least, my friends who were doing this project with me and sharing our

ideas and knowledge. We were helping each other so we cancomplete our project without any

problems

OBJECTIVES

The objectives of this project work are:

Apply and adapt a variety of problem-solving strategies to solve problems.

Develop mathematical knowledge through problem solving in a way that

increases students’ interest and confidence.

Develop positive attitude towards mathematics.

Improve thinking skills and creativity.

Promote efficiency of mathematical communication.

Provide learning environment that stimulates and enhances effective learning

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

INTRODUCTIONDEFINITION

A real-valued function f defined on a domain X has a global (or absolute) maximum point at x∗ if f(x∗) ≥ f(x) for all x in X. Similarly, the function has a global (or absolute) minimum point at x∗ if f(x∗) ≤ f(x) for allx in X. The value of the function at a maximum point is called the maximum value of the function and the value of the function at a minimum point is called the minimum value of the function.

If the domain X is a metric space then f is said to have a local (or relative) maximum point at the point x∗ if there exists some ε > 0 such that f(x∗) ≥ f(x) for all x in X within distance ε of x∗. Similarly, the function has a local minimum point at x∗ if f(x∗) ≤ f(x) for all x in X within distance ε of x∗. A similar definition can be used when X is a topological space, since the definition just given can be rephrased in terms of neighbourhoods. Note that a global maximum point is always a local maximum point, and similarly for minimum points.

In both the global and local cases, the concept of a strict extremum can be defined. For example, x∗ is a strict global maximum point if, for all x in X with x ≠ x∗, we have f(x∗) > f(x), and x∗ is a strict local maximum point if there exists some ε > 0 such that, for all x in X within distance ε of x∗ with x ≠ x∗, we have f(x∗) > f(x). Note that a point is a strict global maximum point if and only if it is the unique global maximum point, and similarly for minimum points.

A continuous real-valued function with a compact domain always has a maximum point and a minimum point. An important example is a function whose domain is a closed (and bounded) interval of real numbers.

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

FINDING FUNCTIONAL MAXIMA AND MINIMA

Finding global maxima and minima is the goal of mathematical optimization. If a function is

continuous on a closed interval, then by the extreme value theorem global maxima and minima

exist. Furthermore, a global maximum (or minimum) either must be a local maximum (or

minimum) in the interior of the domain, or must lie on the boundary of the domain. So a

method of finding a global maximum (or minimum) is to look at all the local maxima (or

minima) in the interior, and also look at the maxima (or minima) of the points on the boundary,

and take the largest (or smallest) one.

Local extrema of differentiable functions can be found by Fermat's theorem, which states that

they must occur at critical points. One can distinguish whether a critical point is a local

maximum or local minimum by using the first derivative test, second derivative test, or higher-

order derivative test, given sufficient differentiability.

For any function that is defined piecewise, one finds a maximum (or minimum) by finding the

maximum (or minimum) of each piece separately, and then seeing which one is largest (or

smallest).

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

EXAMPLES

The function x2 has a unique global minimum at x = 0. The function x3 has no global minima or maxima. Although the first derivative (3x2) is 0

at x = 0, this is an inflection point.

The function has a unique global maximum at x = e.

The function x-x has a unique global maximum over the positive real numbers at x = 1/e. The function x3/3 − x has first derivative x2 − 1 and second derivative 2x. Setting the first

derivative to 0 and solving for x gives stationary points at −1 and +1. From the sign of the second derivative we can see that −1 is a local maximum and +1 is a local minimum. Note that this function has no global maximum or minimum.

The function |x| has a global minimum at x = 0 that cannot be found by taking derivatives, because the derivative does not exist at x = 0.

The function cos(x) has infinitely many global maxima at 0, ±2π, ±4π and infinitely many global minima at ±π, ±3π

The function 2 cos(x) − x has infinitely many local maxima and minima, but no global maximum or minimum.

The function cos(3πx)/x with 0.1 ≤ x ≤ 1.1 has a global maximum at x = 0.1 (a boundary), a global minimum near x = 0.3, a local maximum near x = 0.6, and a local minimum near x = 1.0. (See figure at top of page.)

The function x3 + 3x2 − 2x + 1 defined over the closed interval (segment) [−4,2] has a local maximum at x = −1−√15⁄3, a local minimum at x = −1+√15⁄3, a global maximum at x= 2 and a global minimum at x = −4.

