Selected Problem- A Marching Strip

PROBLEM SUMMARY:

A King has a rectangular courtyard with the dimensions of 63 rows and 90 columns. All tiles are the same, with the exception of the diagonal path from one corner of the courtyard to the other corner (for example; the front left corner to the back right corner). The tiles that are along the diagonal path are more expensive. He wants to know how many tiles he will need that will cover the diagonal path of his courtyard.

PROCESS:

In finding the solution to this problem, I simply started drawing out smaller courtyard with small dimensions to see if I could find a pattern of the amount of tiles on the diagonal path. First, I looked at a courtyard that has a dimension of 2 x 2:





When looking at this small courtyard, I noticed that the area of this courtyard is 4, after multiplying the rows (2) by the columns (2).

I then tried a larger courtyard. I tried using the dimensions of 6 columns and 4 rows. When finding the area of this courtyard, I got a total area of 24. I couldn't see any type of pattern in how the areas are increasing, so I decided to look back at my dimensions:






When looking at 4 and 6, the greatest common divisor of both numbers is 2:








The total amount of expensive tiles it took to create the diagonal path is 8, as counted from the diagram above. Then I realized that you can add the number of rows and columns, and subtract the greatest common divisor to get your number of expensive tiles needed for the diagonal path: 4 (rows) + 6 (columns)= 10... 10 - 2 (greatest common divisor of 4 & 6)= 8. I tried this same method for the smaller courtyard above:
2(rows) + 2(columns)= 4. 4- 2 (greatest common divisor of 2)= 2. This number matched the amount of expensive tiles needed to create the diagonal path since there were only two tiles on the path.


The last courtyard measurement I tried was 12 columns and 10 rows, since I could easily draw it to check my findings:







After drawing the diagram, I confirmed that there were 20 tiles needed to create the diagonal path. When looking at 10 and 12, the greatest common factor between the two of them is 2:








After finding the greatest common divisor of 2, I plugged the values into the function I found above:

10 (rows) + 12 (columns)= 22. 22- 2 (greatest common divisor)= 20.
Alas, the function worked because 20 (as taken from the drawing)= 20 (as found from the function).


I then realized that if a King were to have a courtyard with 63 rows and 90 columns, then I can use my function to find how many expensive tiles are needed for the diagonal. When factoring out 63 and 90, I got 9 as the greatest common divisor:








I then added 63 (rows) + 90 (columns)= 153. 153- 9 (greatest common divisor) = 144. This means that the king needs 144 expensive tiles to create his diagonal path.



If you were given any amount of rows or columns, you can find the tiles needed for the diagonal path from the following: ((rows)+ (columns))- (greatest common divisor of the rows and columns)= total amount of tiles needed.

Statement of Personal Growth

The Bakers Choice book was something I have never seen or worked with before. There were several times where I was stuck on a couple problems throughout the unit, but I was able to persevere and get through it. Typical math prompts only provide one answer without any deep thinking involved. Bakers Choice provides open ended questions that allow for complex thinking and most importantly the application of already understood mathematical concepts. Before starting this unit I of course knew how to graph inequalities and linear functions. I even knew how to spot the feasible region, without realizing that it was actually called a feasible region. Before Baker's Choice, I never actually knew what linear programing entailed. I now realize that linear programming is only an umbrella term that allows students to develop constraints within regular linear functions and inequalities, which are another form of a linear function. Typical linear programming problems involve maximizing profit, which is mainly what we encountered in this unit. Everyone wants to know how to maximize profit, which is why maximizing profits can greatly be applied to math and gain interest in the subject matter as well. In the profitable pictures problem, there were only 2 constraints that were used. This resulted in two linear functions that represented inequalities. The two lines had one intersection point, which coincidentally was the location that satisfied the constraints of the problem while yielding the highest amount of profit. This problem made a lot of sense to me because it is obvious that where two lines intersect should yield your best value, which in this case it did. This combination yielded the highest amount of profit. This problem was very meaningful to me because I was able to apply what I already knew about looking for a place of intersection and seeing why the intersection point of two linear inequalities represents the best possible combination. It was almost as if everything clicked in my brain. This Baker's Choice unit can be helpful in any Algebra class or integrated math class that incorporates the investigations of linear equations and linear inequalities. This unit really gives students the opportunity to think in a complex manner that will deepen their understanding. I also think it is very important to present this unit as a group activity, because multiple opinions and conjectures are needed to effectively understand these problems. This unit is also more complex, and it will be important to use group collaboration to ensure that the students who are behind don't get further behind in class.

