Applications of the First Derivative

1
4
Applications of the Derivative

• Applications of the First Derivative
• Applications of the Second Derivative
• Curve Sketching
• Optimization I
• Optimization II

2
4.1

• Applications of the First Derivative

3
Increasing and Decreasing Functions

• A function f is increasing on an interval (a, b)
  if for any two numbers x1 and x2 in (a, b),
• f(x1) lt f(x2) wherever x1 lt x2.

y
f(x2) f(x1)
x
a
b
x2
x1
4
Increasing and Decreasing Functions

• A function f is decreasing on an interval (a, b)
  if for any two numbers x1 and x2 in (a, b),
• f(x1) gt f(x2) wherever x1 lt x2.

y
f(x1) f(x2)
x
a
b
x2
x1
5
Theorem 1

• If f '(x) gt 0 for each value of x in an interval
  (a, b), then f is increasing on (a, b).
• If f '(x) lt 0 for each value of x in an interval
  (a, b), then f is decreasing on (a, b).
• If f '(x) 0 for each value of x in an interval
  (a, b), then f is constant on (a, b).

6
Example

• Find the interval where the function f(x) x2 is
  increasing and the interval where it is
  decreasing.
• Solution
• The derivative of f(x) x2
• is f '(x) 2x.
• f '(x) 2x gt 0 if x gt 0
• and f '(x) 2x lt 0 if x lt 0.
• Thus, f is increasing on the
• interval (0, ?) and decreasing
• on the interval ( ?, 0).

y
f(x) x2
Example 1, page 245
7
Determining the Intervals Where a Function
is Increasing or Decreasing

• Find all the values of x for which f '(x) 0 or
  f ' is discontinuous and identify the open
  intervals determined by these numbers.
• Select a test number c in each interval found in
  step 1 and determine the sign of f '(c) in that
  interval.
• If f '(c) gt 0, f is increasing on that
  interval.
• If f '(c) lt 0, f is decreasing on that
  interval.

8
Examples

• Determine the intervals where the function
• f(x) x3 3x2 24x 32
• is increasing and where it is decreasing.
• Solution
• Find f ' and solve for f '(x) 0
• f '(x) 3x2 6x 24 3(x 2)(x 4) 0
• Thus, the zeros of f ' are x 2 and x 4.
• These numbers divide the real line into the
  intervals ( ?, 2), ( 2, 4),
  and (4, ?).

Example 2, page 246
9
Examples

• Determine the intervals where the function
• f(x) x3 3x2 24x 32
• is increasing and where it is decreasing.
• Solution
• To determine the sign of f '(x) in the intervals
  we found ( ?, 2), ( 2, 4), and (4,
  ?), we compute f '(c) at a convenient test point
  in each interval.
• Lets consider the values 3, 0, and 5
• f '(3) 3(3)2 6(3) 24 27 18 24
  21 gt 0
• f '(0) 3(0)2 6(0) 24 0 0 24
  24 lt 0
• f '(5) 3(5)2 6(5) 24 75 30 24
  21 gt 0
• Thus, we conclude that f is increasing on the
  intervals ( ?, 2), (4, ?), and is
  decreasing on the interval ( 2, 4).

Example 2, page 246
10
Examples

• Determine the intervals where the function
• f(x) x3 3x2 24x 32
• is increasing and where it is decreasing.
• Solution
• So, f increases on ( ?, 2), (4, ?), and
  decreases on ( 2, 4)

y
60 40 20 20 40 60
y x3 3x2 24x 32
x
7 5 3 1 1 3 5 7
Example 2, page 246
11
Examples

• Determine the intervals where is increasing
• and where it is decreasing.
• Solution
• Find f ' and solve for f '(x) 0
• f '(x) 0 when the numerator is equal to zero,
  so
• Thus, the zeros of f ' are x 1 and x 1.
• Also note that f ' is not defined at x 0, so we
  have four intervals to consider ( ?, 1), ( 1,
  0), (0, 1), and (1, ?).

Example 4, page 247
12
Examples

• Determine the intervals where is increasing
• and where it is decreasing.
• Solution
• To determine the sign of f '(x) in the intervals
  we found ( ?, 1), ( 1, 0), (0, 1), and
  (1, ?), we compute f '(c) at a convenient test
  point in each interval.
• Lets consider the values 2, 1/2, 1/2, and 2
• So f is increasing in the interval ( ?, 1).

