Using Mathematics to Solve Real World Problems
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- Silvia Paul
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1 Using Mathematics to Solve Real World Problems
2 Creating a mathematical model:
3 Creating a mathematical model: We are given a word problem
4 Creating a mathematical model: We are given a word problem Determine what question we are to answer
5 Creating a mathematical model: We are given a word problem Determine what question we are to answer Assign variables to quantities in the problem so that you can answer the question using these variables
6 Creating a mathematical model: We are given a word problem Determine what question we are to answer Assign variables to quantities in the problem so that you can answer the question using these variables Derive mathematical equations containing these variables
7 Creating a mathematical model: We are given a word problem Determine what question we are to answer Assign variables to quantities in the problem so that you can answer the question using these variables Derive mathematical equations containing these variables Use these equations to find the values of these variables
8 Creating a mathematical model: We are given a word problem Determine what question we are to answer Assign variables to quantities in the problem so that you can answer the question using these variables Derive mathematical equations containing these variables Use these equations to find the values of these variables State the answer to the problem
9 Creating a mathematical model: We are given a word problem Determine what question we are to answer Assign variables to quantities in the problem so that you can answer the question using these variables Derive mathematical equations containing these variables Use these equations to find the values of these variables State the answer to the problem Today we will do this using straight lines as our equations, and we will solve the problem by drawing these lines (graphing).
10 Creating a mathematical model: We are given a word problem Determine what question we are to answer Assign variables to quantities in the problem so that you can answer the question using these variables Derive mathematical equations containing these variables Use these equations to find the values of these variables State the answer to the problem Today we will do this using straight lines as our equations, and we will solve the problem by drawing these lines (graphing). This process is called Linear Programming and is one of the most powerful mathematical methods used by businesses and companies to solve problems and help them make the best decisions.
11 Creating a mathematical model: We are given a word problem Determine what question we are to answer Assign variables to quantities in the problem so that you can answer the question using these variables Derive mathematical equations containing these variables Use these equations to find the values of these variables State the answer to the problem Today we will do this using straight lines as our equations, and we will solve the problem by drawing these lines (graphing). This process is called Linear Programming and is one of the most powerful mathematical methods used by businesses and companies to solve problems and help them make the best decisions. Operations Research is the profession that applies mathematical methods like this to problems arising in industry, healthcare, finance, etc.
12 A problem:
13 A problem: A furniture manufacturer produces two sizes of boxes (large, small) that are used to make either a table or a chair.
14 A problem: A furniture manufacturer produces two sizes of boxes (large, small) that are used to make either a table or a chair. table large block small block chair
15 A problem: A furniture manufacturer produces two sizes of boxes (large, small) that are used to make either a table or a chair. A table makes $3 profit and a chair makes $5 profit. table $3 large block small block chair $5
16 A problem: A furniture manufacturer produces two sizes of boxes (large, small) that are used to make either a table or a chair. A table makes $3 profit and a chair makes $5 profit. If M small blocks and N large blocks are produced, how many tables and chairs should the manufacturer make in order to obtain the greatest profit? table $3 N large block M small block chair $5
17 Large block Small block Table 1 large block 1 small block Chair 1 large block 2 small blocks
18 Problem: Given M small blocks and N large blocks, how many tables and chairs should we make to obtain the most profit? Profit: $3 $5 Table Chair
19 Example: small blocks and large blocks
20 Example: small blocks and large blocks We can make 4 tables and 4 chairs: 4 4