Selected problem: Broken Eggs

The Broken Egg problem states that a farmer is taking her eggs to the market but she hits a pothole and every egg gets broken. She is trying to figure out the amount of eggs that she had before they all were broken. There are certain things that she remembers. She knows that when she puts the eggs in groups of 2, 3, 4, 5, or 6 she has one egg left over as a remainder every time. However, when she put her eggs into a group of 7, there was no eggs left over. Our task is to find possible numbers that satisfy all of our constraints. I decided to create a chart of possible numbers of eggs to see if I could eventually find a combination that works.

Here is a picture of my work:





I was able to come up with viable combinations the more that I tested out numbers because I could recall some patterns of numbers relating to divisibility:

1. When numbers are divisible by 2, the last digit in the resulting number MUST be EVEN. With one remainder left over, this means that all combinations that must work has to end in an odd number.
2. When numbers are divisible by 3, you can add up every number in the digit the resulting sum is also divisible by 3. Because there has to be a remainder of 1, I needed to find numbers that were divisible by 3 and then simply add 1. For example; 301 was the first possibility that I found. 300-1=300. 3+0+0=3, and 3 is of course divisible by 3.
3. When a number is divisible by 5, the last digit in the number must be a 0 or a 5. Knowing that there has to be a remainder of 1, the digits must end in 6 or 1.

The constraints that I knew allowed me to only guess and check certain numbers, while eliminating a lot of other possibilities. Overall, I was able to come up with 4 viable solutions to the amount of eggs that the farmer might have had; 301,721,2401, or 4501. It is helpful to notice that each number that was found has a last digit of 1.

Linnear Programming- Profitable Pictures

Profitable Pictures represents another problem that incorporates linear programming. This problem only presents two constraints, as opposed to four constraints in the Cookies problem. The artist Hassan is selling pastels and watercolors. The pastels cost 5$ each to make while the watercolors cost him 15$ to make. He overall has 180$ to spend on materials. He can only make at most 16 pictures and he wants to find the best combination of pastels and watercolors to get him the most profit.
There are two constraints:

1. p+w is less than or equal to 16 (represents the number of pictures)
2. 5p + 15w is less than or equal to 180 (represents the money that he can spend on materials)

In solving this problem, I again created different combinations of pastels and watercolors that are possible and I found the resulting profit:

Pastels (y)
5
10
0
6

Watercolors (x)
8
6
10
10

Profit
1000$
1000$
1000$
1240$

After using these combinations, I felt that the highest amount of profit that Hassan can make is 1240$. It is also important to note that (10,6) is the solution to the systems of inequalities, meaning that it is the intersection of both inequalities. This represents the best combination because it is the only point that is at the highest point of the feasible region that involves both inequalities.

Using Desmos once again, you can see the constraints on the graph and the feasible region that can be used:




Below are pictures of the work that I created for this problem. It is important to note that any stars on the graph represent possible combinations:









The last picture of my work represents combinations that can be made that yield a profit of 500$ and 600$.

Linnear Programming- Picturing Cookies part 1 and part 2

There are two typed of cookies to work with: plain and icing. The Woo's want to make the highest amount of profit possible with their bakery regarding plain and iced cookies. The Woos have several constraints. The only have 110 pounds of cookie dough and they only have 32 pounds of icing. They also only have enough room to only bake 140 dozen cookies total. The also only have 15 hours to prepare the cookies. The constraints can be represented as follows:

(x represents the number of plain cookies while y represents the number of iced cookies)
Dough: (1)x + (0.7)y is less than or equal to 110 pounds (Plain cookies need 1 pound of dough while cookies with icing only requires 0.7 pounds of dough)
Icing: (0)x + (0.4)y is less than or equal to 32 pounds of icing. (The plain cookies do not require any icing, which is why they are represented with a constant of 0.)
Time: (.1)x + (.15)y is less than or equal to 15 total hours.
Oven: x+y is less than or equal to 140 total dozen cookies

The key part to this problem is the profit and cost of each cookie. Plain cookies are sold for 6$ a dozen while iced cookies are sold for 7$. It costs 4.50$ per dozen of plain cookies. It costs 5.00$ per dozen to make iced cookies.

