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Straight Lines and Linear functions

Straight Lines and Linear functions. The Cartesian Coordinate System. y. y axis. A point can be represented in a plane by using the Cartesian Co-ordinates system. The vertical line is called the y- axis while the horizontal line is called x-axis .

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Straight Lines and Linear functions

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  1. Straight Lines and Linear functions

  2. The Cartesian Coordinate System y y axis • A point can be represented in a plane by using the Cartesian Co-ordinates system. • The vertical line is called the y- axis while the horizontal line is called x-axis. • The intersection point is called the Origin. • Similar to the number line the the axis have a scale. origin x o x axis Lamar University 2

  3. The Cartesian Coordinate System y P(x,y) • A point can be represented by (x, y). ‘x’ is called the x-coordinate or abscissa, ‘y’ is called the y-coordinate or ordinate. o x Lamar University 3

  4. Distance Formula • The distance between any two points (x1,y1) and (x2,y2) is given by Lamar University 4

  5. Problem • Find the Distance between points (-3,4) and (6,2). Solution: we have x1 = -3x2= 6 y1 = 4y2 = 2 Using the distance formula we have Lamar University 5

  6. Slope of a Line • Let L represent a straight line that passes through 2 distinct points (x1,y1) and (x2,y2). • If x1=x2, then the line L is a vertical line and the slope is undefined. • If x1=x2, we define the slope of L as follows • When m is -ve the line is said to have a negative slope( the line falls) and when m is +ve it is said to have a positive slope (the line rises). Lamar University 6

  7. Problem • Find the slope that passes through points (-2,5) and (3,5) Solution: The slope is given by Lamar University 7

  8. Problem • Sketch the straight line that passes through the point (-2,5) and has a slope –4/3. Solution: Plot the point (-2, 5) Now remember that a slope of –4/3 indicates that an increase of 1 unit in the x direction produces a decrease of 4/3 units in the y direction.using this information we plot (1,1) and draw the line through the two points. Lamar University 8

  9. Point- Slope form • Two distinct lines are said to b parallel if and only if their slopes are equal or their slopes are undefined. Point-Slope form • The equation of the line that passes through point and has slope m is given by Lamar University 9

  10. Problem • Find the equation of the line that passes through the point (1,3) and has slope 2 Solution: Here x1 = 1 and y1= 3 Substituting in the formula we get y - 3 = 2( x - 1) or 2x - y +1= 0 Which is the equation of the line. Lamar University 10

  11. Problem • Find the equation of the line that passes through the points (-3,2) and (4,-1) Solution: Slope of the line is given by Substituting in the formula we get y - 2 =-3/7( x – (-3)) or 3x + 7y - 5 = 0 Which is the equation of the line. Substitute any of the points in the equation, the equation should be satisfied. Lamar University 11

  12. Perpendicular Lines • If L1and L2 are two non distinct non vertical lines that have slopes m1 and m2 then L1is perpendicular to L2 if and only if Lamar University 12

  13. Problem • Find an equation of the line that passes through the point (3,1) and is perpendicular to the line with slope 2. Solution: Since the slope of the perpendicular line is 2, the slope of the required line is -1/2 Using this slope and the point-slope equation we get Which is the required equation of the line passing through point (3,1) and is perpendicular to the line with slope 2. Lamar University 13

  14. Slope –Intercept Form • The equation of the line that has slope m and intersects the y axis at the point (0,b) is given by y = mx + b Lamar University 14

  15. Problem • Find an equation of the line that has a slope of 3 and y intercept –4 Solution:Using the Slope intercept form with m = 3 and b = -4 the equation of the line is y = 3x – 4 Lamar University 15

  16. Problem Determine the slope and y intercept of the line whose equation is 3x – 4y = 8 Solution: Rewriting the equation in the slope intercept form Comparing with y = mx + b We can see the slope is ¾ and the y intercept is -2 Lamar University 16

