Straight Lines and Linear functions

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Straight Lines and Linear functions
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The Cartesian Coordinate System
y

• 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.

y axis
origin
x
o
x axis
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The Cartesian Coordinate System

• A point can be represented by (x, y). x is
  called the x-coordinate or abscissa, y is
  called the y-coordinate or ordinate.

y
P(x,y)
o
x
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Distance Formula

• The distance between any two points (x1,y1) and
  (x2,y2) is given by

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Problem

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

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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).

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Problem

• Find the slope that passes through points (-2,5)
  and (3,5)
• Solution
• The slope is given by

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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.

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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

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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.

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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.

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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

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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.

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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

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Problem

• Find an equation of the line that has a slope of
  3 and y intercept 4
• SolutionUsing the Slope intercept form with
  m 3 and b -4
• the equation of the line is
• y 3x 4

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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

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http//dept.lamar.edu/industrial/Classes/INEN2301.
htm
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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 know that 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.

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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.

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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)
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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

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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 compounded m times a year is
  given by
• On the other hand the size of a cancer tumor may
  be approximated by the volume of a sphere

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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.

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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.

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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

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Problemcontd

• 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.

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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.

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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

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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.
• SolutionLet x denote the number of units
  produced and sold. Then,
• C(x) 20x20,000
• R(x) 30x
• P(x) (30-20)x 20,000
• 10x 20,000

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Linear Demand and Supply Curves

• In a free market economy the consumers 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
• pf(x)
• Where p is the unit price and x is the number of
  units .

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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 p12, x 32 and the
  point (32,12) also lies on the curve.
• The slope of the is

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Problemcontd.

• Now using point-slope form with the point (32,12)

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Problemcontd.

• The Demand curve is as follows

20
10
80
20
40
60
Units of thousands
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Problemcontd.

• Now if the quantity demanded is 40,000 units (
  x40) from the equation we get
• Now if the price is 14 the demand from the
  equation is

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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)
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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)
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Problem

• Find the point of intersection of the straight
  lines that have equations
• y3x - 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

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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)

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Problem

• Prescott, Inc. manufactures its products at a
  cost of 4 per unit and sells them for 10 per
  unit. If the firms fixed cost is 12,000 per
  month , determine the firms break-even point.
• Solution
• The cost function C and the revenue function R
  are given by
• C(x) 4x12,000
• And R(x)10x
• Equating the cost function and the revenue
  function we get
• 
  10x4x12,000
• 
  6x12,000
• 
  x2,000
• Using this in the revenue function, the break
  even revenue is 20,000

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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
• 
  9,00012,000 -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

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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.

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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-30p45 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

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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

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Problem

• We conclude that the equilibrium quantity is
  2500 and the unit price is 5.83

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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 VCRs 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 VCRs 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).

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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

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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

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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

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