After using these constraints, I simply used the guess and check method to find the amount of plain and iced cookies that satisfies all of the constraints while yielding the highest amount of profit. I actually created a chart to demonstrate the possible combinations of iced and plain cookies and the profit that they will result:


PLAIN
30
110
50
0
75

ICED
80
0
30
80
50

PROFIT
205$
165$
180$
160$
212.50$

Using Desmos, you can see the graphs of the different constraints and how they form a feasible region:



Below are pictures of my work

Selected Problem: POW : Kick It!

This problem allows students to work with different combinations of numbers. It is especially important to note that properties of even numbers, odd numbers, and a combination with even and odd numbers are very important to understand. This problem will help students to explore relationships with even and odd numbers.

POW: KICK IT
Problem Statement:
The Free Thinkers Football League is able to score 5 points for every field goal and 3 points for every touchdown. The football league can score with a combination of a field goal and a touchdown, or just one event at a time (one field goal or one touchdown). One team member on the Free Thinkers Football League notices that not every score is possible. For example, a score of 1 or 2 is not possible because a touchdown yields 3 points while a field goal yields 3 points. The Free Thinkers Football League wants to know if there is a score that represents the highest impossible score they won’t be able to get. Is there a highest impossible score?
The Football League then wants you to come up with other scoring systems using whole numbers and see if there are scores that are impossible to make. Is there a highest impossible score with each scoring system that you created? Is there ever a situation where there is not a highest impossible score? Are you able to find a rule or condition that explains what scoring combinations yield highest impossible scores and what combinations do not?
In combinations that do yield a highest impossible score, can you find any patterns or rules to help people figure out what the highest impossible score is?







Process:
When I went to solve this problem, I first identified that a field gold represents 5 while a touchdown represents 3. After stating my givens, I created a chart to help document my findings. I then started a tally of impossible scores that I found while I was writing out possible combinations in my chart:


At first I thought that 17 was the highest impossible value because I thought to myself; “5+5+5=15, and I can’t add 3 to that to make 17”. I did not think about the combination of multiplying 3 by 4 to yield 12, and then adding 5 to get a score of 17. I did not come to this conclusion until after I presented my findings to the class. After realizing that 17 does work out, I realized that 7 was my highest impossible score that can be achieved with the given constraints. I was able to realize that 7 was the highest impossible score because with every score after 7, you can keep adding 10 to get all possible whole numbers greater than 7.

I then created an additional scoring tally which the instructions request:














After creating this diagram, I realized 1 was my highest impossible score because I can still add 10 (represents 5 touchdowns) like the first score combination. I also realized that in my work I noted that 3 was an impossible score, not taking in to account that 3 can represent one field goal.