  17. http://dept.lamar.edu/industrial/Classes/INEN2301.htm Lamar University 17

  18. Revision In the previous class we have covered • The distance formula • The slope of a line ‘m’ • The Point Slope form of a line • The Point Intercept form of a straight line y = mx + b • Also we knowthat if the lines are parallel they have the same slope, that is the ‘m’ is the same for both the equations, for perpendicular lines the product of the slopes is -1. Lamar University 18

  19. General Equation of a Line • The equation Ax + By + C = 0 Where A , B and C are constants and A and B are not both zero, is called the general form of a linear equations in the variables x and y. We can say that an equation of a straight line is a linear equation; conversely, every linear equation represents a straight line. Lamar University 19

  20. Problem • Sketch the straight line represented by the equation 4x – 3y - 12 = 0 Since every straight line is determined by two distinct points, we need to find any two points through which this line passes in order to sketch it. Setting x to zero we get y = -4 so the line crosses the y axis at (0,-4) Now set y to zero we get x = 3 so the line crosses the x axis at (3,0) (3,0) (0,-4) Lamar University 20

  21. Problem For wages less than the maximum taxable wage base, Social Security contributions by employees are 7.65 % of the employees wages. Find an equation that expresses the relation ship between wages earned (x) and Social Security Taxes paid (y) by an employee who earns less than the maximum taxable wage base. Solution: y = 0.0765 x Lamar University 21

  22. Linear Functions & Mathematical Models • A mathematical model is a representation of a real world problem in mathematical terms. It may make an exact representation or then an acceptable representation is made. For example , the accumulated amount A at the end of t years when a sum of P dollars is deposited in a fixed bank account and earns interest at the rate of r percent per year compoundedmtimes a year is given by • On the other hand the size of a cancer tumor may be approximated by the volume of a sphere Lamar University 22

  23. Functions • A function f is a rule that assigns to each value of x one and only one value of y • An example of a function may be drawn from the relationship between the area of the circle and its radius. Let x and y denote the radius and area of a circle respectively. Then we know This function gives us a value y for every value of x . Here y is the dependent variable, while x is the independent variable. The set of all values that may be assumed by x is called the domain of the function f and the set comprising all the values assumed by y = f(x) as x takes all the values in its domain is called the range. Lamar University 23

  24. Linear Function The function f defined by f(x) = mx + b where m and b are constants , is called a linear function Linear functions play an important role in quantitative analysis of business and economic problems The problems arising in quantitative analysis are usually linear and hence can be formulated in to linear functions. Lets look at some applications using linear functions. Lamar University 24

  25. Problem • A Printing Machine has an original value of $100,000 and is to be depreciated linearly over a period of 5 years with a $30,000 scrap value. Find an expression giving the book value at the end of the year t? what will be the book value of the machine at the end of the second year? What is the rate of depreciation of the machine? Solution: Let V be the value of the machine. We know that the machine has a value of $100,000 at time 0. Also at the end of 5 yrs the value of the machine is $30,000, we can say that the line passes through (0, 100,000) and (5, 30,000). We now find the slope of the line which is given by Lamar University 25

  26. Problem…contd Using Point-Slope form of the equation of a line with the point (5, 30,000) and slope –14,000 We have The book value at the end of the second year is given by The rate of depreciation is given by the negative slope of the depreciation line. Here the slope is –14,000. The rate of depreciation is $14,000. Lamar University 26

  27. Linear Cost, Revenue, And Profit Functions Let x denote the number of units of a product manufactured or sold. Then, the total cost function is C(x) = Total cost of manufacturing x units of the product. The revenue function is R(x)= total revenue realized from the sale of x units of the product. The profit function is P(x) = Total function realized from manufacturing and selling x units of the product. Cost are usually classified as fixed cost and variable costs. Fixed cost are costs that remain constant , regardless of the company activities, for example RENT and EXECUTIVE SALARIES.Variable costs are cost that vary with production and sales, example are wages, cost of raw material etc. Lamar University 27

