Meadows or Malls
Write Up
To solve this problem, we went through several units identifying the constraints and methods we would need to know to solve this problem. We began our research for this problem by being introduced to the corner point principle. The corner point principle is the process of graphing constraints to create corner points within a graph, which will create a feasible region. The feasible region is the area within a graph where all of the possible answers will be. Anything outside of this region is not a possible answer, or it is not the best answer. We learned this so we could set a feasible region for the meadows or malls problem and find which combination of land usage would have the lowest cost. We had several equations to follow in creating a region on the graph and we had to go through several different combinations of equations and constraints to see which ones held true. The corner point principle also leads us into the Inconsistent systems of equations, which are systems of equations that have no answer or an infinite answer. We were also taught how to identify independent and dependent equations. A dependent solution has an infinite answer/s, while an independent equation has one answer. To solve the unit problem, we could eliminate any answer we got if it was not concise or one answer; or if it even had an answer because sometimes equations and contriant combos wouldn’t be possible to solve. But back to the equations and constraints, we were given 4 mandatory equations to solve and had to find the bust combination of two constraints to go along with the system of equations. And if the answers failed to follow the constraints, then we would know that it was faulty. But now we were at a crossroads, where we knew how to find a feasible region, apply constraints to an equation/solution, but we needed a way to chart all of this. Which is when we were introduced to matrices. A matrix is like a grid or chart of numbers used to express statistics or collections of data. Matrices are useful because they are almost like a chart and a number graph in one. In algebra, they are used to chart and graph numbers for data. A matrix has two parts, columns and rows. The rows represent different equations, and the columns represent values, variables, and constraints. There could also be several matrices, one to represent the system of equations, and another to represent the answers. This is applicable to our unit problem given that we have equations and constraints yet need to apply both of them in one chart to be manipulated. We had 4 equations and 2 constraints in a full system; a total of 6, which would be the 6 rows in our equation. Now instead of making two separate charts for equations and constraints, we can have it in one graph/chart.
The reason we were introduced to applying matrices instead of simple charts is because we needed to find the inverse of an equation/constraint combo for our answer. Traditionally, we were taught that it does not matter which way we multiply two numbers because we will always get the same answer. This is not the case because we need to multiply everything in a specific placement. The answer to our system of equations when put in a matrix is the inverse of that matrix. For instance, we charted matrix A which is the system of equations, and matrix B which are the answers both have 6 rows, for 6 equations/constraints, and six answers to them. To get our answer we need to find the inverse of matrix A (trust me Julian, I’m not just explaining the unit problem, I’m going somewhere with this). To do that we multiply ‘A’ the matrix, by ‘-A’ in a matrix calculator.
So what a matrix does for us is two things; one, it finds the inverse of each variable in each equation/constraint which can be done by hand, but it is very difficult and time consuming. And it takes those inverses and multiplies them by each individual variable. Everything can not be multiplied together because we have more than one answer, which is why we use a matrix instead of charts, or one ridiculously long equation.
We learned several different skills and units throughout the past 8 weeks leading up to this problem, with each being as important as the last. And in the end, they all tied together to help us solve this complex problem in a smarter, more efficient manner.
Reflection
Obviously I'm not perfect when it comes to math and it’s rigorous units. But in the recent year and especially with this project, I've noticed a shift in perseverance and advocacy that I hadn’t utilized or have attempted before, especially in this subject. Because typically, when I don’t understand something or am having a hard time with it I would go dead silent and hope that eventually I would understand from just listening. But I’m asking questions, asking for help when I need it, letting my group mates know that I want to help and telling them what work I want to do. I’ve been seeing improvements in my collaborative skills and advocacy this year, and this project greatly improved them even further.
Dating back almost two years ago, covid nearly broke me as a learner. When we went fully online, I stopped showing up for class, didn’t do the assignments, or I would do just barely enough so that my teachers knew that I was still there. And once I seeked guidance from my parents and teachers, then I noticed a change in my attitude and view on school. Even with help, the bulk of covid was a test of perseverance and responsibility in myself. And once we started to attend school in person again, I saw instantaneous change in my academic behavior. Asking questions, persevering, collaborating, being present mentally. And now that we have reached a point as of now where life is almost if not back to the way things were over two years ago, I finally get to capitalize and showcase the skills I have gained during covid.