Solution:
(First combination, field goal is 5 points and touchdown is 3 points)
After creating the chart, I stopped at the combination of 30. I stopped here because I realized that when you look at an increment of 10 that is possible, let’s say from 20-30, I concluded that you can add 10 to each number in the increment to make the next set of 10. You can keep repeating this process to get every single other whole numerical possibility. I chose to add 10 because 10 is a possible point combination for this constraint since 10 is composed of two field goals, or in equation terms 5+5=10. For example:
1. 20+10= 30 (30 is composed of 6 field goals)
2. 21+10= 31 (31 can be created with 7 touchdowns, yielding 21, and two field goals, yielding 10. 21+10= 31)
3. 22+10= 32 (32 can be created with 4 touchdowns, yielding 12, and 4 field goals, yielding 20. 20+12=22)
4. 23+10= 33 (33 can be created with 1 touchdown, yielding 3, and 6 field goals, yielding 30. 30+3=33)
5. 24+10= 34 (34 can be created with 8 touchdowns, yielding 24, and 2 field goals, yielding 10. 24+10= 34)
6. 25+10= 35 (35 can be created with 7 touchdowns, yielding 35)
7. 26+10= 36 (36 can be created with two touchdowns, yielding 6 points, and 6 field goals, yielding 30. 30+6=36)
8. 27+10= 37 (37 can be created with 9 touchdowns, yielding 27 points, and two field goals, yielding 10. 27+10=37)
9. 28+10= 38 (38 can be created with 1 touchdown, yielding 3, and 7 field goals, yielding 35. 35+3=38)
10. 29+10= 39 (39 can be created with 3 touchdowns, yielding 9 points, and 6 field goals, yielding 30. 30+9= 39.
11. 30+10= 40 (40 can be created with 5 touchdowns, yielding 15, and 5 field goals, yielding 25. 25+15=40
Every single number in this process just added 10, which represents two field goals. This process works because 10 is a possible combination, being two field goals.
(Second combination, field goal is 3 points and a touchdown is 2 points)
After using this score combination, I realized that I could again add 10 within a 10 increment like in the scoring above because in this scoring system, 10 represents 5 touchdowns. I also realized though that instead of just using an increment of 10 like the first combination, I can use a smaller increment. I can take an increment of 6, for example, because 6 represents a possible score, which translates to 2 field goals or 3 touchdowns.
(Other possible combinations)
After going over two separate combinations, I realized that if both the touchdown and the field goal yield the same score, then there is no highest possible amount because all possible scores have to be divisible by the chosen score. For example, let’s say that a field goal represents 2 points while a touchdown represents 2 points as well. Using 2 means that all score possibilities will be even numbers. This is because any number multiplied by 2 will yield an even result. This means that all odd numbers represent impossible scores and there is no definite highest impossible score that can be calculated. All possible combinations of field goals and touchdowns that use even numbers will never yield a highest impossible score. This is because anytime an even number is multiplied, added, or subtracted with another even number, the answer will always yield an even number. This leave the odd numbers to represent impossible scores. I was able to use my prior knowledge to help come to this conclusion. Odd numbers can result when two even numbers are divided, but this problem is not asking for division. This means that again, with the 2 and 2 combination, all odd numbers will represent the highest impossible scores and thus there is no highest impossible score that can be recorded. I realized that when you are looking at the combinations of any score, you can simply look at the highest score value, whether touchdown or field goal, and multiply that by two. The product will yield a testable increment with your score combinations. (For example: Touchdown is 3 points and Field Goal is 5 points. The increment used was 10 because the highest score was 5, and 5 multiplied by 2 yields 10. After 7, the increment of 10 from 8-18 can be proven possible because you can add 10, which is again composed of two field goals, to every single number to get your desired value. You can keep adding 10 to get every number after 7. When using the combination of the touchdown representing 2 points and the field goal represents 3 points, I used an increment of 6 because 3 multiplied by 2 equals 6. The highest impossible value was 1 because the increment of 6 after 1, from 2-7, was able to be calculated. I was able to take this increment and add 6 to each number to yield the next increment of 6. You can keep repeating this process to get every single number above your highest impossible value of 1. )