  28. Linear Cost, Revenue, And Profit Functions • Let a firm have fixed cost F, production cost of c dollars, and a selling price of s dollars per unit. The the total cost function C(x) = cx + F revenue function R(x)= sx profit function P(x)=R(x) - C(x) = sx-(cx + F) =(s-c)x – F Here the functions C, R and P are linear functions of x Lamar University 28

  29. Problem • Puriton, a manufacturer of water filters , has a monthly fixed cost of $20,000 a production cost of $20 per unit and a selling price of $30 per unit . Find the cost function, the revenue function and the profit function for Puritron. Solution:Let x denote the number of units produced and sold. Then, C(x) = 20x+20,000 R(x) = 30x P(x) = (30-20)x –20,000 = 10x – 20,000 Lamar University 29

  30. Linear Demand and Supply Curves In a free market economy the consumer’s demand for a commodity depends on the unit price. The demand equation expresses this relation ship between the unit price and the quantity demanded. The corresponding graph is called the demand curve.In general the quantity demanded of a commodity decreases as the price increases. Accordingly the demand function is given by p=f(x) Where p is the unit price and x is the number of units . Lamar University 30

  31. Problem The quantity demanded of the Sentinel alarm clock is 48,000 units when the unit price is $8. At $12 per unit, the quantity demanded drops to 32,000 units. Find the demand equation, assuming that it is linear. What is the unit price corresponding to a quantity demanded of 40,000unts ? What is the unit price corresponding to a quantity demanded if the unit price is $14 . Solution: Let p denote the unit price of an alarm clock (in dollars) and let x (in units of 1000) denote the quantity demanded when the unit price of the clocks $p When p= 8, x =48 thus the point(48,8) lies on the demand curve.Similarly when p=12, x= 32 and the point (32,12) also lies on the curve. The slope of the is Lamar University 31

  32. Problem…contd. Now using point-slope form with the point (32,12) Lamar University 32

  33. Problem…contd. The Demand curve is as follows 20 10 80 20 40 60 Units of thousands Lamar University 33

  34. Problem…contd. Now if the quantity demanded is 40,000 units ( x=40) from the equation we get Now if the price is $14 the demand from the equation is Lamar University 34

  35. Intersection of Straight Lines We have two straight lines L1 represented by y =m1x + b1 and L2 represented by y =m2x + b2 Where m1 b1 , m2 b2 are constants that intersect at a point (x,y) L2 L1 P(x,y) Lamar University 35

  36. Intersection of Straight Lines The point P(x,y) lies on both the lines L1 and L2 and so satisfies both the equations. Thus to find the point of intersection of two lines we need to solve the system composed of the 2 equations y =m1x + b1 and y =m2x + b2 for x and y L2 L1 y =m2x + b2 y =m1x + b1 P(x,y) Lamar University 36

  37. Problem Find the point of intersection of the straight lines that have equations y=3x - 8 and y = -6x + 19 Solution: Substituting the value of y we get 3x - 8 = -6x + 19 9x = 27 or x =3 Now substitute the value of x in y we get y = 3(3)-8 or we get y = -6(3) +9 y = 1 or y = 1 Lamar University 37

  38. Break Even Analysis Consider a firm with linear cost function C(x), revenue function R(x) and profit function P(x) given by C(x) = cx + F R(x)= sx P(x)=R(x) - C(x) = sx-(cx + F) =(s-c)x – F Where c is the unit cost of production, s denotes the selling price per unit, F the fixed cost incurred by the firm, and x denotes level of production and sales. The level of production at which the firm neither makes a profit or sustains a loss is called the break-even level of operation C(x) = R(x) Lamar University 38

