This ejournal will be based on the topics of algebra and graphs. I’ll be showing my Kognity investigations, teaching you how to pronounce some mathematical expressions as well as sharing international mindedness and the Cambridge learner profiles. The digital tools I used include Applet, Desmos, and Microsoft Excel.
Fun Facts
Pronouncing mathematical expressions and symbols correctly can be very important. You wouldn’t want to embarrass yourself, would you? So, here are the pronunciations of some expressions and symbols that I’ve learnt.
x≡y — x is equivalent to (or identical with) y
n! — n factorial
f (x) — fx/f of x/the function f of x
A-1 — A inverse / the inverse of A
x-n — x to the (power) minus n
∴ — therefore
dv — the derivative of v
Kognity Investigations : “2.4.11 Graphs of Function”
Investigation 1 : “Graphs of Linear Function”
1. Set a to a positive value and b to 0. Explain, by thinking about what the function does, why the graph of this function rises from left to right. (Hint: what happens to the output of the function as the input increases?)
If a = 2 and b = 0, the graph of the function rises from left to right since the gradient is positive. When the input of a increases, it’ll rise. When the input of b increases, it’ll move further to the right.
2. Set a to a negative value. Explain why the graph of this function falls from left to right.
The graph of this function falls from left to right as the output will increase whenever the input does too. So, as a (and b) decreases here, the output decreases as well and causes the graph to fall from left to right.
3. Keeping a constant, explore the effect of changing b. Describe and explain how the graph changes as b increases.
Changing b would change the y intercept, but the gradient will stay the same. As b increases, the graph will move further to the right. So, consequently, when b decreases, the graph will move further to the left.
Investigation 2 : “Graphs of Quadratic Function”
1. Set a=1, b=0 and c=0. Explain, by thinking about what the function does, why the graph of this function falls, reaches a minimum, and then rises (as you look from left to right).
f(x)=(y=x2 )
Whatever value of x we put, the result will always be a positive number (x2=positive). In this case, y = x2. There is no chance of x2 being a negative number. So, the minimum is 0 since 02 = 0. The graph rises up because, again, x2 couldn’t be a negative number.
2. Change the value of a to −1. Describe and explain how this changes the graph.
f(x)=(y=-x2 )
Whatever value of x we put, the result will always be a negative number (negative multiplied by positive will equal to negative) as it’s -x2. So, the maximum is 0 since -02 = 0. The graph will fall back down as -x2 couldn’t be a positive number.
3. Explore the effect of changing c. Describe and explain how this affects the graph.
If c changes, the y intercept will also change since it’s equivalent to c. So as c increases, the y intercept increases as well. Hence, the graph rises.
4. (Challenge) Set a=1 and c=0. Switch between b=1 and b=0. Describe and explain the difference this makes. (Hint: the function x2+x is the sum of the functions x2 and x.)
If b changes, the the vertex/turning point will also change. As b increases, the parabola or vertex will move left and downwards more.
Investigation 3 : “Graphs of Reciprocal Functions and Functions of the Form a/x2“
Q: Explore the effect of changing n from 1 to 2. Describe your observations and try to explain them.
A:
When n=1, the function stays as a/xn and the graph’s lines face each other diagonally (with one line being under the x axis while the other is above said axis). When n=2, the x in the function is squared and the entire graph stays on top of the x axis as the y intercept doesn’t go below 0.
Investigation 4 : “Graphs of Cubic Functions”
Q: Explore the effects of changing the values of a, b, c and d. Describe your observations and try to explain some of them. (Hint: start by setting the values of b, c and d to 0. Then explore the effect of changing one at a time.)
A:
If we change a, the steepness of the cubic “s” curve will change as well. As a gets larger, the curve gets steeper and ‘narrower’. When a is negative, it slopes downwards to the right.
If we change b, the curvature of the parabolic element will also change. As b gets larger, the parabola gets steeper and ‘narrower’. When b is negative, it slopes downwards each side of the vertex.
If we change c, the graph’s slope will change. The slope increases when c decreases, and decreases when c increases.
Changing d moves the graph up and down. When d increases, the graph will move up. Hence, when d decreases, the graph will move down.
Investigation 5 : “Graphs of Exponential Functions”
Q: Explore the effects of changing the value of a. Describe your observations and try to explain them.