Example 4, page 247
13
Examples

• Determine the intervals where is increasing
• and where it is decreasing.
• Solution
• To determine the sign of f '(x) in the intervals
  we found ( ?, 1), ( 1, 0), (0, 1), and
  (1, ?), we compute f '(c) at a convenient test
  point in each interval.
• Lets consider the values 2, 1/2, 1/2, and 2
• So f is decreasing in the interval ( 1, 0).

Example 4, page 247
14
Examples

• Determine the intervals where is increasing
• and where it is decreasing.
• Solution
• To determine the sign of f '(x) in the intervals
  we found ( ?, 1), ( 1, 0), (0, 1), and
  (1, ?), we compute f '(c) at a convenient test
  point in each interval.
• Lets consider the values 2, 1/2, 1/2, and 2
• So f is decreasing in the interval (0, 1).

Example 4, page 247
15
Examples

• Determine the intervals where is increasing
• and where it is decreasing.
• Solution
• To determine the sign of f '(x) in the intervals
  we found ( ?, 1), ( 1, 0), (0, 1), and
  (1, ?), we compute f '(c) at a convenient test
  point in each interval.
• Lets consider the values 2, 1/2, 1/2, and 2
• So f is increasing in the interval (1, ?).

Example 4, page 247
16
Examples

• Determine the intervals where is increasing
• and where it is decreasing.
• Solution
• Thus, f is increasing on ( ?, 1) and (1, ?),
  and decreasing on ( 1, 0) and(0, 1)

y
4 2 2 4
x
4 2 2 4
Example 4, page 247
17
Relative Extrema

• The first derivative may be used to help us
  locate high points and low points on the graph of
  f
• High points are called relative maxima
• Low points are called relative minima.
• Both high and low points are called relative
  extrema.

Relative Maxima
y
y f(x)
Relative Minima
x
18
Relative Extrema

• Relative Maximum
• A function f has a relative maximum at x c if
  there exists an open interval (a, b) containing c
  such that f(x) ? f(c) for all x in (a, b).

Relative Maxima
y
y f(x)
x
x1
x2
19
Relative Extrema

• Relative Minimum
• A function f has a relative minimum at x c if
  there exists an open interval (a, b) containing c
  such that f(x) ? f(c) for all x in (a, b).

y
y f(x)
Relative Minima
x
x3
x4
20
Finding Relative Extrema

• Suppose that f has a relative maximum at c.
• The slope of the tangent line to the graph must
  change from positive to negative as x increases.
• Therefore, the tangent line to the graph of f at
  point (c, f(c)) must be horizontal, so that f
  '(x) 0 or f '(x) is undefined.

f '(x) gt 0
y
f '(x) 0
f '(x) lt 0
x
c
b
a
21
Finding Relative Extrema

• Suppose that f has a relative minimum at c.
• The slope of the tangent line to the graph must
  change from negative to positive as x increases.
• Therefore, the tangent line to the graph of f at
  point (c, f(c)) must be horizontal, so that f
  '(x) 0 or f '(x) is undefined.

y
f '(x) gt 0
f '(x) lt 0
f '(x) 0
x
c
b
a
22
Finding Relative Extrema

• In some cases a derivative does not exist for
  particular values of x.
• Extrema may exist at such points, as the graph
  below shows

Relative Maximum
y
Relative Minimum
x
b
a
23
Critical Numbers

• We refer to a number in the domain of f that may
  give rise to a relative extremum as a critical
  number.
• Critical number of f
• A critical number of a function f is any number
  x in the domain of f such that f '(x)
  0 or f '(x) does not exist.

24
Critical Numbers

• The graph below shows us several critical
  numbers.
• At points a, b, and c, f '(x) 0.
• There is a corner at point d, so f '(x) does not
  exist there.
• The tangent to the curve at point e is vertical,
  so f '(x) does not exist there either.
• Note that points a, b, and d are relative
  extrema, while points c and e are not.

y
Relative Extrema
Not Relative Extrema
x
a
b
c
d
e
25
The First Derivative Test

• Procedure for Finding Relative Extrema of a
  Continuous Function f

• Determine the critical numbers of f.
• Determine the sign of f '(x) to the left and
  right of each critical point.
• If f '(x) changes sign from positive to negative
  as we move across a critical number c, then f(c)
  is a relative maximum.
• If f '(x) changes sign from negative to positive
  as we move across a critical number c, then f(c)
  is a relative minimum.
• If f '(x) does not change sign as we move across
  a critical number c, then f(c) is not a relative
  extremum.

