HANOI PEDAGOGICAL UNIVERSITY 2 DEPARTMENT OF MATHEMATICS NGUYEN PHUONG ANH OPTIMIZATION PROBLEM IN 10TH GRADE MATH PROGRAM DEPARTMENT OF MATHEMATICS GRADUATION THESIS Major Applied mathematics Hanoi –[.]
HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS NGUYEN PHUONG ANH OPTIMIZATION PROBLEM IN 10TH GRADE MATH PROGRAM DEPARTMENT OF MATHEMATICS GRADUATION THESIS Major: Applied mathematics Hanoi – 2023 MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY ——————–o0o——————— NGUYEN PHUONG ANH OPTIMIZATION PROBLEM IN 10TH GRADE MATH PROGRAM DEPARTMENT OF MATHEMATICS GRADUATION THESIS Major: Applied mathematics Supervisor PhD Ngoc Chi Le Hanoi – 2023 CONFIRMATION I hereby declare that this thesis is my own research work, not copied by anyone else, which I have researched, read, translated, synthesized, and performed by myself Theoretical content in the thesis I have used some references as presented in the references section The data and results in the thesis are honest and have not been published in any other works Ha Noi, April 2023 Student Nguyen Phuong Anh ACKNOWLEDGMENT The topic "Optimal problem in 10th grade math program" is the content that I researched and made my graduation thesis after studying at the Faculty of Mathematics, Hanoi Pedagogical University During the research process In researching and completing my thesis, I have received a lot of attention and help from teachers, colleagues, family, and friends For the success of this thesis, I would like to express my sincere thanks to: The Department of Mathematics, Hanoi Pedagogical University created a very good learning and training environment, providing me with useful knowledge and skills to help me apply and facilitate my thesis Tutor PhD Ngoc Chi Le is a passionate teacher who has dedicatedly guided and helped me throughout the process of researching and implementing the topic You have exchanged ideas and suggestions for me to complete this research thesis well I would also like to thank the management board and the teaching staff of Hanoi Pedagogical University for giving me the opportunity to work at the school to gain practical knowledge and experience for useful information useful for the thesis Finally, I would like to thank my family and friends for always encouraging and creating the best conditions for me to strive to complete the research well Thank you sincerely! Ha Noi, April 2023 Student Nguyen Phuong Anh CONTENTS Page INTRODUCTION Preparatory knowledge 1.1 1.2 Optimization problem Linear programming problem 10 Methods for solving linear programming problems 14 2.1 Simplex method 14 2.2 Graphical Method 23 Optimization problem in 10th grade math program 28 3.1 How to determine the experimental domain 28 3.2 Method of solving the real problem is reduced to the system of first-order inequalities with two unknowns 29 3.3 Some practical examples 30 CONCLUSION 38 INDEX 39 REFERENCES 40 INTRODUCTION Rationale of the thesis In all areas of life, we are always interested in the problem of finding the best way to achieve the desired goal under certain constraints Optimal methods are a powerful tool to help decision makers come up with the best quantitative and qualitative solutions In recent years, optimization methods have been widely and effectively applied in economics, engineering, information technology and other sciences One of the first classes of optimization problems that has been fully studied in both theory and algorithm is the linear programming problem Linear programming since its inception (in the late 30s - 40s of the twentieth century) has occupied an important position in optimization The linear model is a very common model in practice, and at the same time, the linear fork dependence is a simple dependency and is easy to study mathematically Because optimization problem is a branch of mathematics with many applications in life and economics, in some economic or pedagogical (undergraduate) disciplines there is a subject on this problem For high school students, just consider the simple form of a optimization problem problem presented in the 10th grade Algebra program With the current organization of the national high school exam in the form of multiple choice, in my opinion, optimization problem is an important problem and it is very likely that it will appear in the national high school exam because this is a form of math that comes from the essential needs of life Aims of the thesis The thesis presents about the