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

FUNCTIONS OF MORE THAN ONE VARIABLE

For functions of more than one variable, similar conditions apply. For example, in the (enlargeable) figure at the right, the necessary conditions for a local maximum are similar to those of a function with only one variable. The first partial derivatives as to z (the variable to be maximized) are zero at the maximum (the glowing dot on top in the figure). The second partial derivatives are negative. These are only necessary, not sufficient, conditions for a local maximum because of the possibility of a saddle point. For use of these conditions to solve for a maximum, the function z must also be differentiable throughout. The second partial derivative test can help classify the point as a relative maximum or relative minimum. In contrast, there are substantial differences between functions of one variable and functions of more than one variable in the identification of global extrema. For example, if a bounded differentiable function f defined on a closed interval in the real line has a single critical point, which is a local minimum, then it is also a global minimum (use the intermediate value theorem and Rolle's theorem to prove this by reduction ad absurdum). In two and more dimensions, this argument fails, as the functionshows. Its only critical point is at (0,0), which is a local minimum with ƒ(0,0) = 0. However, it cannot be a global one, because ƒ(2,3) = −5.

-The global maximum is the point at the top -Counterexample

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

MAXIMA OR MINIMA OF A FUNCTIONAL

If the domain of a function for which an extremum is to be found consists itself of functions, i.e. if an extremum is to be found of a functional, the extremum is found using the calculus of variations.

IN RELATION TO SETS

Maxima and minima can also be defined for sets. In general, if an ordered set S has a greatest element m, m is a maximal element. Furthermore, if S is a subset of an ordered set T and m is the greatest element ofS with respect to order induced by T, m is a least upper bound of S in T. The similar result holds for least element, minimal element and greatest lower bound.

In the case of a general partial order, the least element (smaller than all other) should not be confused with a minimal element (nothing is smaller). Likewise, a greatest element of a partially ordered set (poset) is an upper bound of the set which is contained within the set, whereas a maximal element m of a poset A is an element of A such that if m ≤ b (for any b in A) then m = b. Any least element or greatest element of a poset is unique, but a poset can have several minimal or maximal elements. If a poset has more than one maximal element, then these elements will not be mutually comparable.

In a totally ordered set, or chain, all elements are mutually comparable, so such a set can have at most one minimal element and at most one maximal element. Then, due to mutual comparability, the minimal element will also be the least element and the maximal element will also be the greatest element. Thus in a totally ordered set we can simply use the terms minimum and maximum. If a chain is finite then it will always have a maximum and a minimum. If a chain is infinite then it need not have a maximum or a minimum. For example, the set of natural numbers has no maximum, though it has a minimum. If an infinite chainS is bounded, then the closure Cl(S) of the set occasionally has a minimum and a maximum, in such case they are called the greatest lower bound and the least upper bound of the set S, respectively.

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

PART 1

(a)

I. Mathematicaloptiminization deals with the problem of finding numerically minimums

(or maximums or zeros) of a function. In this context, the function is called cost function

or objective function or energy.

II. In mathematicalanalysis, the maxima and minima (the plural of maximum and

minimum) of a function, known collectively as extrema (the pluralof extremum), are the

largest and smallest value of the function within the entire domain of a function (the

global or absolutely extrema). We say that f (x) has an absolutely (or global) maximum

at x=c iff (x)≤ f (c) for every x in domain we are working on.

III. In mathematical amalysis, the maxima and minima (the plural of maximum and

minimum0 of a function, known collectively as extrema (the plural of extremum) are the

largest and smallest value of the function within a given range (the local or relative

extrema).

We say that f (x) has a relative (or local) maximum at x=c iff (x)≤ f (c) for every x in

some open interval around.

We say that f (x) has a relative (or local) minimum at x=c iff (x)≥ f ¿c) for every x in

some open internal around.

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

(b)

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The maximum or minimum value of

the quadratic function

y = ax2 + bx + cDecide whether you are going to find the maximum value or

minimum value.It Is either one or the other, you're not going to find both.The maximum or minimum value of a

quadratic function occurs at its vertex.For y = ax2 + bx + c,

(c - b2/4a) gives the y-value (or the value of the function) at its vertex.

y = a(x-h)2 + kFor y = a(x-h)2 + k,

k is the value of the function at its vertex.k gives us the

maximum or minimum value of the quadratic accordingly as a is negative or positive

respectively.

y = ax^2 + bx + cDifferentiate y with respect to

x. dy/dx = 2ax + bDetermine 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

ADDITIONAL MATHEMATIC PROJECT WORK 2015

PART 2(a) y

200cm fences

x

Let height be x and width be y.

Total amount of fencing required

x+x+x+x+ y+ y

¿4 x+2 y=200

Area of pen¿ xy (length × width)

A=xy

y=100−2 x

A=x(100– 2x )

A=100 x−2x2

dAdx

=100−4 x

100−4 x=0

4 (25−x )=0

25−x=0

x=25

When x=25

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x x x

ADDITIONAL MATHEMATIC PROJECT WORK 2015

y=100−2(25)

y=50

∴The dimension

25m

5m

max area ¿50m×25m

¿1250m2⋕

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25m 25m 25m

ADDITIONAL MATHEMATIC PROJECT WORK 2015

(b)

30cm30−2h

h

30−2h

Let the side of the square to be cut off be hcm .