26
Examples

• Find the relative maxima and minima of

• Solution
• The derivative of f is f '(x) 2x.
• Setting f '(x) 0 yields x 0 as the only
  critical number of f.
• Since f '(x) lt 0 if x lt 0
• and f '(x) gt 0 if x gt 0
• we see that f '(x) changes sign from negative to
  positive as we move across 0.
• Thus, f(0) 0 is a relative minimum of f.

y
f(x) x2
f '(x) gt 0
f '(x) lt 0
Relative Minimum
Example 5, page 252
27
Examples

• Find the relative maxima and minima of
• Solution
• The derivative of f is f '(x) 2/3x1/3.
• f '(x) is not defined at x 0, is continuous
  everywhere else, and is never equal to zero in
  its domain.
• Thus x 0 is the only critical number of f.

• Since f '(x) lt 0 if x lt 0
• and f '(x) gt 0 if x gt 0
• we see that f '(x) changes sign from negative to
  positive as we move across 0.
• Thus, f(0) 0 is a relative minimum of f.

y
4 2
f(x) x2/3
f(x) gt 0
f(x) lt 0
x
4 2 2 4
Relative Minimum
Example 6, page 252
28
Examples

• Find the relative maxima and minima of
• Solution
• The derivative of f and equate to zero
• The zeros of f '(x) are x 2 and x 4.
• f '(x) is defined everywhere, so x 2 and x 4
  are the only critical numbers of f.

Example 7, page 253
29
Examples

• Find the relative maxima and minima of
• Solution
• Since f '(x) gt 0 if x lt 2 and f '(x) lt 0
  if 0 lt x lt 4,
• we see that f '(x) changes sign from positive to
  negative as we move across 2.
• Thus, f(2) 60 is a
• relative maximum.

Relative Maximum
y
60 40 20 20 40 60
f(x)
x
7 5 3 1 1 3 5 7
Example 7, page 253
30
Examples

• Find the relative maxima and minima of
• Solution
• Since f '(x) lt 0 if 0 lt x lt 4 and f '(x) gt
  0 if, x gt 4
• we see that f '(x) changes sign from negative to
  positive as we move across 4.
• Thus, f(4) 48 is a
• relative minimum.

Relative Maximum
y
60 40 20 20 40 60
f(x)
x
7 5 3 1 1 3 5 7
Relative Minimum
Example 7, page 253
31
4.2

• Applications of the Second Derivative

32
Concavity

• A curve is said to be concave upwards when the
  slope of tangent line to the curve is increasing
• Thus, if f is differentiable on an interval (a,
  b), then
• f is concave upwards on (a, b) if f ' is
  increasing on (a, b).

y
x
33
Concavity

• A curve is said to be concave downwards when the
  slope of tangent line to the curve is decreasing
• Thus, if f is differentiable on an interval (a,
  b), then
• f is concave downwards on (a, b) if f ' is
  decreasing on (a, b).

y
x
34
Theorem 2

• Recall that f ?(x) measures the rate of change of
  the slope f '(x) of the tangent
  line to the graph of f at the point (x, f(x)).
• Thus, we can use f ?(x) to determine the
  concavity of f.

1. If f ?(x) gt 0 for each value of x in (a, b), then
   f is concave upward on (a, b).
2. If f ?(x) lt 0 for each value of x in (a, b), then
   f is concave downward on (a, b).

35
Steps in Determining the Concavity of f

• Determine the values of x for which f ?is zero or
  where f ? is not defined, and identify the
  open intervals determined by these numbers.
• Determine the sign of f ? in each interval found
  in step 1.
• To do this compute f ?(c), where c is any
  conveniently chosen test number in the interval.
• If f ?(c) gt 0, f is concave upward on that
  interval.
• If f ?(c) lt 0, f is concave downward on that
  interval.

36
Examples

• Determine where the function
• is concave upward and where it is concave
  downward.
• Solution
• Here, and
• Setting f ?(c) 0 we find
• which gives x 1.
• So we consider the intervals ( ?, 1) and (1, ?)
• f ?(x) lt 0 when x lt 1, so f is concave downward
  on ( ?, 1).
• f ?(x) gt 0 when x gt 1, so f is concave upward on
  (1, ?).