optimization problem in the 10th grade math program, specifically the real problem that is reduced to the system of first order inequalities with two unknowns Object and scope of thesis 3.1 Object: The optimization problem in the 10th grade math program 3.2 Scope: We will present about linear programming, optimization problems and solving methods optimization problem in 10th grade math programming Structure of the thesis In addition to the introduction, conclusion and list of references, the structure of the thesis consists of chapters: Chapter 1: Preparatory knowledge Present the optimization problem and linear programming problems Chapter 2: Methods for solving linear programming problems Presenting simplex methods and Graphical method to solve linear programming problems Chapter 3: The optimization problem in the 10th grade math program Presenting solving methods the real problem that is reduced to the system of first order inequalities with two unknowns Chapter Preparatory knowledge 1.1 Optimization problem General optimization problem Minimum or maximum of the function f (x) Provided that g (x) ≤, =, ≥ b ; i = 1, m i i x ∈ X ⊂ Rn In that: • f (x): objective function with n variables • gi (x) , i = 1, m: Constraints Each inequality is a constraint 2.2 Graphical Method We will first discuss the steps of the algorithm: Step 1: Formulate the LP (Linear programming) problem We have already understood the mathematical formulation of an LP problem in a previous section Note that this is the most crucial step as all the subsequent steps depend on our analysis here Step 2: Construct a graph and plot the constraint lines The graph must be constructed in n dimensions, where n is the number of decision variables This should give you an idea about the complexity of this step if the number of decision variables increases One must know that one cannot imagine more than 3-dimensions anyway! The constraint lines can be constructed by joining the horizontal and vertical intercepts found from each constraint equation Step 3: Determine the valid side of each constraint line This is used to determine the domain of the available space, which can result in a feasible solution How to check? A simple method is to put the coordinates of the origin (0, 0) in the problem and determine whether the objective function takes on a physical solution or not If yes, then the side of the constraint lines on which the origin lies is the valid side Otherwise it lies on the opposite one Step 4: Identify the feasible solution region The feasible solution region on the graph is the one which is satisfied by all the constraints It could be viewed as the intersection of the valid regions of each constraint line as well Choosing any point in this area would result in a valid solution for our objective function Step 5: Plot the objective function on the graph 23 It will clearly be a straight line since we are dealing with linear equations here One must be sure to draw it differently from the constraint lines to avoid confusion Choose the constant value in the equation of the objective function randomly, just to make it clearly distinguishable Step 6: Find the optimum point Optimum Points An optimum point always lies on one of the corners of the feasible region How to find it? Place a ruler on the graph sheet, parallel to the objective function Be sure to keep the orientation of this ruler fixed in space We only need the direction of the straight line of the objective function Now begin from the far corner of the graph and tend to slide it towards the origin • If the goal is to minimize the objective function, find the point of contact of the ruler with the feasible region, which is the closest to the origin This is the optimum point for minimizing the function • If the goal is to maximize the objective function, find the point of contact of the ruler with the feasible region, which is the farthest from the origin This is the optimum point for maximizing the function Step 7: Calculate the coordinates of the optimum point This is the last step of the process Once you locate the optimum point, you’ll need to find its coordinates This can be done by drawing two perpendicular lines from the point onto the coordinate axes and noting down the coordinates Otherwise, you may proceed algebraically also if the optimum point is at the intersection of two constraint lines and find it by solving a set of simultaneous linear equations The Optimum Point gives you the values