  39. Problem Prescott, Inc. manufactures its products at a cost of $4 per unit and sells them for $10 per unit. If the firm’s fixed cost is $12,000 per month , determine the firm’s break-even point. Solution: The cost function C and the revenue function R are given by C(x) = 4x+12,000 And R(x)=10x Equating the cost function and the revenue function we get 10x=4x+12,000 6x=12,000 x=2,000 Using this in the revenue function, the break even revenue is $20,000 Lamar University 39

  40. Problem Using the data from the previous problem #What is the loss sustained by the firm if only 1500 units are produced and sold per month? The profit function P is given by P(x)=R(x) - C(x) = 10x –(4 x +12,000) = 6x – 12,000 With 1500 units produced the profit will be = 6 (1500) –12,000 = 9,000=12,000 =-3000 a loss of $3000 per month What is the profit if 3000 units are produced and sold per month? Substituting in the above equation 6(3,000) – 12,000 = 6,000 or a profit of $ 6,000 Lamar University 40

  41. Market Equilibrium A market equilibrium is said to prevail if the quantity produced is the equal to the quantity demanded. The quantity produced at the market equilibrium is called the equilibrium quantity and the corresponding price is called the equilibrium price. Lamar University 41

  42. Problem The management of the Thermo- Master company, which manufactures an indoor-outdoor thermometer in its Mexico subsidiary, has determined that the demand equation for its product is 5x + 3p – 30 = 0 where p is the price of the thermometer in dollars and x is the quantity demanded in units of a thousand. The supply equation for these thermometers is 52x-30p+45 = 0 where x is the quantity Thermo-Master will make available in the market at p dollars each. Find the equilibrium quantity and price. Solution: we need to solve the system of equations 5x + 3p – 30 = 0 I 52x - 30p + 45 = 0 II Taking equation I and re arranging in terms of p we get 3p = - 5x + 30 Dividing both sides by 3 p = - 5/3 x + 10 Lamar University 42

  43. Problem Substituting the value of p in equation II we get 52x – 30(-5/3x +10) +45 = 0 Solving 52 x +50x – 300 + 45 = 0 102x – 255 = 0 x = 5/2 = 2.5 Substituting p = - 5/3 x + 10 p = - 5/3(2.5) +10 = -12.5/3 + 10 = 5.83 Lamar University 43

  44. Problem We conclude that the equilibrium quantity is 2500 and the unit price is $5.83 Lamar University 44

  45. Problem The quantity demanded of a certain model of videocassette recorder(VCR) is 8000 units when the price is $260. At a unit price of $200, the quantity demanded increases to 10,000 units. The manufacturer will not market any VCR’s if the price is $100 or less. However for each $50 increase in the unti price above $100, the manufacturer will market an additional 1000 units. Both the demand and supply equations are known to be linear. • Find the demand equation • Find the supply equation • Find the equilibrium quantity and price. Solution: Let p denote the unit price in hundreds of dollars and let x denote the number of units of VCR’s in thousands. Demand equation: Since the demand function is linear, the demand curve is a straight line passing through points (8, 2.6) and (10,2). Lamar University 45

  46. Problem Its slope is using point( 10,2) slope – 0.3 the required equation is p – 2 = - 0.3(x-10) p = -0.3x + 5 I Supply equation: The supply curve is a straight line passing through then points (0,1) and (1, 1.5). Its slope is m = (1.5- 1)/(1-0) m = 0.5 Lamar University 46

  47. Problem Using point (1,0) and slope 0.5 in the point-slope form of the equation of a line we get p - 1 = 0.5(x-0) p = 0.5x + 1 II Equilibrium quantity and price: To find the market equilibrium , we solve simultaneously both I & II the demand equation and the supply equation p = -0.3x + 5 p = 0.5x + 1 -0.3x + 5 = 0.5x + 1 0.8x = 4 x = 5 Lamar University 47

  48. Problem Substituting in the second equation p = 0.5(5) + 1 = 3.5 Thus we see the equilibrium quantity is 5000 units and the price is $350 Lamar University 48

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