Example 1, page 265
37
Examples

• Determine where the function
• is concave upward and where it is concave
  downward.
• Solution
• The graph confirms that f is concave downward on
  ( ?, 1)
• and concave upward on (1, ?)

y
60 40 20 20 40 60
f(x)
x
7 5 3 1 1 3 5 7
Example 1, page 265
38
Examples

• Determine the intervals where the function
  is concave upward and
  concave downward.
• Solution
• Here, and
• So, f ?cannot be zero and is not defined at x
  0.
• So we consider the intervals ( ?, 0) and (0, ?)
• f ?(x) lt 0 when x lt 0, so f is concave downward
  on ( ?, 0).
• f ?(x) gt 0 when x gt 0, so f is concave upward on
  (0, ?).

Example 2, page 266
39
Examples

• Determine the intervals where the function
• is concave upward and concave downward.
• Solution
• The graph confirms that f is concave downward on
  ( ?, 0)
• and concave upward on (0, ?)

Example 2, page 266
40
Inflection Point

• A point on the graph of a continuous function
  where the tangent line exists and where the
  concavity changes is called an inflection point.
• Examples

y
Inflection Point
x
41
Inflection Point

• A point on the graph of a continuous function
  where the tangent line exists and where the
  concavity changes is called an inflection point.
• Examples

y
Inflection Point
x
42
Inflection Point

• A point on the graph of a continuous function
  where the tangent line exists and where the
  concavity changes is called an inflection point.
• Examples

y
Inflection Point
x
43
Finding Inflection Points

• To find inflection points
• Compute f ?(x).
• Determine the numbers in the domain of f for
  which f ?(x) 0 or f ?(x) does not exist.
• Determine the sign of f ?(x) to the left and
  right of each number c found in step 2.
• If there is a change in the sign of f ?(x) as we
  move across x c, then (c, f(c)) is an
  inflection point of f.

44
Examples

• Find the points of inflection of the function
• Solution
• We have and
• So, f ? is continuous everywhere and is zero for
  x 0.
• We see that f ?(x) lt 0 when x lt 0, and f ?(x) gt 0
  when x gt 0.
• Thus, we find that the graph of f

• Has one and only inflection point at f(0) 0.
• Is concave downward on the interval ( ?, 0).
• Is concave upward on the interval (0, ?).

y
f(x)
4 2 2 4
Inflection Point
x
2 2
Example 3, page 268
45
Examples

• Find the points of inflection of the function
• Solution
• We have and
• So, f ? is not defined at x 1, and f ? is never
  equal to zero.
• We see that f ?(x) lt 0 when x lt 1, and f ?(x) gt 0
  when x gt 1.
• Thus, we find that the graph of f

y

• Has one and only inflection point at f(1) 0.
• Is concave downward on the interval ( ?, 1).
• Is concave upward on the interval (1, ?).

3 2 1 1 2 3
f(x)
Inflection Point
x
2 1 1 2
Example 4, page 268
46
Applied Example Effect of Advertising on Sales

• The total sales S (in thousands of dollars) of
  Arctic Air Co., which makes automobile air
  conditioners, is related to the amount of money x
  (in thousands of dollars) the company spends on
  advertising its product by the formula
• Find the inflection point of the function S.
• Discuss the meaning of this point.

Applied Example 7, page 270
47
Applied Example Effect of Advertising on Sales

• Solution
• The first two derivatives of S are given by
• Setting S ?(x) 0 gives x 50.
• So (50, S(50)) is the only candidate for an
  inflection point.
• Since S ?(x) gt 0 for x lt 50, and S ?(x) lt 0 for x
  gt 50, the point (50, 2700) is, in fact, an
  inflection point of S.
• This means that the firm experiences diminishing
  returns on advertising beyond 50,000
• Every additional dollar spent on advertising
  increases sales by less than previously spent
  dollars.

Applied Example 7, page 270
48
The Second Derivative Test

• Maxima occur when a curve is concave downwards,
  while minima occur when a curve is concave
  upwards.
• This is the basis of the second derivative test
• Compute f '(x) and f ?(x).
• Find all the critical numbers of f at which f
  '(x) 0.
• Compute f ?(c), if it exists, for each critical
  number c.
• If f ?(c) lt 0, then f has a relative maximum at
  c.
• If f ?(c) gt 0, then f has a relative minimum at
  c.
• If f ?(c) 0, then the test fails (it is
  inconclusive).