of the decision variables necessary to optimize the objective function To find out the optimized objective function, one can simply put in the values of these parameters in the equation of the objective function 24 Example A health-conscious family wants to have a very well-controlled vitamin C-rich mixed fruit breakfast which is a good source of dietary fiber as well; in the form of fruit servings per day They choose apples and bananas as their target fruits, which can be purchased from an online vendor in bulk at a reasonable price Bananas cost 30 rupees per dozen (6 servings) and apples cost 80 rupees per kg (8 servings) Given: banana contains 8.8 mg of Vitamin C and 100-125 g of apples i.e serving contains 5.2 mg of Vitamin C Every person of the family would like to have at least 20 mg of Vitamin C daily but would like to keep the intake under 60 mg How many fruit servings would the family have to consume on a daily basis per person to minimize their cost? Solution We begin step-wise with the formulation of the problem first The constraint variables – x = number of banana servings taken and y = number of servings of apples taken Let us find out the objective function now Cost of a banana serving = 30 rupees = rupees Thus, the cost of x banana servings = 5x rupees Cost of an apple serving = 80 rupees = 10 rupees Thus the cost of y apple servings = 10y rupees Total Cost C = 5x + 10y Constraints: x ≥ 0;y ≥ (non-negative number of servings) Total Vitamin C intake: 8.8x + 5.2y ≥ 20 25 (2.1) 8.8x + 5.2y ≤ 60 (2.2) Let plot a graph with the constraint equations To check for the validity of the equations, put x = 0, y = in (2.1) Clearly, it does not satisfy the inequality Therefore, we must choose the side opposite to the origin as our valid region Similarly, the side towards the origin is the valid region for equation (2.2) Feasible Region: As per the analysis above, the feasible region for this problem would be the one in between the red and blue lines in the graph! For the direction of the objective function; let us plot 5x+10y = 50 Now take a ruler and place it on the straight line of the objective function Start sliding it from the left end of the graph We want the minimum value of the cost i.e the minimum value of the optimum function C Thus we should slide the ruler in such a way that a point is reached, which: Lies in the feasible region 26 Is closer to the origin as compared to the other points This would be our Optimum Point I’ve marked it as A in the graph It is the one which you will get at the extreme right side of the feasible region here Now we must calculate the coordinates of this point To this, just solve the simultaneous pair of linear equations: y=0 8.8x + 5.2y = 20 We will get the coordinates of A as (2.27, 0) This implies that the family must consume 2.27 bananas and apples to minimize their cost and function according to their diet plan 27 Chapter Optimization problem in 10th grade math program Optimization problem in 10th grade math program is a real problem that is reduced to s system of first-order with two unknowns Students only learn how to solve problems using the graphical method 3.1 How to determine the experimental domain Inequality of first order two unknowns In the coordinate plane, the line is straight (d) : ax+by +c = divide the plane into two halves of the plane, one of which (excluding the (d)) consists of points whose coordinates satisfy the inequality ax + by + c > 0, the other half of the plane (excluding (d)) consists of points whose coordinates satisfy the inequality ax + by + c < • To determine the solution domain of inequality ax + by + c < 0, we have the following rules of practice for the representation of experimental geometry (or representation of the solution domain): Step Draw a line (d) : ax + by + c = Step Consider a point M (x0 , y0 ) not located on (d) – If ax + by + c < then half-plane (excluding (d)) contains the point M which is the solution domain of inequality ax+by+c < – If ax+by +c > then half-plane (excluding (d)) does not contain the point M as the solution domain of any equation ax+by+c > 28 The system of first-order inequalities with two unknowns In the coordinate plane, we call the set of points whose coordinates satisfy all inequalities in the system the solution domain of the system So the solution domain of the system is the assignment of solution domains of inequalities in the system To determine the solution domain of the system, we use the