Review

1
Review
2
Introduction to Linear Functions (Equations)

• A function is a rule (process or method) that
  produces a correspondence between two set of
  elements such that to each element in the first
  set (x-values) there corresponds one and only one
  element in the second set (y-values). The set all
  possible x-values is called the domain, and the
  set of all possible y-values is called the range.

3
Linear Equation
Range
Domain

• Y 2x 4 y-intercept (when x
  0)
• Independent Variable
• Slope
• Dependent Variable

X
Y
Y
-2 -1 0 1 2
0 2 4 6 8
4
X
-2
0
4
The Cartesian Coordinate System 

• Two perpendicular lines one of which is
  horizontal 
• The lines are called the coordinate axes and
  their point of intersection the origin

Y Axis
(,) Quadrant I
Quadrant II (-,)
X Axis
IV (,-)
III (-,-)
5
The Cartesian Coordinate System 

• The horizontal axis is usually called the x-axis
• The vertical axis is called the y-axis 
• A point in the plane can now be represented
  uniquely in this coordinate system by an ordered
  pair of numbers - that is, a pair (x,y) where x
  is the first number and y the second 
• Sketch the straight line that passes through the
  point (-2,1) (3,4). 

(3 , 4)
1
3
-2
6
Evaluating a function

• For any element x in the domain of the function
  , the symbol (x) represents the element in the
  range of corresponding to x in the domain of .
  If x is an input value, then (x) is the
  corresponding output value.
• For example if (x) 5x 5 evaluate (1), (-1),
  (0) and then make a table of values including
  ordered pairs.
• (1)5(1)510 (1, 10)
• (-1)5(-1)50 (-1, 0)
• (0) 5(0)55 (0, 5)

7
Slope-intercept form of a linear Equation

• Slope-intercept form of a linear Equation
•  Y mx b
• b the y-intercept or the value of y when x 0
• m slope 
• The concept of slope is extremely important in
  applications to business since it is a measure of
  the rate of change in Y (dependent Variable)
  given a one unit increment in X (independent
  Variable)

8
SLOPE

• Marginal Revenue Change in Total Revenue /
  Change in Quantity 
• The slope of a line is used to measure the
  steepness of a line Tells how much line rises or
  falls in response to a one unit increment along
  horizontal axis 
• The slope of a straight line is a measure of the
  rate of change of Y with respect to X
•  m (y2 - y1) / ( x2 - x1) vertical change /
  horizontal change rise / run
•  A positive slope slants upward from left to
  right ( Y increases as x increases ) Whereas,
  line with negative slope slants downward from
  left to right ( Y decreases as X increases)

9

Y
X
a. The line rises (mgt0)
Y
X
b. The line falls (mlt0)
10
Point-Slope Form

• Recall the Slope Formula can be stated as 
• m (y2 - y1) / (x2 - x1) 
• which is equal to 
• y2 - y1 m (x2 - x1) or 
• y - y1 m ( x - x1) This is the point-slope
  form of a linear equation
•  Point-Slope Form
• An Equation of the line that has slope m and
  passes through the point ( x1, y1) is given by
• y - y1 m (x - x1)

11
Point Slope Form

• Find an equation of the line that passes through
  the point (1, 3) and has slope 2
• y-32(x-1) y-32x-2 y2x1
•  

12
SLOPE FORMULA

• Find an equation of the line that passes through
  the point
• (-3, 2) and (4, -1)
• Using the point slope form of the equation of a
  line with the point (4,-1 and a slope m-3/7 we
  have

13
DEMAND APPLICATIONS

• Demand Equation
• Suppose consumer will demand 40 units of a
  product when the price is 12 per unit and 25
  units when the price is 18 each. Find the
  demand equation assuming that is linear. Find the
  price per unit when 30 units are demanded
•  

14
Demand Equation Solution

• The Points are
• (40,12) and (25,18)
• Slope Formula
• Hence an equation of the line is
• P-12-2/5 (Q-40) or P-2/5Q28

15
Demand Equation Solution

• If Q0 then P28
• If P0 then Q70
• When Q30 P16

P
28
16
30
Q
70
16
SUPPLY APPLICATIONS

• Supply Equation
• Suppose a manufacturer of shoes will place on the
  market 50 (thousand pairs) when the price is 35
  (dollars per pair) and 35 when the price is 30.
  Find the supply equation, assuming that price p
  and quantity q are linearly related

17
Supply Equation Solution

• Solution
• Hence the equation of the line is
• P-351/3(Q-50) or

18
Cost Equation

• Cost Equation
• Suppose the cost to produce 10 units of a product
  is 40 and the cost of 20 units is 70. If cost
  c is linearly related to output q, find a linear
  equation relating c and q. Find the cost to
  produce 35 units.