A:
When a is 1, the graph will form a straight line. When a is more than 1, the graph will slowly move closer towards the y axis as a increases. We can confirm this by comparing the graphs in figure (24) and figure (25). The graph in figure (25) is much closer to the y axis since its a value is double of figure (24)’s.
Kognity Investigations : “2.7.4 Linear Programming”
Q: Research the uses of linear programming in the real world and write a short (500 words) article.
A: Linear programming is an optimization technique for a system of linear constraints and a linear objective function. This program is used for obtaining the most optimal solution for a problem with given constraints. In linear programming, we formulate our real-life problem into a mathematical model or, in simpler words, graphs (which is especially helpful for visual learners).
But, to be exact, how do we actually solve Linear Programming questions?
- Determine the objective function, which must be maximized (for example : cost or profit), and express it in the terms x and y, which would then become the decision variables.
- State the inequalities which are imposed on each side of the decision variables.
- Using x and y axes, sketch the areas defined by the inequality statements (shade the unwanted region).
- Find the co-ordinates of the vertices (corner points) of the unwanted region.
- Find the solution of the problem by calculating the maximum/minimum profit (or cost, budget, etc) for each corner point of the unwanted region, the wanted region as well as the objective function.
These instructions can be very confusing to many. So, to make things easier and more understandable for the readers, here are some questions with full and correct working.
Additionally, I know that this topic can be boring and dreadful to a lot of people. With this kept in mind, I decided to make these questions based on fun games such as Pokemon and Clash of Clans! I truly hope these questions can make solving linear programming clearer and more enjoyable for everyone.
Q1 : A Pokeball factory that distributes Pokeballs across the Pokemon community makes two types of balls : greatballs and ultraballs. Each greatball requires 5 hours from the fabricating department and 2 more hours from the finishing department. Each ultraball requires 4 hours from the fabricating department and 3 hours from the finishing department.
The maximum amount of of production hour available per week in the fabricating department and the finishing department are 200 hours and 108 hours respectively. If Niantic (the company) makes a profit of $40 on each ultraball and $50 on each greatball, how many Pokeballs of each type should be produced each week to maximize the total weekly profit (assuming that all Pokeballs can and will be sold)? Additionally, what is the maximum profit?
A :
1. Objective function and decision variables :
We have to find the maximum profit of the company, given that it makes a profit of $40 on each ultraball and $50 on each greatball. Thus, our decision variables will be the ultraball and the greatball.
x = ultraball, y = greatball
Therefore, the objective function is c=40x+50y
2. Inequalities :
From the table above, we can conclude :
5x+4y ≤ 200
2x+3y ≤ 108
x ≥ 0
y ≥ 0
Now, we have to find both x and y :
x intercept = (5x+4(0) = 200)
5x = 200
x=40
y intercept = (5(0)+4y = 200)
4y = 200
y = 50
Therefore, the first equation is 5x+4y ≤ 200
x intercept = (2x+3(0) = 108)
2x = 108
x = 54
y intercept = (2(0)+3y = 108)
3y = 108
y = 36
Therefore, the second equation is 2x+3y ≤ 108
3. Graph :
According to the graph, the wanted region is the area which is enclosed by the quadrilateral (which is, to be more specific, a trapezium) OABC. The vertices are : A(0, 36), B(24, 20), C(40, 0), O(0, 0). B is the intersection between the two inequalities whereas O is the origin.
4. Find the maximum value from the objective function :
Make a table so that this can be more understandable :
We got the results of c = 40x+50y by substituting the corner points of x and y into this very equation. By looking at the results, we can then conclude that the maximum profit that Niantic can earn is $1960.
Q2 : A training camp in Clash of Clans is training troops for the clan war. There are two troops trained : dragons and balloons. Each dragon takes 10 minutes to be trained and 20 minutes to upgrade whereas each balloon takes 20 minutes to be trained and 30 minutes to upgrade.
The maximum training hours a week for dragons and balloons are 800 hours and 640 hours respectively. If the cost of training for dragons is 600 elixir while the cost for balloons is 300 elixir, what is the maximum amount of elixir that a trainer should send?