49
Example

• Determine the relative extrema of the function
• Solution
• We have
• so f '(x) 0 gives the critical numbers x 2
  and x 4.
• Next, we have
• which we use to test the critical numbers
• so, f(2) 60 is a relative maximum of f.
• so, f(4) 48 is a relative minimum of f.

Example 9, page 272
50
4.3

• Curve Sketching

51
Vertical Asymptotes

• The line x a is a vertical asymptote of the
  graph of a function f if either
• or

52
Finding Vertical Asymptotes of Rational Functions

• Suppose f is a rational function
• where P and Q are polynomial functions.
• Then, the line x a is a vertical asymptote of
  the graph of f if Q(a) 0 but P(a) ? 0.

53
Example

• Find the vertical asymptotes of the graph of the
  function
• Solution
• f is a rational function with P(x) x2 and
  Q(x) 4 x2.
• The zeros of Q are found by solving
• giving x 2 and x 2.
• Examine x 2
• P(2) (2)2 4 ? 0, so x 2 is a vertical
  asymptote.
• Examine x 2
• P(2) (2)2 4 ? 0, so x 2 is also a
  vertical asymptote.

Example 1, page 285
54
Horizontal Asymptotes

• The line y b is a horizontal asymptote of the
  graph of a function f if either

55
Example

• Find the horizontal asymptotes of the graph of
  the function
• Solution
• We compute
• and so y 1 is a horizontal asymptote.
• We compute
• also yielding y 1 as a horizontal asymptote.

Example 2, page 287
56
Example

• Find the horizontal asymptotes of the graph of
  the function
• Solution
• So, the graph of f has two vertical asymptotes x
  2 and x 2, and one
  horizontal asymptote y 1

y
x 2
x 2
x
y 1
f(x)
Example 2, page 287
57
Asymptotes and Polynomials

• A polynomial function has no vertical asymptotes.
• To see this, note that a polynomial function P(x)
  can be written as a rational function with a
  denominator equal to 1.
• Thus,
• Since the denominator is never zero, P has no
  vertical asymptotes.

58
Asymptotes and Polynomials

• A polynomial function has no horizontal
  asymptotes.
• If P(x) is a polynomial of degree greater or
  equal to 1, then
• are either infinity or minus infinity that is,
  they do not exist.
• Therefore, P has no horizontal asymptotes.

59
A Guide to Sketching a Curve

1. Determine the domain of f.
2. Find the x- and y-intercepts of f.
3. Determine the behavior of f for large absolute
   values of x.
4. Find all horizontal and vertical asymptotes of f.
5. Determine the intervals where f is increasing and
   where f is decreasing.
6. Find the relative extrema of f.
7. Determine the concavity of f.
8. Find the inflection points of f.
9. Plot a few additional points to help further
   identify the shape of the graph of f and sketch
   the graph.

60
Examples

• Sketch the graph of the function
• Solution
• The domain of f is the interval ( ?, ?).
• By setting x 0, we find that the y-intercept is
  2.
• (The x-intercept is not readily obtainable)
• Since
• we see that f decreases without bound as x
  decreases without bound and that f increases
  without bound when x increases without
  bound.
• Since f is a polynomial function, there are no
  asymptotes.

Example 3, page 288
61
Examples

• Sketch the graph of the function
• Solution
• Setting f '(x) 0 gives x 1 and x 3 as
  critical points.
• Testing with different values of x we find
  that f '(x) gt 0 when x lt 1, f '(x) lt 0 when
  1 lt x lt 3, and f '(x) gt 0 when x gt 3.
• Thus, f is increasing in the intervals ( ?, 1)
  and (3, ?),
• and f is decreasing in the interval (1, 3).

Example 3, page 288
62
Examples

• Sketch the graph of the function
• Solution
• 6. f ' changes from positive to negative as we
  go across x 1, so a relative maximum of f
  occurs at (1, f(1)) (1, 6).
• f ' changes from negative to positive as we go
  across x 3, so a relative minimum of f
  occurs at (3, f(3)) (1, 2).