21 Example: small blocks and large blocks We can make 4 tables and 4 chairs: large 4 small 4 large 8 small 8 large small
22 Example: small blocks and large blocks We can make 4 tables and 4 chairs: large 4 small 4 large 8 small 8 large small Profit = ($3) x 4 + ($5) x 4 = $32
23 small blocks large blocks 4 tables and 4 chairs: 4 4 $3 $5 Used: 8 large blocks small blocks (4 large blocks left)
24 small blocks large blocks 4 tables and 4 chairs: 4 4 $3 $5 Used: 8 large blocks small blocks (4 large blocks left) I can make 2 more tables if I make 1 less chair; 3 chairs and 6 tables increase my profit! (1 chair 2 tables, profit goes up by $1) Profit = ($3) x 6 + ($5) x 3 = $33
25 small blocks large blocks 6 tables and 3 chairs: 6 3 $3 $5 Used: 9 large blocks small blocks (3 large blocks left)
26 small blocks large blocks 6 tables and 3 chairs: 6 3 $3 $5 Used: 9 large blocks small blocks (3 large blocks left) I can do it again; change one chair into 2 tables; 8 tables and 2 chairs Profit = ($3) x 8 + ($5) x 2 = $34
27 small blocks large blocks 8 tables and 2 chairs: 8 2 $3 $5 Used: 10 large blocks small blocks (2 large blocks left)
28 small blocks large blocks 8 tables and 2 chairs: 8 2 $3 $5 Used: 10 large blocks small blocks (2 large blocks left) I can do it again; change one chair into 2 tables; 10 tables and 1 chair Profit = ($3) x 10 + ($5) x 1 = $35
29 small blocks large blocks 10 tables and 1 chair: 10 1 $3 $5 Used: 11 large blocks small blocks (1 large block left)
30 small blocks large blocks 10 tables and 1 chair: 10 1 $3 $5 Used: 11 large blocks small blocks (1 large block left) I can do it again; change one chair into 2 tables; tables and 0 chairs Profit = ($3) x + ($5) x 0 = $36
31 small blocks large blocks tables and 0 chairs: $3 Profit = $36 Used: large blocks small blocks (no blocks left) Is this the best?
32 Now you try: 20 small blocks 20 large blocks How many tables and chairs? $3 $5
33 Another: 25 small blocks 20 large blocks How many tables and chairs? $3 $5
34 table $3 chair $5 small = large = tables chairs profit Not many small blocks small = 20 large = tables chairs profit Not many small blocks small = 25 large = tables chairs profit Many small blocks
35 Summary M = # small blocks N = # large blocks M N Table Chair
36 Two cases: M = # small blocks N = # large blocks M N Table Chair
37 Two cases: 1. Many small blocks: M >= 2N M = # small blocks N = # large blocks M N Table Chair
38 Two cases: 1. Many small blocks: M >= 2N M = # small blocks N = # large blocks M N make N chairs, 0 tables Table Chair
39 Two cases: 1. Many small blocks: M >= 2N M = # small blocks N = # large blocks M N make N chairs, 0 tables Table 2. Not many small blocks: M < 2N Chair
40 Two cases: 1. Many small blocks: M >= 2N M = # small blocks N = # large blocks M N make N chairs, 0 tables Table 2. Not many small blocks: M < 2N Chair mixture of tables and chairs....
41 Two cases: 1. Many small blocks: M >= 2N M = # small blocks N = # large blocks M N make N chairs, 0 tables Table 2. Not many small blocks: M < 2N Chair mixture of tables and chairs.... What is the magic number of tables and chairs??
42 Two cases: 1. Many small blocks: M >= 2N M = # small blocks N = # large blocks M N make N chairs, 0 tables Table 2. Not many small blocks: M < 2N Chair mixture of tables and chairs.... What is the magic number of tables and chairs?? Let s make a mathematical model to find out
43 M = # small blocks N = # large blocks M N Table Chair
44 = # tables built Y = # chairs built Our variables M = # small blocks M N = # large blocks N Table Chair
45 = # tables built Y = # chairs built How many can I build? M = # small blocks N = # large blocks M N Table Chair
46 = # tables built Y = # chairs built How many can I build? M = # small blocks N = # large blocks M N + 2Y <= M; I have only M small blocks + Y <= N; I have only N large blocks Table Chair Our equations
47 = # tables built Y = # chairs built How many can I build? M = # small blocks N = # large blocks M N + 2Y <= M; I have only M small blocks + Y <= N; I have only N large blocks Table Chair Our equations (Note: these are inequalities, not equalities!)
48 = # tables built Y = # chairs built How many can I build? M = # small blocks N = # large blocks M N + 2Y <= M; I have only M small blocks + Y <= N; I have only N large blocks Table Chair Let s plot all possible choices for and Y for a given M and N, and then we ll pick the,y that gives the greatest profit.
49 = # tables built Y = # chairs built How many can I build? M = # small blocks N = # large blocks M N + 2Y <= M; I have only M small blocks + Y <= N; I have only N large blocks Table Chair Let s plot all possible choices for and Y for a given M and N, and then we ll pick the,y that gives the greatest profit. Solving our equations...