To solve this problem, we went through several units identifying the constraints and methods we would need to know to solve this problem. We began our research for this problem by being introduced to the corner point principle. The corner point principle is the process of graphing constraints to create corner points within a graph, which will create a feasible region. The feasible region is the area within a graph where all of the possible answers will be. Anything outside of this region is not a possible answer, or it is not the best answer. We learned this so we could set a feasible region for the meadows or malls problem and find which combination of land usage would have the lowest cost. We had several equations to follow in creating a region on the graph and we had to go through several different combinations of equations and constraints to see which ones held true. The corner point principle also leads us into the Inconsistent systems of equations, which are systems of equations that have no answer or an infinite answer. We were also taught how to identify independent and dependent equations. A dependent solution has an infinite answer/s, while an independent equation has one answer. To solve the unit problem, we could eliminate any answer we got if it was not concise or one answer; or if it even had an answer because sometimes equations and contriant combos wouldn’t be possible to solve. But back to the equations and constraints, we were given 4 mandatory equations to solve and had to find the bust combination of two constraints to go along with the system of equations. And if the answers failed to follow the constraints, then we would know that it was faulty. But now we were at a crossroads, where we knew how to find a feasible region, apply constraints to an equation/solution, but we needed a way to chart all of this. Which is when we were introduced to matrices. A matrix is like a grid or chart of numbers used to express statistics or collections of data. Matrices are useful because they are almost like a chart and a number graph in one. In algebra, they are used to chart and graph numbers for data. A matrix has two parts, columns and rows. The rows represent different equations, and the columns represent values, variables, and constraints. There could also be several matrices, one to represent the system of equations, and another to represent the answers. This is applicable to our unit problem given that we have equations and constraints yet need to apply both of them in one chart to be manipulated. We had 4 equations and 2 constraints in a full system; a total of 6, which would be the 6 rows in our equation. Now instead of making two separate charts for equations and constraints, we can have it in one graph/chart.
The reason we were introduced to applying matrices instead of simple charts is because we needed to find the inverse of an equation/constraint combo for our answer. Traditionally, we were taught that it does not matter which way we multiply two numbers because we will always get the same answer. This is not the case because we need to multiply everything in a specific placement. The answer to our system of equations when put in a matrix is the inverse of that matrix. For instance, we charted matrix A which is the system of equations, and matrix B which are the answers both have 6 rows, for 6 equations/constraints, and six answers to them. To get our answer we need to find the inverse of matrix A (trust me Julian, I’m not just explaining the unit problem, I’m going somewhere with this). To do that we multiply ‘A’ the matrix, by ‘-A’ in a matrix calculator.
So what a matrix does for us is two things; one, it finds the inverse of each variable in each equation/constraint which can be done by hand, but it is very difficult and time consuming. And it takes those inverses and multiplies them by each individual variable. Everything can not be multiplied together because we have more than one answer, which is why we use a matrix instead of charts, or one ridiculously long equation.
We learned several different skills and units throughout the past 8 weeks leading up to this problem, with each being as important as the last. And in the end, they all tied together to help us solve this complex problem in a smarter, more efficient manner.
Reflection
Obviously I'm not perfect when it comes to math and it’s rigorous units. But in the recent year and especially with this project, I've noticed a shift in perseverance and advocacy that I hadn’t utilized or have attempted before, especially in this subject. Because typically, when I don’t understand something or am having a hard time with it I would go dead silent and hope that eventually I would understand from just listening. But I’m asking questions, asking for help when I need it, letting my group mates know that I want to help and telling them what work I want to do. I’ve been seeing improvements in my collaborative skills and advocacy this year, and this project greatly improved them even further.
Dating back almost two years ago, covid nearly broke me as a learner. When we went fully online, I stopped showing up for class, didn’t do the assignments, or I would do just barely enough so that my teachers knew that I was still there. And once I seeked guidance from my parents and teachers, then I noticed a change in my attitude and view on school. Even with help, the bulk of covid was a test of perseverance and responsibility in myself. And once we started to attend school in person again, I saw instantaneous change in my academic behavior. Asking questions, persevering, collaborating, being present mentally. And now that we have reached a point as of now where life is almost if not back to the way things were over two years ago, I finally get to capitalize and showcase the skills I have gained during covid.