The volume of open box is

V=h¿

V=h(900−120h+4 h2)

V=900h−120h2+4h3

To find the maximum value

dVdh

=900−240h+12h2

¿12 (75−20h+h2 )

¿12 (h−150 ) (h−5 )

h−15=0 @ h−5=0

h=15h=5

¿is rejected, not belong to domain ofV ¿

V=h¿

V=900 (5 )−120 (5 )2+4 (5 )3

V=2000 cm3⋕

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30−2hcm

domain of V=0<x<15

ADDITIONAL MATHEMATIC PROJECT WORK 2015

PART 3

(i) Based on the equation , a table has been constructed where t represent the number of hours

and P represent the number of people.

T 0 1.5 3 4.5 6 7.5 9 10.5 12

P(t) 0 527 1800 3073 3600 3073 1800 527 0

(ii) The mall reaches its peak hours 6 hours after opening at 9.30 a.m which is at 3.30 p.m. the

number of people is 3600.

(iii) The number of people in the mall at 7.30 p.m is about 940.

(iv) Based on the graph, assuming that the malls open during business hours, the mall reaches

2570 peoples at t=3.8 ,8.1. It is 1.18 p.m and 5.36 p.m when the number of people reaches

2570.

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

*graph

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

FURTHER EXPLORATION(i)(b)

x : y ≥2:3

xy≥ 2

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∴2 y ≤3 x

Let 2 y=3 x

When x=0 , y=0 so (0 ,0)

When x=6 , y=9 so (6 ,9)

100 x+200 y ≤1400

∴ x+2 y≤14

Letx+2 y ≤14

When x=0 , y=7so (0, 7¿

When y=0 , x=14so(14,0¿

The 3 inequalities which statisfy all the constraints.

2 y≤3 x x+2 y ≤14 3 x+4 y ≤36

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3 x≥2 y

0.6 x+0.8 y≤7.2⟶(×10)

6 x+8 y≤72⟶ (÷2 )

∴3 x+4 y≤36

Let 3 x+4 y ≤36

When x=0 , y=9so(0 ,9 )

When y=0 , x=12so (12 ,0 )

3

2

1

Cabinet X Cabinet Y

Cost(RM) 100 200

Space(m2) 0.6 0.8

Volume(m3 ) 0.8 1.2

ADDITIONAL MATHEMATIC PROJECT WORK 2015

*graph

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

(ii)

METHOD 1

Volume ¿0.8 x+1.2 y

∴The equation is0.8 x+1.2 y=1.92

When x=0 ,

0.8 (0 )+1.2 y=1.92

1.2 y=1.92

y=1,921.2

y=1.6so(0 ,1.6)

When y=0

0.8 x+1.2(0)=1.92

0.8 x=1.92

x=2.4so(2.4 ,0)

From the graph, the maximum storage value ¿8 (8.0 )+3 (1.2 )

¿10m3#

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Maximum point (8, 3)

0.8×1.2×2=1.92

ADDITIONAL MATHEMATIC PROJECT WORK 2015

METHOD 2

Point Cabinetx Cabinet y Total Volume ¿0.8 x+1.2 y

(4 ,5) 4 5 0.8 (4 )+1.2 (5 )=9.2

(5 ,4) 5 4 0.8 (5 )+1.2 (4 )=8.8

(6 ,4 ) 6 4 0.8 (6 )+1.2 (4 )=9.6

(7 ,3) 7 3 0.8 (7 )+1.2 (3 )=9.2

(8 ,3) 8 3 0.8 (8 )+1.2 (3 )=10

(9 ,2) 9 2 0.8 (9 )+1.2 (2 )=9.6

(10 ,1) 10 1 0.8 (10 )+1.2 (1 )=9.2

From the table above,the maximum storage volume is 0.8 (8 )+1.2 (3 )=10m3#

(iii)

Cabinet x Cabinet y Total Cost (RM)4 6 16005 5 15006 4 14007 3 13008 3 14009 2 1300

(iv) If I was Aaron, I will choose 6 units combination of cabinet x and 2 units combination of cabinet y because its cost is below allocation which is RM1400.

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

REFLECTION

Many years ago I say, maths 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,

addmaths came into my life and "cut" me like a knife.

It's interesting oh yes it is, this cannot be denied,

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.

Addmaths, I have a question to ask, should I love you or should i hate you?

Cause 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,

Ok friends and everyone, now that I have said.

I'll just do my Addmaths homework now, and see what my brain will store.

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ADDITIONAL MATHEMATIC PROJECT WORK 2015

REFERENCES

1. https://en.wikipedia.org/wiki/Mathematical_optimization

2. https://en.wikipedia.org/wiki/Maxima_and_minima

3. https://www.mathsisfun.com/calculus/maxima-minima.html

4. Addmath’s Textbook

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