50 Intermission: A primer on linear equations
51 Intermission: A primer on linear equations A linear equation: a + by = c, or Y = m + b, a,b,c,m numbers,,y variables
52 Intermission: A primer on linear equations A linear equation: a + cy = d, or Y = m + b, a,b,c,d,m numbers,,y variables Y If we plot all the,y that satisfy a linear equation, it forms a line:
53 Intermission: A primer on linear equations A linear equation: a + cy = d, or Y = m + b, a,b,c,d,m numbers,,y variables Y If we plot all the,y that satisfy a linear equation, it forms a line: b run rise Slope of a line: rise run = m
54 Intermission: A primer on linear equations A linear equation: a + cy = d, or Y = m + b, a,b,c,d,m numbers,,y variables Y If we plot all the,y that satisfy a linear equation, it forms a line: b run rise m=-1 m=-2 m=2 m=1 Slope of a line: rise run = m m=-1/2 m=1/2 m=0
55 Back to our problem...
56 Let s say M =, N = + 2Y <= Y = -(1/2) + 6 slope = -1/2 = # tables Y = # chairs M = # small blocks N = # large blocks + Y <= Y = -(1) + slope = -1
57 Let s say M =, N = + 2Y <= Y = -(1/2) + 6 slope = -1/2 = # tables Y = # chairs M = # small blocks N = # large blocks + Y <= Y = -(1) + slope = -1 We want to find all possible and Y that satisfy these two equations.
58 Let s say M =, N = + 2Y <= Y = -(1/2) + 6 slope = -1/2 = # tables Y = # chairs M = # small blocks N = # large blocks + Y <= Y = -(1) + slope = -1 We want to find all possible and Y that satisfy these two equations. First draw the equality lines; Y + Y = 6 + 2Y = slope = -1 slope = -1/2
59 Now the inequalities: + Y <= which side of the line is this region? Y + Y = 6
60 Now the inequalities: + Y <= which side of the line is this region? Let s check one point: Is (0,0) in this region? Y + Y = 6
61 Now the inequalities: + Y <= which side of the line is this region? Let s check one point: Is (0,0) in this region? Yes: When =0 and Y=0 then <= Y + Y = 6
62 Now the inequalities: + Y <= which side of the line is this region? Let s check one point: Is (0,0) in this region? Yes: When =0 and Y=0 then <= Y + Y = So the allowed values of and Y are below this line (feasible points) 6
63 + 2Y <= which side of the line is this region? Y 6 + 2Y =
64 + 2Y <= which side of the line is this region? Y Region is also below the line 6 + 2Y =
65 So the allowed region for both inequalities is the common region (the intersection) Y Only and Y in this region are allowed (for M=, N=) 6 + Y <= AND + 2Y <=
66 Now, for and Y in this region, which one gives the highest profit? Profit; P = 3 + 5Y Y = -(3/5) + P/5, slope = -3/5 Y 6 + Y <= AND + 2Y <=
67 Now, for and Y in this region, which one gives the highest profit? Profit; P = 3 + 5Y Y = -(3/5) + P/5, slope = -3/5 Y Let s plot the profit line P = 3 + 5Y 6 + Y <= AND + 2Y <=
68 Now, for and Y in this region, which one gives the highest profit? Profit; P = 3 + 5Y Y = -(3/5) + P/5, slope = -3/5, Y intercept = P/5 Along the profit line, all (,Y) give the same profit P Y Slope = -1 Slope = -3/5 Slope = -1/2 All and Y along this line give the same profit P P/5 6 Slope = -3/5 Slope = -1/2
69 Now, for and Y in this region, which one gives the highest profit? Profit; P = 3 + 5Y Y = -(3/5) + P/5, slope = -3/5, Y intercept = P/5 Along the profit line, all (,Y) give the same profit P Y Slope = -1 Slope = -3/5 Slope = -1/2 higher profit 6 lower profit
70 Now, for and Y in this region, which one gives the highest profit? Profit; P = 3 + 5Y Y = -(3/5) + P/5, slope = -3/5, Y intercept = P/5 Y What is the highest profit for those,y in the feasible region? higher profit 6 lower profit
71 Now, for and Y in this region, which one gives the highest profit? Profit; P = 3 + 5Y Y = -(3/5) + P/5, slope = -3/5, Y intercept = P/5 higher profit Y We keep moving the profit line upwards until the last feasible point is touched 6 lower profit
72 Now, for and Y in this region, which one gives the highest profit? Profit; P = 3 + 5Y Y = -(3/5) + P/5, slope = -3/5, Y intercept = P/5 Y We keep moving the profit line upwards until the last feasible point is touched 6 It is this line!