following method of geometric representation: • For each inequality in the system, we define its solution domain and cross out (color) the remaining domain • After doing the above in turn for all inequalities in the system on the same coordinate plane, the remaining domain that is not crossed out (colored) is the solution domain of the given inequalities system 3.2 Method of solving the real problem is reduced to the system of first-order inequalities with two unknowns The problem of finding the solution domain of the system of firstorder inequalities is closely related to the problem of optimal production planning and the problem of linear planning To solve the real problem is reduced to the system of first-order inequalities with two unknowns, we put the problem information in the form of finding the maximum (minimum) value of the expression f (x, y) on the solution domain of the inequality system To find the maximum (minimum) value of an expression f (x, y) on the solution domain of an inequality system we the following: • Step 1: Determine the solution domain of the given system of inequalities 29 • Step 2: Calculate the values of the function f (x, y) with (x, y)is the coordinate of the vertices of the solution domain • Step 3: Compare the calculated values with each other, which is the maximum (minimum) value is the maximum (minimum) value of f (x, y)on the solution domain of the given system of inequalities 3.3 Some practical examples Example In a bartending competition, each team can use up to 24 g of flavorings, liters of water, and 210 g of sugar to prepare orange juice and apple juice For the preparation of liter of orange juice requires 30 g of sugar, liter of water, and g of flavoring; the preparation of liter of apple juice requires 10 g of sugar, liter of water, and g of flavoring Each liter of orange juice receives 60 bonus points, and each liter of apple juice receives 80 bonus points Ask how many liters of juice each needs to be prepared to get the largest number of bonus points Solution Call x, y are the number of liters of orange and apple juice of a bartending team, respectively (x, y ≥ 0) The bonus points of this team are: f (x, y) = 60x + 80y The number of grams of sugar to consume is: 30x + 10y The number of liters of water required is: x + y The number of grams of flavoring to use is: x + 4y Since in the bartending competition, each team uses a maximum of 24 g of flavorings, liters of water and 210 g of sugar, we have an inequality 30 system: 3x + y ≤ 21 30x + 10y ≤ 210 x+y ≤9 x+y ≤9 (∗) ⇔ x + 4y ≤ 24 x + 4y ≤ 24 x, y ≥ x, y ≥ The problem becomes finding the maximum value of a function f (x, y) on the solution domain of the inequality system (*) The solution domain of the system of inequalities (*) is the pentagonal OABCD (including boundary) Function f (x, y) = 60x + 80y will reach the maximum value on the solution domain of the inequality system (*) when (x, y) is the coordinate of one of the vertices O (0; 0) , A (7; 0) , B (6; 3) , C (4; 5) , D (0; 6) We have: f (0; 0) = 60.0 + 80.0 = 0; f (7; 0) = 60.7 + 80.0 = 420; 31 f (6; 3) = 60.6 + 80.3 = 600; f (4; 5) = 60.4 + 80.5 = 640; f (0; 6) = 60.0 + 80.6 = 480 Deduce that is the maximum value of the function f (x, y) on the solution domain of the system (*) Thus, to get the largest number of bonus points, it is necessary to prepare liters of orange juice and liters of apple juice Example It is planned to use two types of raw materials to extract at least 140 kg of substance A and kg of substance B From each ton of class I materials costing million VND, 20 kg of substance A and 0.6 kg of substance B can be extracted From each ton of grade II materials priced at million VND, 10 kg of substance A and 1.5 kg of substance B can be extracted Ask use as many tons of raw materials as possible so that the cost of purchasing materials is minimal, knowing that the material supplier can only supply no more than 10 tons of grade I materials and no more than tons of grade II materials Solution Call x, y the number of tons of grade I and class II raw materials, respectively ≤ x ≤ 9, ≤ y ≤ 10 Then the amount to buy raw materials is: f (x, y) = 4x + 3y From x tons of Class I raw materials, 20xkg of substance A and 0, 6x kg of substance B are extracted 32 From y tons of Class II raw materials extracted 10y kg of substance A and 1, 5y kg of substance B Inferred from x tons of type I raw materials and y tons of extracted class II materials 20x + 10y kg of