19
Cost Equation Solution

• The line passing through (10,40) and (20,70) has
  slope
• So an equation for the line is
• C-403(Q-10) or C3Q10
• If Q35 then C3(35)10115

20
Some Definitions from Microeconomics

• Fixed Cost or FC is the component of cost that is
  independent of output
• Variable Cost or VC is that component of cost
  that varies with output
• Total Cost or TC is the sum of fixed cost and
  variable cost or TC FC VC
• Average Variable Cost is the variable cost per
  unit of output or AVC VC/q
• Average Fixed Cost is the Fixed Cost per unit of
  output or AFC FC/q
• Average Total Cost is the TC (VC FC) per unit
  of output or ATC TC /q or (FCVC)/q
• Marginal Cost is the Extra cost associated with
  producing an additional unit of output.
• MC Change in TC / Change in q
• TR Price Quantity

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COST-OUTPUT IN A LINEAR FRAMEWORK

• TC3003q

VC
600
FC
If Q100 C600
VC
AVC(100)VC/Q300/1003
ATCTC/Q600/1006
300

FC
MC3(Slope of TC)
100
22
Quadratic Function

• Although many business relationships are linear
  or can be approximated by a linear function, in
  some cases we must look for alternative
  (nonlinear) functional forms to express business
  relationships.
• At this point we want to discuss quadratic
  functions (text material is on pp. 10-12 in the
  Algebra Refresher and 67-70)A quadratic function
  is a function of the following form
• where a, b, and c are constants and not 0
• The y-intercept is c
• The x-intercepts are found by setting y0 and
  solving for x

23

• A quadratic function graphs as a parabola and
  looks like
•  

Vertex
Vertex
If a gt 0
If a lt 0
24
Vertex

• If agt0 the vertex is the lowest point on the
  parabola. This means that f(x) has a minimum
  value at this point.
• If alt0 the vertex is the maximum value of f(x)
• The Vertex is always

25

• A quadratic equation is also referred to as
• a second-degree equation
• an equation of degree two
• Solving quadratic equations
• Solution by factoring
• Example Solve
• thus (x-6)(x-6)0
• pr x6 note only 1 root

26

• Solution by quadratic formula

27

• Example Solve
• 1. Could factor (2x-3)(x5)0
• so 2x-30 or x3/2
• and x50 or x-5
• 2. Could use Quadratic Formula
• note a2, b7, and c-15

-5
3/2
28

• Suppose we have the following quadratic profit
  function
•  
• We can use quadratic formula to find roots and
  the vertex is always located at the point -b/2a
• The vertex will occur where Adv (the x variable
  in general) is equal to -b/2a
• Where a and b are as defined in the quadratic
  equation.

29

• So for our specific function
•  
• a -2 b 20 and c 400
• So the vertex occurs where Adv -b/2a
  -20/2(-2) 5
• If Adv 5
•  
•  

30
Quadratic Formula

• What are the roots (Adv values where Profit0)?
•  

31

• Profit

Profit
450
Adv
5
-10
20
32
Maximum Revenue

• The demand function for a manufacturers product
  is p1000-2q, where p is the price (in dollars)
  per unit when q units are demanded (per week) by
  consumers. Find the level of production that will
  maximize the manufacturers Total Revenue and
  determine this revenue.
• TRPQ(1000-2q)q1000q-2q2
• Since alt0 the parabola of this quadratic function
  opens downward, TR is maximum at the vertex
• The maximum Revenue 1000(250)-2(250)125,000

33
Continued

• Thus the maximum revenue that the manufacturer
  can receive is 125,000 which occurs at a
  production level of 250 units

TR1000q-2q2
TR
q
250
500
34
BREAK-EVEN ANALYSIS

• Break-Even
• Means that a firm has a profit of zero
• Means TR TC, since Profit TR-TC
• Provides bench mark for profit planning