73 Now, for and Y in this region, which one gives the highest profit? Profit; P = 3 + 5Y Y = -(3/5) + P/5, slope = -3/5, Y intercept = P/5 Y We keep moving the profit line upwards until the last feasible point is touched 6 It is this line! And this is the only point; (, 0)
74 Now, for and Y in this region, which one gives the highest profit? Profit; P = 3 + 5Y Y = -(3/5) + P/5, slope = -3/5, Y intercept = P/5 Y So the greatest profit is achieved when = and Y=0; Profit = = 3 + 5Y
75 Now let s look at Case 1: M >=2N N< M/2 + 2Y <= M Y = -(1/2) + M/2 slope = -1/2 = # tables Y = # chairs M = # small blocks N = # large blocks + Y <= N Y = -(1) + N slope = -1 M/2 Y N + Y = N + 2Y = M slope = -1/2 slope = -1 N
76 Now let s look at Case 1: M >=2N N< M/2 + 2Y <= M Y = -(1/2) + M/2 slope = -1/2 = # tables Y = # chairs M = # small blocks N = # large blocks + Y <= N Y = -(1) + N slope = -1 We want to find all possible and Y that satisfy these two equations. Y M/2 N + Y = N + 2Y = M slope = -1/2 slope = -1 N
77 Now let s look at Case 1: M >=2N N< M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks We want to find all possible and Y that satisfy these two equations. M/2 Y N + Y = N + 2Y = M Feasible region slope = -1 slope = -1/2 N
78 Now let s look at Case 1: M >=2N N< M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Now plot the profit line: 3 + 5Y = P M/2 Y N + Y = N + 2Y = M Feasible region slope = -1 slope = -1/2 N
79 Now let s look at Case 1: M >=2N N< M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Now plot the profit line: 3 + 5Y = P (slope = -3/5) M/2 Y N + Y = N + 2Y = M slope = -1/2 profit line slope = -1 N
80 Now let s look at Case 1: M >=2N N< M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Move the profit line up until it last touches the feasible region M/2 Y N + Y = N + 2Y = M Greatest profit line slope = -1 slope = -1/2 N
81 Now let s look at Case 1: M >=2N N< M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks M/2 Y Point (0,N) N + Y = N + 2Y = M Greatest profit line slope = -1 slope = -1/2 N
82 Now let s look at Case 1: M >=2N N< M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks So the greatest profit is achieved when = 0, Y = N M/2 Y Point (0,N) N + Y = N + 2Y = M Greatest profit line slope = -1 slope = -1/2 N
83 Now let s look at Case 1: M >=2N N< M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Point (0,N) So the greatest profit is achieved when = 0, Y = N M/2 N Y + Y = N + 2Y = M Make N chairs 0 tables Greatest profit line slope = -1 slope = -1/2 N
84 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M Y = -(1/2) + M/2 slope = -1/2 = # tables Y = # chairs M = # small blocks N = # large blocks + Y <= N Y = -(1) + N slope = -1 Y N + Y = N M/2 + 2Y = M slope = -1 slope = -1/2 N
85 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Y The feasible region is: N + Y = N M/2 + 2Y = M slope = -1 slope = -1/2 N
86 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Y The feasible region is N M/2 + Y = N + 2Y = M Note there is a new corner; R R slope = -1 slope = -1/2 N
87 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Y Now let s add the profit lines N + Y = N M/2 + 2Y = M R slope = -1 slope = -1/2 N
88 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Y Now let s add the profit lines The highest one touches R! N + Y = N M/2 + 2Y = M R N
89 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Y Now let s add the profit lines The highest one touches R! N M/2 + Y = N + 2Y = M So R is the point (,Y) that has the greatest profit R N
90 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks Y What is (,Y) at point R? N + Y = N M/2 + 2Y = M R N
91 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks N Y + Y = N What is (,Y) at point R? It is where the two lines +Y=N and +2Y=M meet. M/2 + 2Y = M R N
92 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks N Y + Y = N What is (,Y) at point R? It is where the two lines +Y=N and +2Y=M meet. M/2 Y=M-N + 2Y = M R We find that = 2N - M Y = M - N =2N-M N
93 Now let s look at Case 2: M < 2N N > M/2 + 2Y <= M + Y <= N = # tables Y = # chairs M = # small blocks N = # large blocks N Y + Y = N So we have our answer: If M < 2N then we make = 2N-M tables and Y = M N chairs M/2 + 2Y = M Y=M-N R =2N-M N
94 table $3 chair $5 Y M = N = M < 2N tables chairs profit = 2N M = Y = M N = 0 M = 20 N = M < 2N tables chairs profit = 2N M = 4 Y = M N = 8 M = 25 N = M > 2N tables chairs profit = 0 Y = N =