PreCalc Year Reflection
In recent years, I have been getting better at advocating for struggles in math. And in return, I have been seeing improvement in my mathematical skills and learning pace. The biggest habit I’ve been working towards and still do, is advocacy. I have always struggled with math, and never advocating shoved me down in a pit even further. But since I started advocating more and more, I’ve been seeing improvement in my grades. And I’m not just turning in work, I’m actually understanding what is being taught to me. As opposed to repeating or copying what Julian is telling me. When you let things slip away when you don’t understand them, it will hurt you in the long run. I used to think that if I just let this one unit go because I don’t understand it, then I would do better in the next unit and focus even more attention on that; it doesn’t work like that. To take 5 steps you need to take the first 4.
The second biggest habit I've been working on is refinement. I will be the first to say that I HATE going back to refine my work with a passion. I want to get something done and then have it be over with. But I found that if I do not refine my work when it’s heavily flawed, that it will still leave me not understanding what we did. If I wanted to truly understand what is taught to me, then I had to go back and correct my mistakes. Otherwise, my learning can’t show through; I might as well have guessed everything the whole unit.
One major thing that kind of falls behind the scenes is how much it helps to take affirmative action on your struggles. So many students push off work, push off their problems because they don’t know how it feels once their problems are solved and once they’ve seeked help. Cause ask anyone who's seeked support including myself, and they will praise all day how much they’ve changed as a learner. And I mean obviously I'm not cured of procrastination and my no mathematical brain, but I've gotten better tremoundacley through the habits of heart and mind that I’ve practiced.
The second biggest habit I've been working on is refinement. I will be the first to say that I HATE going back to refine my work with a passion. I want to get something done and then have it be over with. But I found that if I do not refine my work when it’s heavily flawed, that it will still leave me not understanding what we did. If I wanted to truly understand what is taught to me, then I had to go back and correct my mistakes. Otherwise, my learning can’t show through; I might as well have guessed everything the whole unit.
One major thing that kind of falls behind the scenes is how much it helps to take affirmative action on your struggles. So many students push off work, push off their problems because they don’t know how it feels once their problems are solved and once they’ve seeked help. Cause ask anyone who's seeked support including myself, and they will praise all day how much they’ve changed as a learner. And I mean obviously I'm not cured of procrastination and my no mathematical brain, but I've gotten better tremoundacley through the habits of heart and mind that I’ve practiced.
Pow 4: Planning the Platforms
For this problem of the week, we were asked to solve a problem for the increasing rate in height for platforms relevant to the number of said platforms. A band concert will be hosted on the fourth of July, and they plan to have baton twirlers attend the party and perform. Yet Kevin, who is the one planning the party, wants each baton twirler to have their own platform. He hasn’t decided how tall he wants the first platform nor what the difference in height will be for each platform. The reason being is that the platforms will get taller at a constant rate, and the height of the tallest one will depend on the number of platforms. And a separate planner Camilla, is planning to have ribbons hang off of each platform. Although she will be using a similar formula as Kevin to find out the length of ribbon she needs to buy, she can’t get started or apply it until Kevin makes up his mind on the height of the starting platform. My task is to create two equations for the platform rate in increasing height, and for the length of ribbon she needs to buy based on the starting platform height.
When beginning this pow, I had a general idea of where I would go with the equation. Because there is a starting value, an unknown number of platforms, and an increasing rate of change. So I thought I could start by using the equation Y=Mx+B; where Mx is the increasing platform height relevant to the number of platforms there are, and B is the height of the first platform or the starting value. Just as I was getting started Victor pointed out that although I was heading towards the right formula, there were a couple of variables that I wasn’t accounting for. It was him who suggested the revised equation Pn=b+m(n-1) ; where n is the number of platforms whatever it may be, Pn is the height of all of the platforms combined, b is the height of the first platform, and m is the interval of added height or how much the height is increasing with each rising platform. And out of the few equations that I was suggested and worked with, with my peers this one seemed to be the most accurate. If we break it down, each component makes sense. B+M, represents a constant increase of height for each platform, going off of the height of the first platform. And n-1 is to eliminate the starting value/length. As for the second equation, I saw the first as being applicable to the length of the ribbon.