substance A and 0, 6x + 1, 5y kg of substance B Since we must extract at least 140 kg of substance A and kg of substance B, we have the following system of inequalities: 20x + 10y ≥ 140 0, 6x + 1, 5y ≥ 0≤x≤9 ⇔ ≤ y ≤ 10 2x + 5y ≥ 30 2x + y ≥ 14 0≤x≤9 ≤ y ≤ 10 (∗) The problem becomes to find the minimum value of a function f (x, y) on the solution domain of the inequality system (*) The solution domain of the system of inequalities (*) is the quadrangle ABCD (including boundary) Function f (x, y) = 4x + 3y will reach the minimum value on the solution domain of the inequality system (*) when (x, y) is the coordinate 33 of one of the vertices A (5; 4) , B (10; 2) , C (10; 9) , D 2; We have: f (5; 4) = 32; f (10; 2) = 46; f (10; 9) = 67; f 2; = 37 Inferred f (x, y) minimum when (x; y) = (5; 4) Thus, in order for the cost of purchasing raw materials to be at least tons of class I raw materials and tons of class II materials Example 3: A family needs at least 900 units of protein and 400 units of dietary lipids per day Each kilogram of beef contains 800 units of protein and 200 units of lipids Each kilogram of pork contains 600 units of protein and 400 units of lipids Knowing that this family only buys a maximum of 1.6 kg of beef and 1.1 kg of pork; The price of kg of beef is 45 thousand VND, kg of pork is 35 thousand VND Ask the family how many kilograms of meat each has to buy so that the money spent is the least? Solution Call x and y the number of kilograms of beef and pork the family buys each day, respectively (0 ≤ x ≤ 1, 6; ≤ y ≤ 1, 1) Then the cost to buy the above meat is: f (x; y) = 45x + 35y thousand VND In x kg of beef contains 800x protein units and 200y lipid units In y kg of pork contains 600x protein units and 400y lipid units Deduce the number of protein units and the number of single lipids as 34 800x + 600y unit and 200x + 400y unit Since this family needs at least 900 units of protein and 400 units of lipids in its feed per day, we have the following system of inequalities: 8x + 6y ≥ 800x + 600y ≥ 900 x + 2y ≥ 200x + 400y ≥ 400 (∗) ⇔ ≤ x ≤ 1, ≤ x ≤ 1, 0 ≤ y ≤ 1, ≤ y ≤ 1, The problem becomes to find the minimum value of a function f (x; y) on the solution domain of the inequality system (*) The solution domain of the system of inequalities (*) is the quadrangle ABCD (including boundary) Function f (x; y) = 45x + 35y will reach the minimum value when (x; y) is the coordinate of one of the vertices A (1, 6; 1, 1) , B (1, 6; 0, 2) , C (0, 6; 0, 7) , D Which we have: f (1, 6; 1, 1) = 110, 5; f (1, 6; 0, 2) = 79; f (0, 6; 0, 7) = 51, 5; f (0, 3; 1, 1) = 52 Inferred f (x; y) minimum when (x; y) = (0, 6; 0, 7) Therefore, this family needs to buy 0.6 kg of beef and 0.7 kg of pork so 35 that the money spent is minimal Some other examples Example 4: A scientist studies the combined effects of vitamin A and B vitamins on the human body Accordingly, a person per day can receive no more than 600 units of vitamin A and no more than 500 units of vitamin B; a person per day needs between 400 and 1000 units of vitamins both A and B Due to the combined action of the two vitamins, every day, the number of units of B vitamins is not less 12 the number of units of vitamin A but not more than times the number of units of vitamin A The price of a unit of vitamin A is VND, the price of a unit of vitamin B is 7.5 VND Ask how much money you need to spend at least each day to get enough of both vitamins Example 5: A company needs to rent a car to carry 140 people and tons of cargo The rental car has two types of cars A and B, of which car type A has 10 units and car type B has units A Class A car for rent for VND million, a Class B car for rent for VND million Know that each Class A vehicle can carry up to 20 people and 0.6 tons of cargo; each Class B vehicle can carry up to 10 people and 1.5 tons of cargo Ask how many cars to rent each so that the cost is minimal Example 6: A workshop produces two types of products For the production of each kilogram of class I products requires kg of raw materials and 30 hours; to 36 produce each kilogram of Class II products requires kg of raw materials and 15 hours This workshop has 200 kg of raw materials and can operate for 50 days continuously Know that each kilogram of Class I products makes a profit of 40 thousand VND, each kilogram of Class II products makes a profit of 30 thousand VND Ask how many products each type should be produced so that the profit obtained is the greatest 37