Extensions:
This problem can be extended to further knowledge regarding properties of numbers. This problem required knowledge of combinations of numbers, specifically what happens when you have combinations of two even numbers, two odd numbers, and both an even and odd number. Possible strategies to help extend thinking include:
1. Are there any score combinations that yield a highest possible score of 10 or more? (Allows students to work with and experiment with greater numbers)
2. List three separate score combinations. 1 score combination needs to represent an even and an odd number, 1 score combination needs to represent two even numbers, and 1 score combination needs to represent two odd numbers. (This prompt is helpful to drive students right to the catch that two combinations of even numbers can’t yield a highest impossible score and the other two combinations can yield a highest impossible score.)
Evaluation:
Option B: Personal Reaction to the Problem
When I first encountered this problem I thought it was a mouthful. I was first unsure of how to approach the problem. Because I am a visual learner, I decided to start creating a chart of score combinations that were possible and hopefully I will be able to see the highest impossible score, if any. This problem was educationally worthwhile because it reminded me of how important it is to understand the properties of odd and even numbers and the possible combinations they can make. For this problem a combination of two even numbers, two odd numbers, and a combination with one even number and one odd number was very important to understand, in regards to addition and multiplication. I was able to access and apply my prior knowledge of these different combinations to help come to my conclusion in this problem. I enjoyed working on this problem because it challenged me to recall prior knowledge and it allowed me to apply that prior knowledge. This problem was enjoyable to me because again it only proves that mathematics constantly builds on itself, which is very enjoyable to me. I found that even though in my first chart I wrote numerical combinations a lot higher than what I needed to, I enjoyed doing it because it was something I understood. It was almost like a puzzle to me in trying to find the missing piece, as the highest impossible score represented that missing piece. Eventually I was able to stop and try to look for patterns, which allowed me to conclude that the highest impossible value I found was in fact true. This problem at first was hard for me because it was not a problem that could be solved using one or two steps. After taking the problem piece by piece I was able to break it down and make it easier. I did this through creating the diagram and listing the possible score combinations. This is a great problem to use in math classrooms because it encompasses several different math practices; persevering in solving problems, attending to precision, and looking for structure and regularity in repeated reasoning. (retrieved from: http://www.corestandards.org/Math/Practice )

Cover Letter

The Bakers Choice unit covers problems that allow students to see the effects and implications of linear equations and linear inequalities and how they can be applied to linear programming. It is important for students to be able to understand how to graph linear equations and inequalities. Using the graphs of linear equations and inequalities, students will be able to compare the graphs to develop meaningful solutions.
Picturing Cookies was the first problem for students to tackle regarding linear programming. The problem required students to investigate how many combinations of plain and iced cookies can be created with given constraints. Students were given four constraints to work with; the amount of dough that can be used, the amount of icing that can be used, the amount of time that they are given to make the cookies, and the amount of oven space. These four constraints can be represented using inequalities:

Dough: (1)x + (0.7)y is less than or equal to 110.
Icing: (0)x + (0.4)y is less than or equal to 32
Time: (.1)x + (.15)y is less than or equal to 15
Oven: x+y is less than or equal to 140.
Using these inequalities, x represents the amount of plain cookies while y represents the amount of iced cookies.
Graphing these constraints will give present a region that satisfies all 4 equations, which is also known as a feasible region. From there, students can experiment with the amount sold and the amount of profit to help determine the best combination that will yield the highest profit with the least amount of cost.

Profitable Pictures was another problem to help demonstrate linear programming and utilizing given constraints. Profitable pictures provides a problem with less constraints in comparison to Picturing Cookies (Part 1 and 2). Profitable pictures provides students with two separate pictures to sell; pastels and watercolors. The pastels and watercolors present different profits. Pastels can yield a profit of 40$ while watercolors can yield a profit of 100$. The cost of each pastel is 5$ and each watercolor costs 15$ to make. There are two constraints in this problem: Only 16 pictures can be made and there is only 180$ to spend on materials. The two constraints can be written as follows:

Number of pictures: p+w is less than or equal to 16
Money for materials: 5p + 15w is less than or equal to 180.
Using these constraints, p represents the pastels and w represents the watercolors. After graphing these two inequalities students will again see a feasible region that can represent possibilities for how many pastels and how many watercolors can be created. After this feasible region is defined, students can experiment with the feasible region to find possibilities of the amount of pastels and watercolors that can be made. Students can then further their thinking by testing out possible values to find the combination that will yield the highest profit overall.

Linear programming problems are important not only in mathematics but in everyday life. The problems in Bakers choice center on finding the best profit region. This can be applied to everyday life, in the fact that everyone wants to spend the least amount of money possible to gain the highest profit in return. Linear programming is an effective method to help determine cost constraints to ensure that you are gaining the highest amount of profit possible.