As stated before, our goal was to find an equation for the rising platforms and the length of ribbon needed. And out of the several equations that my peers and I have tested, the equation harper, victor and I worked through seems to be the moist fit; at least to our current understanding.
While solving this problem, I tried to step out of my comfort zone a little and ask for help from my peers. And to my surprise, I actually learned a lot and they were very useful. Victor gave me a run down…in his own words. And Harper helped me solidify my understanding of the problem. And although my answer may not be fully correct, I do believe I have the right concepts for solving the problem and others like this.
If we were to find an equation that can be applied to any increasing rate and other similar variables, what would it be? And more so, is there a limit to the amount of constraints one of these problems could have to where the equation could not be applied?
When beginning this pow, I had a general idea of where I would go with the equation. Because there is a starting value, an unknown number of platforms, and an increasing rate of change. So I thought I could start by using the equation Y=Mx+B; where Mx is the increasing platform height relevant to the number of platforms there are, and B is the height of the first platform or the starting value. Just as I was getting started Victor pointed out that although I was heading towards the right formula, there were a couple of variables that I wasn’t accounting for. It was him who suggested the revised equation Pn=b+m(n-1) ; where n is the number of platforms whatever it may be, Pn is the height of all of the platforms combined, b is the height of the first platform, and m is the interval of added height or how much the height is increasing with each rising platform. And out of the few equations that I was suggested and worked with, with my peers this one seemed to be the most accurate. If we break it down, each component makes sense. B+M, represents a constant increase of height for each platform, going off of the height of the first platform. And n-1 is to eliminate the starting value/length. As for the second equation, I saw the first as being applicable to the length of the ribbon.
As stated before, our goal was to find an equation for the rising platforms and the length of ribbon needed. And out of the several equations that my peers and I have tested, the equation harper, victor and I worked through seems to be the moist fit; at least to our current understanding.
While solving this problem, I tried to step out of my comfort zone a little and ask for help from my peers. And to my surprise, I actually learned a lot and they were very useful. Victor gave me a run down…in his own words. And Harper helped me solidify my understanding of the problem. And although my answer may not be fully correct, I do believe I have the right concepts for solving the problem and others like this.
If we were to find an equation that can be applied to any increasing rate and other similar variables, what would it be? And more so, is there a limit to the amount of constraints one of these problems could have to where the equation could not be applied?
Orchard PortfolioAsher Rehman
Unit Problem Portfolio Last Line of Sight In the past two to three months, we have been studying the properties and patterns of a tree orchard. And how to find the last line of sight given any radius for an orchard. We used a multitude of functions to solve this unit problem. One of them being the distance formula to calculate the distance of a coordinate from a midpoint. To wrap up our unit and solve the problem, we are left with one final question to solve. We have a radius 50 orchard, meaning that at any point on the circumference of the orchard, that point will be 50 trees out of the center of the orchard. Each tree is planted 10 feet apart on the X and Y axis, and there is an open spot in the center of the orchard for you the observer to look out. The cross sectional area of the trees increase by 1.5 square inches a year and each have a circumference of 2.5 inches. And the last/furthest line of sight from the origins (0,0) is (25,0.5). Knowing that this is the last line of sight from the center of the orchard. How long will it take until the trees block out your vision from the center completely? To begin this problem we first need to map out the coordinates on a graph. This will include the radius of (50), the last line of sight from the origin (25, 0.5), and will give us the baseline for finding the starting radius and cross sectional area for the trees. The two green dots rest on the coordinates (1, 0) and (49, 1). These trees are equidistant from the last line of sight, and we are looking for the distance from those trees to the line of sight. We first needed to look for the starting radius of the tree, and we can do this with the circumference of the tree which is 2.5 inches. The formula for finding the radius of a circle given the circumference is C=2r; The circumference is equal to two pie times the radius. How we solve for the radius is by getting R by itself, and we do this by dividing both sides by 2. We know that (two pi times R) divided by (two pi) equals R, we eliminate the variables and are left with the radius. And then we divide our circumference