1. Math 101. Math 101 Website: richard/math101 Section 2 Website: richard/math101/fall06

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1 1. Math 101 Lines and Slope Professor Richard Blecksmith Dept. of Mathematical Sciences Northern Illinois University Math 101 Website: richard/math101 Section 2 Website: richard/math101/fall06 2. Construction Problem A contractor builds two types of homes: the standard model and the deluxe model. The standard model requires one lot, $12,000 capital, 150 labor-days to build, and is sold for a profit of $2400. The deluxe model requires one lot, $32,000 capital, 200 labor-days to build, and is sold for a profit of $3400. The contractor has 150 lots. The bank is willing to loan him $2,880,000 for the project and he has a maximum labor force available of 24,000 labor-days. How many houses should he build to realize the greatest profit? Slope Intercept Form y = mx + b m is the slope b is the y-intercept 3. Two Forms of a Line 1

2 2 Standard Form ax + by = c 4. Point Slope Formula If we know (1) the slope m of a line and (2) a point P = (x 1, y 1 ) on the line then we can use The Point Slope Formula y y 1 = m(x x 1 ) to find the equation of the line 5. Point Slope Example Problem. Find the equation of the line that passes through the point (2,3) and has the slope m = 1 2. Solution. In the Point Slope Formula (1) the slope m = 1 2 and (2) the point P = (2, 3) on the line Plugging these into the Point Slope Formula y y 1 = m(x x 1 ) yields y 3 = 1 2 (x 2) y 3 = 1 2 x 1 y = 1 2 x Slope Problem: How to find the slope of a line: If you know two points P = (x 1, y 1 ) and Q = (x 2, y 2 ) Subtract: y 2 y 1

3 3 Subtract: x 2 x 1 Now divide: slope = y 2 y 1 x 2 x 1 Example: Find the slope of the line through the points P = (1, 2) and Q = (3, 8) (1,2) 7. Graph (3,8) slope = = 6 2 = 3 8. Line thru Two Points Problem. Find the equation of the line passing through the points (1, 2) and (3, 8) Solution. Once we know the slope m = = 6 2 = 3 we can plug m = 3 and P = (1, 2) into the Point Slope Formula

4 4 y y 1 = m(x x 1 ) to get y 2 = 3(x 1) y 2 = 3x 3 y = 3x 1 9. The Distance Formula The distance between two points (x 1, y 1 ) and (x 2, y 2 ) is given by The Distance Formula distance = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 The distance between (1, 2) and (3, 8) is distance = (3 1) 2 + (8 2) 2 = = = Converting Celcius to Fahrenheit Note that Celsius and Centigrade are interchangable They refer to the metric system measurement of temperature The problem in converting temperature is that the 0 points are different in the two systems 0 meters = 0 inches Both signify no length 0 square meters = 0 square yards Both signify no area But 0 degrees centigrade = 32 degrees Fahrenheit

5 5 the freezing point of water 11. Celcius to Fahrenheit Cont d You cannot just multiply centigrade temperature by a conversion factor to get the Fahrenheit temperature You must adjust for the fact that the zero point are different The conversion is represented by a line in the plane The x values are degrees Centigrade The y values are degrees Fahrenheit 12. Celcius to Fahrenheit Cont d We know two points on this line. Freezing point of water 0 Centigrade = 32 Fahrenheit So (0, 32) is a point on the line Boiling point of water 100 Centigrade = 212 Fahrenheit So (100, 212) is another point on the line

6 6 13. Graph of the Line 212 (100,212) Degrees Fahrenheit (0,32) Degrees Centigrade Finding the Equation The slope of this line is m = = = 9 5 The equation for our line is F = 9 5 C + b To determine the constant b, Plug in the point C = 0, F = 32:

7 7 32 = b So b = 32 and F = 9 5 C + 32 F = 9 5 C Converting Temperatures You are visiting Paris, France, and the a sign says the temperature is 16 Since France uses the metric system, you know this temperature is in Centigrade What s the temperature in Fahrenheit? Use the Formula F = = 60.8 or about 61 degrees. 16. A Practical Rule of Thumb The problem with this method is No one remembers the formula. The fractions are hard to compute in your head. Here s an easier method: Use the formula F = 2C + 30 For the previous problem, if C = 16, then F = = 62 This answer is very close to the correct answer of 60.8