35
Application

• Profit Analysis
• We know from microeconomics that profit is the
  difference between total revenue and total cost.
  Therefore, we can write Profit Total Revenue -
  Total Cost or Profit TR - TC
• If a firm can sell all it produces for 5 per
  unit,
• TR 5q, where q is output or units sold
• If this same firm has fixed costs of 150 and
  its variable costs increase by 3 for every unit
  produced, TC 150 3q

36
Application

• TR 5q and TC 150 3q and recall that
  
• and profit is defined as, Profit TR - TC
• Thus, Profit 5q - (150 3q)
• or Profit 2q 150

37
Applications

• a. Graph the following supply and demand
  functions
• SS p0.20q 10
• DD p-0.40q70
• By substituting 0.20q10 for p in the demand
  equation
• we get
• 0.20q10-0.40q 70
• q0.6060 and q100
• Substitute this value in the SS equation to find
  p30

38
Continued

• B. Find the equilibrium point E
• (q,p) (100,30)

p
SS
P0.20q10
70
30
DD
10
P-0.40q70
100
q
39
Continued

• C. Plot the new supply function
• SS p0.25q18(NEW)
• SS p0.20q10(Old)
• DD p-0.40q70
• By substituting 0.25q18 for p in the demand
  equation we get
• 0.25q18-0.40q70
• 0.65q52 OR q80
• Substitute this value in the New SS equation to
  find p
• P0.25(80)1838

40
Continued

• D. Find the new equilibrium point (80,38)

p
New SS
New Equilibrium
Old SS
38
30
18
DD
10
80
100
q
41
Application

• If the DD p-0.40q70 find the Revenue Function.
• Ans TR-.4q270q

42
Calculus

• Derivatives is basically nothing more than a set
  of rules for finding slopes and rates of change.
• In a business context finding the slope of a
  curve at a point can mean among other uses,
• 1.  Finding slope of Total Cost function at
  certain level of output. MC is the slope of TC
  and the derivative of TC
• 2.  Finding slope of Total Revenue function at a
  certain level of out put. MR is the slope of TR
  and the derivative of TR
• Knowing both MC (derivative of TC) and MR
  (derivative of TR) enables the manager to
  determine how much output a firm must produce to
  earn its maximum possible profit

43
RULES FOR DIFFERENTIATION
44
1. CONSTANT RULE

• 1. THE CONSTANT RULE
• The derivative of a constant function is zero.
• That is f ( c ) 0, c is a constant.
• Ex f(x)28 then f (x) 0
• if f(x) -2 then f (x) 0

45
2. THE (SIMPLE) POWER RULE

• f (x) xn then f (x) nx(n-1)
• Ex if f(x) x then
• f (x) 1x(1-1) 1
•  
• if f(x) x8 then
• f (x) 8x(8-1) 8x7
•  
• if f(x) x-10, f (x) -10x-11
• if f(x) x5/2 f (x) 5/2x3/2

46
3. THE CONSTANT MULTIPLE RULE

• The derivative of a constant times a
  differentiable function is equal to the constant
  times the derivative of the function.
• if f (x)3x2 f (x) 23x 6x
• if f(x) 5x3 then f (x)15x2
•  if f(x) 3/ x1/2 f (x) 3x-1/2
• f (x) 3(-1/2x-3/2) -3/2(x-3/2).

47
4. THE SUM AND DIFFERENCE RULES

• The derivative of the sum or difference of two
  differentiable functions is the sum or difference
  of their derivatives.
• Ex f(x) g(x) f (x) g (x)
• and
• f(x) - g (x) f (x) - g (x)
•  

48
Continued

•  1.   f(x) 4x5 3x4 - 8x2 x 3
• f (x) 20x4 12x3 - 16x 1 0
•  

49
Main Concepts

• 1.  A derivative is the slope of a tangent line
  to a function.
• 2.  Evaluating a derivative means finding the
  value of the derivative of the function yf(x) at
  a point.
• 3.  The derivative is an instantaneous rate of
  change
• 4.  If a derivative of a function is positive at
  a point the slope will be positive at the same
  point.