Example 3, page 288
63
Examples

• Sketch the graph of the function
• Solution
• which is equal to zero when x 2.
• Testing with different values of x we find that
  f ?(x) lt 0 when x lt 2 and f ?(x) gt 0 when 2
  lt x.
• Thus, f is concave downward in the interval (
  ?, 2) and concave upward in the interval (2, ?).
• Since f ?(2) 0, we have an inflection point at
  (2, f(2)) (2, 4).

Example 3, page 288
64
Examples

• Sketch the graph of the function
• Solution
• Summarizing, weve found the following
• Domain ( ?, ?).
• Intercept (0, 2).
• Asymptotes None.
• f is increasing in the intervals ( ?, 1) and (3,
  ?), and f is decreasing in
  the interval (1, 3).
• A relative maximum of f occurs at (1, 6).
• A relative minimum of f occurs at (1, 2).
• f is concave downward in the interval ( ?, 2)
  and f is concave upward in
  the interval (2, ?).
• f has an inflection point at (2, 4).

Example 3, page 288
65
Examples

• Sketch the graph of the function
• Solution
• Sketch the graph

y
7 6 5 4 3 2 1
(1, 6)
Relative maximum
(2, 4)
Inflection point
(3, 2)
Relative minimum
Intercept
x
1 2 3 4 5 6
Example 3, page 288
66
Examples

• Sketch the graph of the function
• Solution
• Sketch the graph

y
7 6 5 4 3 2 1
x
1 2 3 4 5 6
Example 3, page 288
67
Examples

• Sketch the graph of the function
• Solution
• 1. f is undefined when x 1, so the domain of f
  is the set of all real numbers other than x 1.
• Setting y 0, gives an x-intercept of 1.
• Setting x 0, gives an y-intercept of 1.

Example 4, page 290
68
Examples

• Sketch the graph of the function
• Solution
• Since
• we see that f(x) approaches the line y 1 as
  x becomes arbitrarily large.
• For x gt 1, f(x) gt 1, so f approaches the line y
  1 from above.
• For x lt 1, f(x) lt 1, so f approaches the line y
  1 from below.
• From step three we conclude that y 1 is a
  horizontal asymptote of f.
• Also, the straight line x 1 is a vertical
  asymptote of f.

Example 4, page 290
69
Examples

• Sketch the graph of the function
• Solution
• So, f '(x) is discontinuous at x 1 and is
  never equal to zero.
• Testing we find that f '(x) lt 0 wherever it is
  defined.
• From step 5 we see that there are no critical
  numbers of f, since f '(x) is never equal to
  zero.

Example 4, page 290
70
Examples

• Sketch the graph of the function
• Solution
• Testing with different values of x we find that
  f ?(x) lt 0 when x lt 1 and f ?(x) gt 0 when 1
  lt x.
• Thus, f is concave downward in the interval (
  ?, 1) and concave upward in the interval (1, ?).
• From point 7 we see there are no inflection
  points of f, since f ?(x) is never equal
  to zero.

Example 4, page 290
71
Examples

• Sketch the graph of the function
• Solution
• Summarizing, weve found the following
• Domain ( ?, 1) U (1, ?).
• Intercept (0, 1) (1, 0).
• Asymptotes x 1 is a vertical asymptote.
• y 1 is a horizontal asymptote.
• f is decreasing everywhere in the domain of f.
• Relative extrema None.
• f is concave downward in the interval ( ?, 1)
  and f is concave upward in
  the interval (1, ?).
• f has no inflection points.

Example 4, page 290
72
Examples

• Sketch the graph of the function
• Solution
• Sketch the graph

y
4 2 2
y 1
x
2 2 4
x 1
Example 4, page 290
73
4.4

• Optimization I

74
Absolute Extrema

• The absolute extrema of a function f
• If f(x) ? f(c) for all x in the domain of f, then
  f(c) is called the absolute maximum value of f.
• If f(x) ? f(c) for all x in the domain of f, then
  f(c) is called the absolute minimum value of f.

75
Examples

• f has an absolute minimum at (0, 0)

y
4 2
x
2 1 1 2
Absolute minimum
76
Examples

• f has an absolute maximum at (0, 4)

y
Absolute maximum
4 2
x
2 1 1 2
77
Examples

• f has an absolute minimum at
• and an absolute maximum at

y
1 1/2 1/2 1
Absolute maximum
x
1 1
Absolute minimum
78
Examples

• f has no absolute extrema

y
7 6 5 4 3 2 1
x
1 2 3 4 5 6
79
Theorem 3

• Absolute Extrema in a Closed Interval
• If a function f is continuous on a closed
  interval a, b, then f has both an absolute
  maximum value and an absolute minimum value on
  a, b.