8 8 17. Comparing the Lines 212 (100,212) Degrees Fahrenheit F = 2C + 30 F = 9 5 C + 32 (0,32) Degrees Centigrade Zooming In to Normal Temps Degrees Fahrenheit (0,32) F =2C+30 F = 9 5 C Degrees Centigrade

9 9 19. Conclusion As can be seen from the graph, for temperatures between 0 and 30 Centigrade, the second formula F = 2C + 30 is accurate to within 2 degrees. 20. Linear Versus Nonlinear Many functions are linear, that is, their graphs are straight lines. Example: You work at an hourly wage of $6.50. Then your weekly salary is linear, as a function of the number of hours worked: On your first hour, you make $6.50. On your second hour, you make $6.50. On your 19th hour, you make $6.50. In general, your salary is Salary = 6.5 hours worked 21. Some Non-Linear Functions The interest you pay on your mortgage is not linear. It is very large at the beginning of your loan and tapers off toward the end. The Dow Jones Average is not linear. It fluctuates from day to day. When you drive a car, the distance you travel is not linear (with respect to time) unless you are using cruise control The temperature in DeKalb is not linear.

10 PT Cruiser Example: Road & Track Magazine tested Chrysler s PT Cruiser. They needed 131 feet to stop the car from a speed of 60 mph. They needed 232 feet to stop the car from a speed of 80 mph. Increasing the speed by 20 mph almost doubles the distance to stop. This is not linear behavior. Moral: Not all functions are linear 23. Sir Issac Newton Sir Issac Newton: inventor of Calculus (and a popular figfilled cookie): Stopping distance increases with the square of the speed. Stopping Distance = K Speed 2 where K is the (braking) constant. 24. Stopping a PT Cruiser Plugging in Distance = 131 and Speed = 60 allows us to solve for K: 131 = K 60 2 Solve for K by dividing by 60 2 : K = =.0364 Stopping Distance =.0364 Speed 2

11 Newton s Prediction Stopping Distance =.0364 Speed 2 This equation predicts that when Speed = 80, the stopping distance will be Stopping Distance = = Compare this will the actual value found by Road & Track s test drivers: Stopping distance at 80 mph was 232 feet. Way to go, Newton. Not bad for a guy who predated the automobile by 200 years. 26. Stopping Distance Table Speed Stopping Distance Speed Stopping Distance Stopping Distance Graph Stopping Distance vs. Speed

12 Driving Considerations If you increase your speed by 5 mph, it requires an additional 4.6 feet to stop if you are going 10 to 15 mph. It requires an additional 30 feet to stop if you are going 80 to 85 mph. This is not a linear function. It take four times longer to stop from 60 mph than from 30 mph. It take nine times longer to stop from 90 mph than from 30 mph. Moral: Don t tailgate on the interstates. 29. Homework and Quiz Homework due next week in Recitation Personal Finance Section V.3 page 301. Note: The homework has been changed; so check the new homework list on line.

13 Miniquiz 3: 1 pt What s your name, Z-number, and Recitation Section Number? 2 pts You borrow $11,000 to buy a car. Your car payments are $275 per month and the length of the loan is 4 years. What is the total amount of interest you are paying? 30. Math Linear Systems of Equation Professor Richard Blecksmith richard@math.niu.edu Dept. of Mathematical Sciences Northern Illinois University Math 101 Website: richard/math101 Section 2 Website: richard/math101/fall Correct Homework List Ch 0.2 page 15. 1acegi, 2abcd, 3ac Ch 0.4 page 27. 1ac, 5abc IV.3 page , 2 III.1 page abcd IV.1 page , 8abcdf, 17, 23, 25, 29 III.2 page , 3, 7, 8, 9, 22

14 14 This homework is due on Wednesday, Sept 27, the day of Test Slope from the equation Suppose you have an equation for the line: Case 1. If the equation looks like y = mx + b then m is the slope. Examples 1. The slope of y = 2x + 3 is 2. Examples 2. The slope of y = 3 4 x is 33. Case 2 Case 2. If the equation looks like ax + by = c Solve for y to find the slope. Example. To find the slope of the line 2x + 3y = 6 Move 2x to right: 3y = 2x + 6 Divide by 3: y = 2x + 6 = x + 2 The slope is 2 3 Given two lines. 34. Connection to Geometry If they have different slopes, then they intersect.