50
5. THE PRODUCT RULE

• The derivative of the product of two
  differentiable functions is equal to the first
  function times the derivatives of the second plus
  the second function times the derivative of the
  first.
•  
• d/dx f(x) g(x) f(x) g (x) g(x) f
  (x)
•  
• Ex f(x) (2x2 - 1)(x33)
•  
• f (x) (2x2 - 1)(3x2) (x3 3) (4x)
• 6x4 -3x2 4x4 12x
• 10x4 - 3x2 12x

51
6. THE QUOTIENT RULE
The derivative of the quotient of two
differentiable functions is equal to the
denominator times the derivative of the numerator
minus the numerator times the derivative of the
denominator, all divided by the square of the
denominator   d/dx f(x)/g(x) g(x) f (x) -
f(x) g(x)/ g (x)2  
52
THE QUOTIENT RULE

• Ex

53
THE QUOTIENT RULE

• Ex

54
Marginal Analysis

• Marginal Cost function MC(x)C(x)
• Marginal Revenue function MR(x)R(x)
• Marginal Profit function MP(x)P(x)
• Average Cost function AC(x) C(x)/x
• Average Profit function AP(x)P(x)/x

55
Applications

• An important use of rates of change is in the
  field of economics. Economists refer to marginal
  profit, marginal revenue and marginal cost as the
  rates of change of the profit, revenue, and cost
  with respect to the number q of units produced or
  sold. An equation that relates these three
  quantities is PR-C
• where P, R and C represent the following
  quantities
• P total profit, RTotal revenue and Ctotal
  cost
• The derivatives of these quantities are called
  the marginal profit, marginal revenue, and
  marginal cost.

56
Continued

• Ex Find the marginal profit for a production
  level of 50 units
• P 0.0002q3 10q
• MP 0.0006q2 10
• 0.0006(502)10
• 11.50 per unit

57
Applications

• Ex Find the marginal revenue
• A fast-food restaurant has determined that the
  monthly demand for their hamburgers is given by
• p (60,000 -q)/20,000
• Find the MR when q20,000
• TR pq
• (60,000q-q2)/20,000
• MR 60,000-2(20,000)/20,000 1 per unit.
•  
•  

58
Applications

• Ex find the marginal cost for a production level
  10 units
• IF TC0.1q2 3
• MC .2(10) 2
• MC is the approximate cost of one additional unit
  of output.

59
Finding Average Cost

• A company estimates that the cost (in dollars) of
  producing x units of a product can be modeled by
• C800 0.04x 0.0002x2
• Find the the average cost function
• Solution
• C represent the total cost, x represents the
  number of units produced, and AC represents the
  average cost per unit.
• AC C/x
• Substituting the given equation for C produces
• AC (8000.04x 0.0002x2)/(x)
• 800/x 0.04 0.0002x

60
7. CHAIN RULE (14.1)

• If yf(u) is a differentiable function of u, and
  g(x) is a differentiable function of x, then
  yf(g(x)) is a differentiable function of x, and
• d/dx f(g(x)) f (g(x))g(x).
• When applying the Chain rule it helps to think of
  the composite function yf(g(x) or yf(u) as
  having two parts an inside and an outside
• yf(g(x))f(u)

Inside
Outside
61
7. CHAIN RULE

• The Chain Rule tells you that the derivative of
  yf(u) is te derivative of the outer function
  times the derivative of the inner function. That
  is yf (u)u

62
General Power Rule

• d/dx un nu(n-1) u

63
Further Applications of the Derivative

• First Derivatives and Graphs
• Second Derivatives and Graphs
• Optimizing Functions on a Closed Interval
• The second Derivative Test and Optimization

64
APPLICATION OF DIFFERENTIAL CALCULUS

• WHAT IS OPTIMIZATION
•   Optimization deals with finding when minimum
  values exist and linking that to variables
  controlled by managers. And one of the most
  useful applications to calculus in management and
  economics is finding the maximum and minimum
  values for a function.

65
Continued

• STATIONARY POINTS
•   The slope of a line tangent to a curve at a
  point is, by definition, the value of the first
  derivative at that point. Hence when the first
  derivative is 0, the tangent line is horizontal.
  Point when this occurs are called
• STATIONARY POINTS
• A stationary point on f(x) is a point where
  f(x)0

66
STATIONARY POINTS

• Ex if
• 1.To find the stationary point we start with the
  first derivative
• and then we set it equal to zero

67
Continued

• The values of x are determined by setting each
  factor of the last expression equal to 0. Thus
  x0 and x4.
• The function values for x0 and x4 are
• The stationary points are P(0,37) and Q(4,5)