which is 2.5 by 2, which will give us an answer of (0.397 = R). Now that we have the radius of the tree, we can solve for the cross sectional area. The cross sectional area of a tree is the surface area of the face of the trunk, which is why the radius is applicable for solving for the area. The formula for finding the area is A=r2. We take the radius of the tree (0.39) times pi, which gives us .5 inches squared for the starting cross sectional area of the tree. Now that we have our starting radius and starting area of the trees, we can go back to the graph and use the Pythagorean theorem to look for the shortest distance from the tree to the line of sight. This will give us the distance/ space that is left between the tree and our vision. This is one of the hardest points in the unit problem because we now are looking for patterns and similarities in the graph. We dipped into it slightly, but the next step for finding the distance from the tree to the line of sight is by using ratios of similar right triangles. If you look at the picture above of the graphed coordinates, it creates a right triangle with an adjacent side length of 50 units, and an opposite side length of 1 unit; we can solve for the hypotenuse. Now you might be wondering how this will help us to find the distance we're looking for, but now we can use this bigger right triangle to make another triangle within it as seen below. This smaller right triangle is similar to the larger one because they both share the same angle coming from the origin, and both triangles have a 90 degree angle. We can now find the ratio of the two triangles and solve for D (the distance from tree to line of sight). (Not drawn to scale) First, we’ll find the hypotenuse for the larger right triangle. Fifty squared is two thousand five hundred, and one by one is just one. Added together gives us the square root of 2,501. We then need to take this distance and convert it to the radius from the coordinate (1,1). We set the equation R= Square root of 2,501 over. We solve for the radius by dividing both sides by 1, where R divided by 1 is R and the square root of 2,501 divided by 1 is 0.0199 units. Convert the units into feet and we get 0.199 ft. And finally, converted into inches gives us a final radius of 2.4in. With our final radius we can solve for the final cross sectional area of the tree using the formula A=r2. Our radius of 2.4 squared will be 5.76, and then multiplied by pi will give us the final area of 18.1in squared. Given that we now have the final area of the tree, we can apply it to the area a tree will gain/grow in a year for our final answer. But first, we need to find the difference between our starting area and our final area. Take 0.5 from 18.1, and you have the max area of 17.6in squared. This is the max amount of cross sectional area the tree will gain before our vision is blocked entirely, but how long will it take to grow to this mass? We simply divide the area by the area growth per year we are given, which is 1.5in squared a year. Giving us 11.7 as our answer to the problem in the years it will take for the last line of sight to be fully blocked out. We now have an answer to the unit problem, in which it will take 11.7 years for the trees to grow and block our final line of sight from the center of the orchard. Discussion To solve the unit problem, we went through and studied a multitude of formulas to forge the tools that we would need to solve for the radius 50 orchard step by step. The first unit we learned would be used in the beginning and the very end of the unit problem, which is coordinate geometry. This unit entailed locating coordinates and calculating the distance in units between the two coordinates. To do this, we learned how to apply the distance formula, which is D=√(x2−x1)+(y2−y1). Where you are adding the two X values from each coordinate, same with the Y values, they are added together and square rooted to find the distance. We used this in the orchard unit to find the distance from coordinates to the origin and prove whether they fell into the orchard's radius or not. And then you have the Midpoint Formula which determines the midpoint between two coordinates. The formula is structured like the distance formula but entails division instead of addition; which is structured as (x1 + x2)/2 + (y1 + y2)/2. A snack in the middle is an excellent worksheet to showcase this. In which we had a problem where two colleagues were working in an orchard or garden, and they were on opposite sides while they gardened. And they need to find the midpoint between both of them to set up a snack table to make it as easy as possible for both of them to get to it. These equations are important because it helps us find the center between two coordinates and structures equidistance, and we can find that midpoint by calculating the distance between the two coordinates as well. Equidistance is where two coordinates are the same distance away from a mid line across a midpoint. We used Equidistance for solving the unit problem through calculating where the LLS(last line of sight) will lie between two trees. And sets us up to solve for the distance from the closest tree to the line of sight. Or also labeled as the shortest distance from a tree to the line of sight. Which brings us to the next unit of discussion. Circles, Pi, and