15 15 If they have the same slope, then they are parallel. Unless they are the same line. See Homework III.1, page 164, #7 e and g. 35. Homework page 164 Problem 7e Determine whether the following two lines intersect, are parallel, or are the same line: { 3x + 5y = 15 x 3y = 9 Find the slope of Line 1. 3x + 5y = 15 can be rewritten 5y = 3x + 15 or y = 3 5 x + 3 So the first line has slope Problem 7e Cont d Find the slope of Line 2. x 3y = 9 can be rewritten 3y = x 9 or y = 1 3 x 3 So the second line has slope 1 3 Since these slopes are different, the lines intersect.

16 16 Line Picture Line Homework page 164 Problem 7g Determine whether the following two lines intersect, are parallel, or are the same line: { y = x 6 3y 3x = 0 Find the slope of Line 1. We see immediately that y = x 6 has slope 1. Find the slope of Line 2. 3y 3x = 0 can be rewritten 3y = 3x or y = x So the second line has slope 1

17 Problem 7g Cont d Since these slopes are the same, but the equations y = x 6 and y = x are different, the lines are parallel. 40. Picture Line 1 Line Where do two Lines Cross? Determine where the following two lines intersect: { 3x + 5y = 15 x 3y = 9

18 18 The Addition Subtraction Method Multiply the two equations by appropriate numbers so that When they are added (or subtracted) One of the variables drops out of the equation. 42. Problem #7e Revisited Step 1. Write the equations. (1) 3x + 5y = 15 (2) x 3y = 9 Step 2. Multiply so that x drops out (1) 3x + 5y = 15 3 (2) 3x 9y = 27 Subtract 0x + 14y = 12 So y = = Problem #7e Cont d Step 3. Plug back into original equation (2) x 3y ( = 9 x 3 6 ) = 9 7 x = 9 x = 9 18 = = Step 4. Write down the solution:

19 x = 45 7, y = 6 7, 19 Line Picture ( 45 7, 6 7) Line Peach Problem #20, Page 166 Brenda decides to sell the peaches from her backyard to the women in the neighborhood. She sells Mrs. Jones 10 large peaches and 15 small peaches for $9.50. She sells Mrs. Williams 15 large peaches and 10 small peaches for $ How much does she charge for a small and a large peach? 46. Problem #20, Page 166 Step 1. Make the variables. L = the price of a large peach S = the price of a small peach Step 2. Write the equations. (1) 10L + 15S = 9.5 (2) 15L + 10S = 10.5 Step 3. Solve for one of the variables. 3 (1) 30L + 45S = (2) 30L + 20S = 21

20 20 Subtract 0L + 25S = 7.5 So S = = Problem #20 Cont d Step 4. Plug back into original equation (1) 10L = L = L = 5.00 L =.50 Step 5. Write down the solution: A large peach costs 50 cents; a small peach costs 30 cents. 48. Math 101 Linear Programming Problems Professor Richard Blecksmith richard@math.niu.edu Dept. of Mathematical Sciences Northern Illinois University Math 101 Website: richard/math101 Section 2 Website: richard/math101/fall Solving Inequalities You solve inequalities the way you solve equations:

21 21 Algebra Rule Equation Inequality 2x 5 = 3 2x 5 3 Add 5 to both sides 2x = 8 2x 8 Divide by 2 x = 4 x 4 With One Important Difference... Consider the following 50. The Inequality Trap Algebra Rule Equation Inequality 3x + 2 = 4 3x Subtract 2 3x = 6 3x 6 Divide by 3 x = 2 x 2 Why did the inequality change from greater than ( ) to less than ( )?

22 The Negative Rule for Inequalities When multiplying or dividing an inequality by a negative number, the direction of the inequality changes. Multiply 2 < 3 by 7: Solution: 14 > below zero is warmer than 21 below zero Divide 3x 6 by 3: Solution: x 2 Check. Try value of x less than 2: x = 1 satisfies 3(1) 6 x = 0 satisfies 3(0) 6 x = 2 satisfies 3( 2) Graphing a Line in Standard Form Graph the line x + 2y = 6 Step a. Find x intercept. When y = 0, x + 0 = 6 = x = 6 So x intercept is (6, 0) Step b. Find y intercept. When x = 0, 0 + 2y = 6 = y = 3 So y intercept is (0, 3) 53. Plot and Graph Step c. Draw line. Mark the points (6, 0) and (0, 3) on your graph paper. Using a ruler, draw a line through these two points.