68
Continued

• Graph of
• Where Plocal maximum. Q is a local minimum.
• For this class Local Maximum and local minimum
  are also Absolute minimum and absolute maximum

P(0,37)
Q(4,5)
69
Critical Number or Critical Value

• Critical Number is a number for x that makes the
  derivative of the function either 0 or
  undefined.
• Ex
• Ex
• x-2 is a critical number because it makes the
  function undefined.
• Ex

70
Continued

• To find all critical numbers
• Take the derivative of a given function
• 2. Set both the numerator and denominator of a
  derivative0 and solve for x

71
Relative Extrema

• We will examine the points at which a function
  changes from increasing to decreasing or
  viceversa. At such point, the function has a
  relative extremum. (The plural of extremum is
  extrema.) The relative extrema of a function
  include the relative minima and the relative
  maxima of the function

72
Relative ExtremaRelative Maximum and Minimum

• To find relative max or min
• 1. Put critical number on a number line
• 2. Pick any number to the left side and
  substitute into the first derivative
• If you get a positive value, the functions graph
  is sloping upward in that section of graph. If
  you get a negative value, the functions graph
  slopes down in that section.
• 3. Pick any number to right side and substitute
  into 1st derivative. If you get a positive value
  the functions graph is sloping upward in that
  section of graph. If you get a negative value,
  the functions graph slopes down in that section

73
Continued

• Ex 1.
• Differentiate original function
• Set derivative equal to 0
• Factor
• Critical x0 and x1

74
Continued
75
Continued

• f(x)

CN x0, x1
(0,0)
Increasing
Increasing
Decreasing
(1, ½)
76
Continued

• Ex 2. Find the critical numbers and the open
  intervals on which the function is increasing or
  decreasing

77
Find the Relative Extrema

• Find the Relative Extrema
• f(x) 2x3 -3x2 -36x14
•  1st find the critical s of f
•  f (x)6x2 -6x -36 Find derivative of f
•  6x2 -6x -36 0 Set derivative equal
  to 0
•  6(x2 -x -6)0 Factor out common
  factor
•  6(x-3)(x2)0 Factor
•  x -2,3 Critical Numbers
•  Using these numbers, you can form three test
  intervals
• (-infinity, -2) (-2,3) and (3, infinity)

78
Continued
79
Continued

• You can conclude that the critical number 2
  yields a relative maximum (f (x) changes sign
  from positive to negative) and the critical
  number 3 yields a relative minimum (f (x)
  changes sign from negative to positive)
• To find the y-coordinates of the relative
  extrema, substitute the x-coordinates into the
  original function. For instance the relative
  maximum is f(-2)58 and the relative minimum is
  f(3)-67

80
Applications

• Find the production level that produces a maximum
  profit for the following profit function.
• Solution
• Begin by setting the MP equal to 0 and solving
  for x
• x24,400 units

81
Continued

• X24,400 correspond to the production level that
  produces a maximum profit. To find the maximum
  profit, substitute x24,000 into the Profit
  function.

82
Cost Function

• A retailer has determined the cost ( C ) for
  ordering and storing x units of a product to be
• The delivery truck can bring at most 200 units
  per order. Find the order size that will minimize
  the cost
• Solution
• Find the critical numbers x?82 units
• Therefore C(82)489.90 which is the minimum

0ltxlt2000
83
Profit

• When a soft drinks sold for .80 per can at
  football games, approximately 6,000 cans were
  sold. When the price was raised to 1 per can,
  the quantity demanded dropped to 5,600. The
  initial cost is 5,000 and the cost per unit is
  .40. Assuming that the demand function is
  linear, what price will yield the maximum profit?
• Solution
• Demand (6,000 , 0.80) (5,600, 1)
• m (1-.80)/(5600-6000) -0.0005
• p -1 -0.0005(x-5600)
• p -0.0005x 3.80
• Cost
• C5000 .40x

84
Continued

•  Profit Revenue - Cost(TOTAL COST)
• Revenue PriceQuantity (instead of q lets use
  x)
• R x(-0.0005x 3.80)
• Profit x(-0.0005x 3.80) - (5000.40x)
• -0.0005x2 3.40x -5000
• P -0.001x 3.400
• Critical number x3,400
•  
• p(3,400) The price that will yield the maximum
  profit is 2.10 per can.
•  