Triangles. One key unit we covered was applying the Pythagorean theorem in solving for coordinates and finding similar triangles within a coordinate plane. The Pythagorean theorem, a2 + b2 = c2, is used to calculate the length of the longest side of a right triangle which is known as the Hypotenuse. We can solve for the side lengths of the biggest right triangle on the plane in the radius 50 orchard, and use that to find the ratio of the right triangle that can be formed with the last line of sight, and the shortest distance from a tree to said line of sight. We know that the two triangles are similar because they both share the same angle created from the LLS and the adjacent side of 50, and both have a right angle. Finding this ratio allowed us to find the distance from the closest tree to the LLS to later determine the time it would take for the trees to grow and reach that point. As far as calculating the area, and radius of a circle, we learned two crucial equations. C=2r: The circumference is equal to the radius by pi, times two. And A=r2: The area is equal to the radius by pi squared. And really we are looking for the cross sectional area, which is the area of the face of the trunk, or circle; calculating that for the whole trunk would be volume. Anyway, when given the circumference, we can calculate the radius of a circle, and then we can use the radius of a circle to find the cross sectional area. We want to know this area so we can have a starting area for the trees. Because later when we go through finding the distance from the tree to the LLS, we’ll get a finishing radius which we can plug into the area equation to get a finishing area. And as one final step, subtract the starting area, because that was the area we were starting off with and this will show us how much the tree will grow before it hits the LLS. As a short recap, we start and finish with the Radius and Cross Sectional Area formula to get a final answer to the unit problem. Reflection Through this unit, I was shown the importance of math through the relationships of algorithms and logic. Because math truly isn’t just numbers and pain. It holds a bond with other factors in the real world that are relevant and beneficial to learn. I personally do not enjoy math or understand it as well as others. But this unit showed me how mathematics as unforgiving as algebra and geometry are applied in the real world and how simply put they are. It’s a dangerous mindset to have when you think you are not good at something and it will not change. I was taught through this unit that half of your skill is your mindset and where your mental baseline rests. If you don’t think you can do something, then you’re going to perform in a way that you are incapable of doing. And it’s a major key insight that I found in myself, that I needed to start breaking that mold and trying to solve problems and equations even if I don’t understand them. The prefix was that I didn’t want to do them, because I didn’t know where to start or how to solve the equation, so it was like “What’s the point?”. But I was shown that I can’t learn anything if I’m not willing to try. And of course you have boundaries for misunderstanding, if you need help, seek it, but don’t let your discomforts and doubts hold you back from learning; is what I found. This unit problem was one of the most interesting and engaging units I have done in my educational career. Because we spent months learning different units and equations to lead up to solving this one problem, which is really cool. I can see a problem like this in the real world if it hasn’t already been used/solved. Which was another dead weight on my mind that is still holding me back from fully engaging in mathematics. With each and every unit we go over such as this one, I want to learn and try my best. But it has always been stuck in the back of my mind thinking “I’m never gonna use this, why are we even learning it?”. And these thoughts and doubts that I have are clipping me from diving into problems straight away. Will I use every piece of mathematics that I learn in my life going forward? almost a guarantee that I won’t; Julian even said that I will not. But does that mean that I should look at all units and algebra that way, absolutely not. I was proven wrong through this unit problem, and I see that it is completely relevant and logical to the real world outside of a sheet of paper. I learn materials the best when I can see examples of units and problems. I can have the best notes in the world; and my notes are already pretty damb good, but I still run into snags when I try to apply the formulas to a problem. If I have the formula, and I have a solid example of that formula being applied in a problem and how that problem is solved. Then I can study that and learn the order of operations for solving any either equation constructed in that same formula. Going forward, I want to try everything to the fullest even if it’s a guaranteed incorrect answer. Because I’m not losing anything if I try and fail, I can only gain experience and practise/refine. I wish I knew the benefits of that from the very beginning of my schooling, but I have learned how to fully utilize my thinking and problem solving. I’m going to be finishing this unit strong, and I’m looking forward to the next. Selection of Work |