23 23 This is the line x + 2y = Graph of the Line (0,3) Line: x + 2y = 6 (6,0) 55. Graphing an Inequality Graph the inequality x + 2y 6 Fact: This inequality is a halfplane, whose edge is the line x + 2y = 6, which we just graphed. Idea: To determine which halfplane it is (above or below the line), try a test point, usually (0, 0). Is ? Yes. So the halfplane x + 2y 6 contains (0, 0) and therefore lies below the line x + 2y = 6.

24 Graph of the Halfplane (0,3) Halfplane: x + 2y 6 (6,0) 57. Systems of Inequalities Sketch the region determined by the following set of inequalities: x + y 16 5x + 2y 50 y 12 x 0, y Halfplane 1: x + y 16 Step a. Find x intercept. When y = 0, x + 0 = 16 = x = 16 So x intercept is (16, 0) Step b. Find y intercept. When x = 0, 0 + y = 16 = y = 16

25 25 So y intercept is (0, 16) 59. Halfplane 1: x + y 16 Step c. Draw the line Mark the points (16, 0) and (0, 16) on your graph paper. Using a ruler, draw a line through these two points. This is Line 1. Step d. Use (0, 0) as a test point to identify the halfplane. Is ? Yes. So the halfplane x + y 16 contains (0, 0) and therefore lies below the line x + y = Graph of Halfplane 1 (0,16) Halfplane 1: x + y 16 (6,10) Line 1: x+y=16 (0,0) (16,0) 61. Halfplane 2: 5x + 2y 50 Step a. Find x intercept.

26 26 When y = 0, 5x + 0 = 50 = x = 10 So x intercept is (10, 0) Step b. Find y intercept. When x = 0, 0 + 2y = 50 = y = 25 So y intercept is (0, 25) 62. Halfplane 2: 5x + 2y 50 Step c. Draw line Mark the points (10, 0) and (0, 25) on your graph paper. Using a ruler, draw a line through these two points. This is Line 2. Step d. Use (0, 0) as a test point to identify the halfplane. Is ? Yes. So the halfplane 5x + 2y 50 contains (0, 0) and therefore lies below the line 5x + 2y = 50.

27 Graph of Halfplane 2 (0,25) Halfplane 2: 5x + 2y 50 Line 2: 5x+2y=50 (0,0) (10,0) 64. Where do Line 1 and Line 2 intersect? Step a. Write the equations: (1) 5x + 2y = 50 (2) x + y = 16 Step b. Solve for one of the variables. original(1) 5x + 2y = 50 2 (2) 2x + 2y = 32 Subtract 3x = 18 So x = 18 3 = Interesection Cont d Step c. Solve for the other variable.

28 28 Plug x = 6 back into original equation (2) x + y = y = 16 implies y = 10 Step d. Write down the solution: The intersection point is (6, 10). 66. Halfplane 3: y 12 Step a. Find y intercept. When x = 0, y = 12 So y intercept is (0, 12) Step b. There is no x intercept, since the horizontal line is parallel to the x axis. Step c. Draw the line Mark the point (0, 12) on your graph paper. Using a ruler, draw a horizontal line through this point. This is Line Halfplane 3: y 12 Step d. Use (0, 0) as a test point to identify the halfplane. Is 0 12? Yes. So the halfplane y 12 contains (0, 0) and therefore lies below the line y = 12.

29 Graph of Halfplane 3 Halfplane 3: y 12 (0,12) Line: y=12 (0,0) 69. Where do Line 1 and Line 3 intersect? Step a. Write the equations: (1) x + y = 16 (2) y = 12 Step b. Plug y = 12 into equation (1) x + 12 = 16 = x = 4 Step c. Write down the solution: The intersection point is (4, 12). 70. Putting it all together Step a. Graph Line 1, using the points (16, 0) and (0, 16) Step b. Graph Line 2, using the points (10, 0) and (0, 25) Step c. Graph Line 3, the horizontal line through the point (0, 12)

30 30 Step d. Label the corner points (6, 10), (4, 12), (0, 16), (10, 0), and (0, 0) Step e. Shade in the intersection of the 3 halfplanes. (0,25) 71. Graph (0,16) (6,10) Line 2: 5x+2y=50 (0,12) (4, 12) Line: y=12 Line 1: x+y=16 (0,0) (10,0) (16,0) 72. Linear Programming Problems A linear programming problems has the form: Maximize an objective function usually profit or revenue subject to certain conditions usually resource or time constraints The objective function and the restraining conditions are all linear, that is no squares or worse.