85
HIGHER ORDER DERIVATIVE

• We know that the derivative of a function yf(x)
  is itself a function, f (x) If we differentiate
  f (x), the resulting function is called
  the second derivative of f at x. It is denoted f
  (x), which is read f double prime of x.
  Similarly, the derivative of the second
  derivative is called the third derivative,
  written f(x). Continuing this way, we get
  higher order derivative.
•  Ex If f(x) 6x3 -12x26x-2 find all higher
  order derivatives.
•  f (x) 18x2 -24x 6

86
HIGHER ORDER DERIVATIVE

• f(x) 36x -24
• f(x)36
• f(4)(x)0

87
Concavity

• We use the second derivative to find the
  intervals when a function is concave up, concave
  down, point of inflection or point of diminishing
  returns.
•  

88
Points of Inflection

• If the tangent line to a graph exists at a point
  at which the concavity changes, then the point is
  a point of inflection.
• Finding Points of inflection
• Discuss the concavity of the graph of f and find
  its points of inflection.
• f(x) x4 x3 - 3x2 1
• Solution
• Begin by finding the second derivative of f
• f (x) 4x3 3x2 - 6x Find first
  derivative
• f (x) 12x2 6x -6 Find second
  derivative
• 6(2x-1)(x1) Factor

89
Continued

• From this you can see that the possible points of
  inflection occur at x1/2 and x-1.
• After testing the intervals
• (-infinity, -1) , (-1, 1/2) and (1/2, infinity),
• you can determine that the graph is concave
  upward in (-infinity, -1), concave downward in
  (-1, 1/2), and concave upward in (1/2,
  infinity). Because the concavity changes at x-1
  and x1/2, you can conclude that the graph has
  points of inflection a these x-values.

90
Continued

• Graph

Concave Upward
Point of inflection
Concave Upward
-1
Concave Downward
1/2
-2
Point of inflection
91
Second Derivative Test for Relative Extrema

• 1. Set f (x)0 and solve for x to determine the
  stationary points
• 2.
• (a) If f (x) gt0 (concave up), then the
  stationary point is a local minimum
• (b) If f (x) lt0 (concave down), then the
  stationary point is a local maximum

92
Continued

• Ex If f(x) x3 -6x2 37
• Find local minimum and local maximum and
  inflection points
• 1. f (x) 3x2 -12x Find the first derivative
• x(3x-12) Factor
• x0 and x4 are the stationary points
• 2. f (x) 6x -12 Find the second
  derivative
• x2 Inflection point
•  (a) f (0) -12
• Because f (0) lt0 (concave down) the stationary
  point is a local maximum
•  (b) f (4) 12
• Because f (4) gt0 (concave up) the stationary
  point is a local minimum

93
Continued

• Graph

Q (0, 37)
Inflection Point
Concave Up
Concave Down
2
Q (4, 5)
94
APPLICATIONS

• In economics, the notions of concavity is related
  to the concept of diminishing returns. (The
  postulate that as more and more units of a
  variable resource are combined with a fixed
  amount of other resources, employment of
  additional units of the variable resource will
  eventually increase output only at a decreasing
  rate. Once diminishing return are reached, it
  will take successively larger amounts of the
  variable factor to expand output by one unit)
• Example By increasing its advertising cost x (in
  thousand of dollars) for a product, a company
  discovers that it can increase the sales y
  (thousands of dollars) according to the model
•  y 1/(10,000)(300x2 -x3) 0x200
• Find the point of diminishing returns for this
  product.

95
Solution

• Begin finding the first derivatives of
•  
• y 1/10000(600x -3x2) First derivative
• y1/10000(600-6x) Second
  derivative
• The second derivative is zero only when x100. By
  testing the intervals (0, 100) and (100,200), you
  can conclude that the graph has a point of
  diminishing returns when x100.