95 A problem for you...
96 A problem for you... hours per kg hours available Columbian Mexican roaster A roaster B grind/package profit/kg 3 2.5
97 A problem for you... hours per kg hours available Columbian Mexican roaster A roaster B grind/package profit/kg Variables: Y
98 <= 4 roaster A
99 <= 4 roaster A Y
100 <= 4 roaster A Y = 4 4
101 <= 4 roaster A 2Y <= roaster B Y Y = 6 6 = 4 4
102 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y Y = 6 6 = Y = 18 slope =
103 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y Y = 6 6 = Y = 18 slope =
104 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y Y = 6 6 (2,6) = 4 (4,3) 3 + 2Y = 18 slope =
105 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y = P profit line, slope = -1.2 Y Y = 6 6 (2,6) = 4 (4,3) 3 + 2Y = 18 slope =
106 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y = P profit line, slope = -1.2 Y Y = 6 6 (2,6) = 4 profit line (4,3) 3 + 2Y = 18 slope =
107 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y = P profit line, slope = -1.2 Y Y = 6 6 (2,6) = 4 profit line (4,3) 3 + 2Y = 18 slope =
108 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y = P profit line, slope = -1.2 profit line Y = 6 6 (2,6) = 4 (4,3) 3 + 2Y = 18 slope =
109 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y = P profit line, slope = -1.2 profit line 6 (2,6) Y = 6 = 4 Best point is (2,6); Solution is = 2 Y = 6 (4,3) 3 + 2Y = 18 slope =
110 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y = P profit line, slope = -1.4 Y Y = 6 6 (2,6) = 4 profit line (4,3) 3 + 2Y = 18 slope =
111 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y = P profit line, slope = -1.4 profit line Y = 6 6 (2,6) = 4 (4,3) 3 + 2Y = 18 slope =
112 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging Y = P profit line, slope = -1.4 profit line 6 (2,6) Y = 6 Best solution is still = 2 Y = 6 = 4 (4,3) 3 + 2Y = 18 slope =
113 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging + 2.5Y = P profit line, slope < -1.5 (steeper than -1.5) Y Y = 6 6 (2,6) = 4 (4,3) 3 + 2Y = 18 slope =
114 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging + 2.5Y = P profit line, slope < -1.5 (steeper than -1.5) Y Y = 6 6 (2,6) = 4 (4,3) 3 + 2Y = 18 slope = profit line
115 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging + 2.5Y = P profit line, slope < -1.5 (steeper than -1.5) Y Y = 6 6 (2,6) = 4 (4,3) 3 + 2Y = 18 slope = profit line
116 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging + 2.5Y = P profit line, slope < -1.5 (steeper than -1.5) 6 Y (2,6) Y = 6 Best solution now is = 4 Y = 3 = 4 (4,3) 3 + 2Y = 18 4 profit line
117 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging + 2.5Y = P profit line, slope < -1.5 (steeper than -1.5) 6 Y (2,6) Y = 6 = 4 Best solution now is = 4 Y = 3 So need (4,3) - < Y = 18 4 profit line
118 <= 4 roaster A 2Y <= roaster B 3 + 2Y <= 18 grinding, packaging + 2.5Y = P profit line, slope < -1.5 (steeper than -1.5) 6 Y (2,6) Y = 6 = 4 Best solution now is = 4 Y = 3 So need (4,3) - < > 3 (2.5/2) = Y = 18 4 profit line
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