80
Example

• The relative minimum of f at x3 is also the
  absolute minimum of f.
• The right endpoint b of the interval a, b gives
  rise to the absolute maximum value f(b) of f.

y
Relative maximum
Absolute maximum
Relative maximum
Absolute minimum
Relative minimum
x
a
x1
x2
x3
b
81
Finding Absolute Extrema

• To find the absolute extrema of f on a closed
  interval a, b.
• Find the critical numbers of f that lie on (a,
  b).
• Compute the value of f at each critical number
  found in step 1 and compute f(a) and f(b).
• The absolute maximum value and absolute minimum
  value of f will correspond to the largest and
  smallest numbers, respectively, found in step 2.

82
Examples

• Find the absolute extrema of the function F(x)
  x2 defined on the interval 1, 2.
• Solution
• The function F is continuous on the closed
  interval 1, 2 and differentiable on the open
  interval (1, 2).
• Setting F ' 0, we get F '(x) 2x 0, so there
  is only one critical point at x 0.
• So, F(1) (1)2 1, F(0) (0)2 0, and
  F(2) (2)2 4.
• It follows that 0 is the absolute minimum of F,
  and 4 is the absolute maximum of F.

Example 1, page 300
83
Examples

• Find the absolute extrema of the function F(x)
  x2 defined on the interval 1, 2.
• Solution

y
4 3 2 1
Absolute maximum
x
2 1 1 2
Absolute minimum
Example 1, page 300
84
Examples

• Find the absolute extrema of the function
• defined on the interval 0, 3.
• Solution
• The function f is continuous on the closed
  interval 0, 3 and differentiable on the open
  interval (0, 3).
• Setting f ' 0, we get
• which gives two critical points at x 2/3 and
  x 2.
• We drop x 2/3 since it lies outside the
  interval 0, 3.
• So, f(0) 4, f(2) 4, and f(3) 1.
• It follows that 4 is the absolute minimum of f,
  and 4 is the absolute maximum of f.

Example 2, page 300
85
Examples

• Find the absolute extrema of the function
• defined on the interval 0, 3.
• Solution

y
4 2 2 4
Absolute maximum
x
2 2 4
Absolute minimum
Example 2, page 300
86
Applied Example Maximizing Profits

• Acrosonics total profit (in dollars) from
  manufacturing and selling x units of their model
  F speakers is given by
• How many units of the loudspeaker system must
  Acrosonic produce to maximize profits?
• Solution
• To find the absolute maximum of P on 0, 20,000,
  first find the stationary points of P on the
  interval (0, 20,000).
• Setting f ' 0, we get
• which gives us only one stationary point at x
  7500.

Applied Example 4, page 301
87
Applied Example Maximizing Profits

• Acrosonics total profit (in dollars) from
  manufacturing and selling x units of their model
  F speakers is given by
• How many units of the loudspeaker system must
  Acrosonic produce to maximize profits?
• Solution
• Evaluating the only stationary point we get
• P(7500) 925,000
• Evaluating the endpoints we get
• P(0) 200,000
• P(20,000) 2,200,000
• Thus, Acrosonic will realize the maximum profit
  of 925,000 by producing 7500 speakers.

Applied Example 4, page 301
88
Applied Example Maximizing Profits

• Acrosonics total profit (in dollars) from
  manufacturing and selling x units of their model
  F speakers is given by
• How many units of the loudspeaker system must
  Acrosonic produce to maximize profits?
• Solution

y
1000 800 600 400 200 200
Maximum Profit
(Thousands of dollars)
x
2 4 6 8 10 12 14 16
(Thousands of speakers)
Applied Example 4, page 301
89
4.5

• Optimization II

90
Guidelines for Solving Optimization Problems

• Assign a letter to each variable mentioned in the
  problem.
• If appropriate, draw and label a figure.
• Find an expression for the quantity to be
  optimized.
• Use the conditions given in the problem to write
  the quantity to be optimized as a function f of
  one variable.
• Note any restrictions to be placed on the domain
  of f from physical considerations of the problem.
• Optimize the function f over its domain using the
  methods of Section 4.4.