31 Solving Linear Programming Problems The feasible region is the set of points satisfying the constraining equations. This region will consist of the intersection of halfplanes, whose respective lines meet in corner points of the region. When the feasible set is finite, the maximum (or minimum) is found by the following Linear Programming Rule: Maximums (and minimums) occur at corner points 74. Linear Programming Example So to find the maximum (or minimum) we simply evaluate the objective function at all the corner points of the feasible region. Maximize P = 7x + 5y subject to the following set of inequalities: x + y 16 5x + 2y 50 y 12 x 0, y 0 We have seen that the corner points are (0, 0), (0, 16), (4, 12), (6, 10), and (10, 0)

32 32 x y P = 7x + 5y Solving the Problem Maximum value of P is 92 and occurs at the point (6, 10) Note: The minimum value is 0 at the point (0, 0) 76. Construction Problem A contractor builds two types of homes: the standard model and the deluxe model. The standard model requires one lot, $12,000 capital, 150 labor-days to build, and is sold for a profit of $2400. The deluxe model requires one lot, $32,000 capital, 200 labor-days to build, and is sold for a profit of $3400. The contractor has 150 lots. The bank is willing to loan him $2,880,000 for the project and he has a maximum labor force available of 24,000 labor-days. How many houses should he build to realize the greatest profit? Let 77. Set the Variables

33 33 x = the number of standard houses built y = the number of deluxe houses built The problem is to find the values of x and y which will maximize the profit and not exceed the resources of lots, capitol, or labor 78. Organizing the Data Resource Standard Deluxe Available Lot Capital 12,000 32,000 2,880,000 Labor ,000 Profit P Resource Inequalities: Lot 1x + 1y 150 Capital 12000x y Labor 150x + 200y Profit Equation: Profit P = 2400x y 79. Simplify the Math Resource Inequalities and Profit:

34 34 Lot 1x + 1y 150 Capital 12000x y Labor 150x + 200y Profit P = 2400x y Math Simplification Divide Capital Inequality by 4000 Divide Labor Inequality by 50 Simplified Equations: Lot x + y 150 Capital 3x + 8y 720 Labor 3x + 4y 480 Profit P = 2400x y (0,150) 80. Graph (0,120) (0,90) Lot x + y = 150 Labor 3x+4y=480 Capitol 3x+8y=720 (0,0) (150,0) (160,0) (240,0) 81. Where do Capitol and Labor Lines Cross? Write the equations:

35 35 Capitol 3x + 8y = 720 Labor 3x + 4y = 480 Subtract 4y = 240 Divide by 4: So y = = 60 Plug y = 60 back into the Labor equation 3x + 4y = Capitol and Labor Line Intersection 3x = 480 3x = 480 3x = = 240 x = = 80 The intersection point is (80, 60). 83. Where do Labor and Lot Lines Intersect? Write the equations: Labor 3x + 4y = 480 Lot x + y = 150

36 36 Multiply Lot equation by 3 Capitol 3x + 4y = 480 Lot 3x + 3y = 450 Subtract y = 30 Plug y = 30 back into the Lot equation x + 30 = 150 or x = 120 The intersection point is (120, 30) (0,150) 84. Graph (0,120) (0,90) Lot x + y = 150 Labor 3x+4y=480 (80,60) Feasible Set (120,30) Capitol 3x+8y=720 (0,0) (150,0) (160,0) (240,0)

37 Solving the Problem x y P = 2400x y $306, $396, $390, $360,000 Maximum Profit 80 standard 60 deluxe $396,000 profit Note: 10 lots are unused 86. Homework and Quiz Homework due next week in Class on Wednesday Turn it in when you hand in your exam. Assigned Sections: 0.2, 0.4, IV.3, III.1, IV.1, III.2 Check the homework list on line. Miniquiz 3: 1 pt What s your name, Z-number, and Recitation Section Number? 2 pts Find the equation of the line through the points (1, 2) and (3, 6). Write your answer in the form y = mx + b.