96
Finding The Maximum Revenue

• A company has determined that its total revenue
  (in dollars) for a product can be modeled by
• R -x3 450x2 52,500x 0x546
• where x is the number of units produced and sold.
  What production level will yield a maximum
  revenue?
•  Solution To maximize the revenue, find the
  critical numbers
• R -3x2 900x 52,5000 Set derivative
  equal to 0.
• -3(x-350)(x50) 0 Factor
• x 350 , -50
  Critical numbers
• The only critical number in the feasible domain
  is x350. Therefore a production level of 350
  units corresponds to a maximum revenue

97
Graph

• Revenue

(350, 30,625,000)
(546,0)
350
98
Finding the Minimum Average Cost

• A company estimates that the cost (in dollars) of
  producing x units of a product can be modeled by
• C800 0.04x 0.0002x2
• Find the production level that minimizes the
  average cost per unit.
• Solution
• C represent the total cost, x represents the
  number of units produced, and AC represents the
  average cost per unit.
• AC C/x

99
Continued

• Substituting the given equation for C produces
• AC (8000.04x 0.0002x2)/(x)
• 800/x 0.04 0.0002x
• You can find the critical numbers as follows
• AC - 800/x2 0.00020 Set the
  derivative to zero.
• x2 800/0.0002
• x2 4,000,000
• x 2,000
• The production level of x2000 minimizes the
  average cost per unit.

100
Continued

• Graph

AC
.84
2000 Units
101
Finding the Maximum Revenue

• A business sells 2,000 units per month at a price
  of 10 each. It can sell 250 more items per month
  for each 0.25 reduction in price. What price per
  unit will maximize the monthly revenue? Solution
  
• 1. Let x represent the number of units sold in a
  month, let p represent the price per unit, and
  let R represent the monthly revenue.
• 2. Because the revenue is to be maximized, the
  primary equation is
• R xp
• 3. A price of p10 corresponds to x 2,000 and
  a price of p9.75 corresponds to x 2250. Using
  this information you can use the point-slope form
  to create the demand equation.

102
CONTINUED

• m (10 - 9.75) / (2000 -2250) Find the slope
• m -.001
• p-10 -0.001(x-2000) Point-slope
  form
• p-0.001x 12
• Substituting this value into the revenue equation
  produces
• R x(-0.001x12)
• -0.001x2 12x
• 4. To maximize the revenue function, find the
  critical numbers
• R 12 -.002x0

103
Continued

• x 6,000 Critical number
•  The price that corresponds to this production
  level is
• p12-0.001x Demand function
• 12 - 0.001(6000) Substitute 6,000 for x
• 6 Price per unit

(6000, 36000)
Revenue
6000
104
Regression Using Excel

• Go to Excel, Select Tools, Choose Data Analysis,
  Choose Regression from the list of Analysis
  tools. Click OK.
• Enter the Y input Range, Enter the Xs range,
  select labels, select confidence levels. Select
  Residuals, Residuals Plot, Standardized
  Residuals.

105
Continued
106
Continued
107
Example Programmer Salary Survey

• Excel Computer Output
• The regression is
• Salary 3.17 1.40 Exper 0.251 Score
• Predictor Coef Stdev
  t-ratio p
• Constant 3.174 6.156 .52 .613
• Exper 1.4039 .1986 7.07 .000
• Score .25089 .07735 3.24 .005
• s 2.419 R-sq 83.4
  R-sq(adj) 81.5

108
Interpreting the Coefficients
b1 1. 404
Salary is expected to increase by 1,404
for each additional year of experience (when
the variable score on programmer attitude test
is held constant).
109
Interpreting the Coefficients
b2 0.251
Salary is expected to increase by 251 for
each additional point scored on the programmer
aptitude test (when the variable years of
experience is held constant).
110
Example Programmer Salary Survey

• Excel Computer Output (continued)
• Analysis of Variance
• SOURCE DF SS MS
  F P
• Regression 2 500.33 250.16 42.76 0.000
• Error 17 99.46 5.85
• Total 19 599.79

111
Example Programmer Salary Survey

• F Test
• Hypotheses H0 ?1 ?2 0
• Ha One or both of the parameters
• is not equal to zero.
• Rejection Rule
• For ? .05 and d.f. 2, 17 F.05
  3.59
• Reject H0 if F gt 3.59.
• Test Statistic
• F MSR/MSE 250.16/5.85 42.76
• Conclusion
• We can reject H0.

112
Example Programmer Salary Survey

• t Test for Significance of Individual Parameters
• Hypotheses H0 ?i 0
• Ha ?i 0
• Rejection Rule DFn-p-1
• For ? .05 and d.f. 17, t.025 2.11
• Reject H0 if t gt 2.11
• Test Statistics
• Conclusions
• Reject H0 ?1 0 Reject H0
  ?2 0