91
Applied Maximization Problem Packaging

• By cutting away identical squares from each
  corner of a rectangular piece of cardboard and
  folding up the resulting flaps, the cardboard may
  be turned into an open box.
• If the cardboard is 16 inches long and 10 inches
  wide, find the dimensions of the box that will
  yield the maximum volume.

10
16
Applied Example 2, page 314
92
Applied Maximization Problem Packaging

• Solution
• Let x denote the length in inches of one side of
  each of the identical squares to be cut out of
  the cardboard.
• The dimensions of the box are (16 2x) by (10
  2x) by x in.

x
10
10 2x
x
16 2x
x
x
16
Applied Example 2, page 314
93
Applied Maximization Problem Packaging

• Solution
• Let V denote the volume (in cubic inches) of the
  resulting box. The volume,
• is the quantity to be maximized.

x
10 2x
16 2x
Applied Example 2, page 314
94
Applied Maximization Problem Packaging

• Solution
• Each side of the box must be nonnegative, so x
  must satisfy the inequalities x ? 0, 16 2x ? 0,
  and 10 2x ? 0.
• All these inequalities are satisfied if 0 ? x ?
  5.
• Therefore, the problem at hand is equivalent to
  finding the absolute maximum of
• on the closed interval 0, 5.

Applied Example 2, page 314
95
Applied Maximization Problem Packaging

• Solution
• f is continuous on 0, 5. Setting f '(x) 0 we
  get
• which yields the critical numbers x 20/3 and x
  2.
• We discard x 20/3 for being outside the
  interval 0, 5.
• We evaluate f at the critical point and at the
  endpoints
• f(0) 0 f(2) 144 f(5) 0
• Thus, the volume of the box is maximized by
  taking x 2.
• The resulting dimensions of the box are 12? ? 6?
  ? 2?.

Applied Example 2, page 314
96
Applied Minimization Problem Inventory Control

• Dixies Import-Export is the sole seller of the
  Excalibur 250 cc motorcycle.
• Management estimates that the demand for these
  motorcycles will be 10,000 for the coming year
  and that they will sell at a uniform rate
  throughout the year.
• The cost incurred in ordering each shipment of
  motorcycles is 10,000, and the cost per year of
  storing each motorcycle is 200.
• Dixies management faces the following problem
• Ordering too many motorcycles at one time
  increases storage cost.
• On the other hand, ordering too frequently
  increases the ordering costs.
• How large should each order be, and how often
  should orders be placed, to minimize ordering and
  storage costs?

Applied Example 5, page 317
97
Applied Minimization Problem Inventory Control

• Solution
• Let x denote the number of motorcycles in each
  order.
• Assuming each shipment arrives just as the
  previous shipment is sold out, the average number
  of motorcycles in storage during the year is x/2,
  as shown below
• Thus, Dixies storage cost for the year is given
  by 200(x/2), or 100x dollars.

x
Average inventory
Inventory Level
Time
Applied Example 5, page 317
98
Applied Minimization Problem Inventory Control

• Solution
• Next, since the company requires 10,000
  motorcycles for the year and since each order is
  for x motorcycles, the number of orders required
  is
• This gives an ordering cost of
• dollars for the year.
• Thus, the total yearly cost incurred by Dixies,
  including ordering and storage costs, is given by

Applied Example 5, page 317
99
Applied Minimization Problem Inventory Control

• Solution
• The problem is reduced to finding the absolute
  minimum of the function C in the interval
  (0,10,000.
• To accomplish this, we compute
• Setting C '(x) 0 and solving we obtain x
  1000.
• We reject the negative for being outside the
  domain.
• So we have x 1000 as the only critical number.

Applied Example 5, page 317
100
Applied Minimization Problem Inventory Control

• Solution
• So we have x 1000 as the only critical number.
• Now we find
• Since C ?(1000) gt 0, the second derivative test
  implies that x 1000 is a relative
  minimum of C.
• Also, since C ?(1000) gt 0 for all x in
  (0,10,000, the function C is concave upward
  everywhere so that also gives the absolute
  minimum of C.
• Thus, to minimize the ordering and storage costs,
  Dixies should place 10,000/1000, or 10, orders
  per year, each for a shipment of 1000 motorcycles.

Applied Example 5, page 317
101
End of Chapter