A :
1. Determine the objective function :
We have to find the maximum amount of elixir, given that the training cost for balloons is 300 elixir and cost for dragons is 600 elixir. Thus, our decision variables will be the balloon and dragon.
x = balloons, y = dragons
Therefore, the objective function is c = 300x + 600y
2. Sort out the inequalities :
From the table above, we can conclude :
20x+30y ≤ 300
10x+20y ≤ 600
x ≥ 0
y ≥ 0
Now, we have to find both x and y :
x intercept = (20x+30(0) = 300)
20x = 300
x=15
y intercept = (20(0)+30y = 300)
30y = 300
y = 10
Therefore, the first equation is 20x+30y ≤ 300
x intercept = (10x+20(0) = 600)
10x = 600
x = 60
y intercept = (10(0)+20y = 600)
20y = 600
y = 30
Therefore, the second equation is 10x+20y ≤ 600
3. Graph :
The Corner Points :
A = (0, 10)
B = (0, 15)
O = (0, 0)
4. Find the maximum value from the objective function :
Make a table so that this can be more understandable :
Therefore, by looking at the results, we can conclude that the maximum amount of elxir a trainer should spend is 9000.
Real Life Applications of Algebra and Graphs
Cooking — When cooking, measurements of ingredients are vital. Putting too much or too little salt can definitely affect the outcome taste. With this, algebraic methods and operations can come in handy. For example, they can be used to convert temperature scales such as from Celsius to Fahrenheit.
Medicine — Linear graphs can be used in medicine and pharmacies to figure out the accurate strength of drugs. These graphs are vital factors, and can be of great help, for the overall study of drugs.
Sports — Quadratic functions and graphs can play a huge part in the technical side of sports. For example, they can be used to find the maximum height of a football (once thrown), to calculate and compare the parabolas of golf balls, etc.
International Mindedness
You might have been wondering why the title of this ejournal is “bonesetting.” Well, the word “algebra” is actually derived from the Arabic word “al-jabr” which means “reunion of broken parts” and “bonesetting.” So it’s not a surprise that algebra had originated in Egypt and Babylon.
Abu Jaafar Mohammad Ibn Mousa Al Khwarizmi and Diophantus are considered the two most famous “Fathers of Algebra.” Khwarizmi developed methods for balancing and reducing algebraic equations and introduced algorithms. Diophantus wrote 13 books entitled ‘Arithmetica’, which contain problems and solutions that have furthered algebraic notation (Hodges & Cena, n.d.).
Cambridge Learner Profiles
Confident : While doing the investigations, I made sure to be confident. While working as a pair, I didn’t want to make my partner feel unsure of my contributions because I wasn’t confident in them. This is a reason why confidence is important in teamwork.
Responsible : Responsibility is always very important. Especially when working as a pair, I knew I needed to be responsible in doing my parts by putting effort and having them ready on time.
Reflective : Reflecting is an always when it comes to doing investigations. When I learn new things and explore topics, I reflect in order to properly register them into my mind.
Innovative : The investigations done include the use of innovative methods all the time. This helps me to really develop critical thinking and creativity, particularly in solving tricky questions.
Engaged : I engage by focusing and paying attention in class, especially when something is being explain. This, in turn, makes doing the investigations easier as I’m already familiar with the topics at hand.
References
13 Examples Of Algebra In Everyday Life. (n.d.). Studious Guy. Retrieved November 20, 2020, from https://studiousguy.com/examples-of-algebra-in-everyday-life/
Doyon, M. (2002, January 24). Parabolic Golf Shot Equations. Math Forum. http://mathforum.org/library/drmath/view/53302.html
Ferguson, T. (2016, June 14). THE SPORT OF SOLVING QUADRATIC EQUATIONS. Thought Hub. https://www.sagu.edu/thoughthub/the-sport-of-solving-quadratic-equations
Hodges, C., & Cena, C. (n.d.). History of Algebra: Lesson for Kids. Study. Retrieved November 20, 2020, from https://study.com/academy/lesson/history-of-algebra-lesson-for-kids.html
Valiaho, H. (1999, February 17). Pronunciation of mathematical expressions. Par.Cse.Nsysu.Edu.Tw. http://par.cse.nsysu.edu.tw/link/math-pronunciation.pdf
