# Finding Maxima and Minima using Derivatives Contributed by: In this section, we will learn: How to find the maximum and minimum values of a function using derivatives.
1. 4
APPLICATIONS OF DIFFERENTIATION
2. APPLICATIONS OF DIFFERENTIATION
applications of derivatives.
However, now that we know the differentiation
rules, we are in a better position to pursue
the applications of differentiation in greater
3. APPLICATIONS OF DIFFERENTIATION
Here, we learn how derivatives affect
the shape of a graph of a function and,
in particular, how they help us locate
maximum and minimum values of functions.
4. APPLICATIONS OF DIFFERENTIATION
Many practical problems require us to
minimize a cost or maximize an area or
somehow find the best possible outcome
of a situation.
 In particular, we will be able to investigate
the optimal shape of a can and to explain
the location of rainbows in the sky.
5. APPLICATIONS OF DIFFERENTIATION
4.1
Maximum and
Minimum Values
In this section, we will learn:
How to find the maximum
and minimum values of a function.
6. OPTIMIZATION PROBLEMS
Some of the most important
applications of differential calculus
are optimization problems.
 In these, we are required to find the optimal (best)
way of doing something.
7. Here are some examples of such problems
that we will solve in this chapter.
 What is the shape of a can that minimizes
manufacturing costs?
 What is the maximum acceleration of a space shuttle?
(This is an important question to the astronauts
who have to withstand the effects of acceleration.)
8. Here are some more examples.
 What is the radius of a contracted windpipe
that expels air most rapidly during a cough?
 At what angle should blood vessels branch
so as to minimize the energy expended by
the heart in pumping blood?
9. OPTIMIZATION PROBLEMS
These problems can be reduced to
finding the maximum or minimum values
of a function.
 Let’s first explain exactly what we mean
by maximum and minimum values.
10. MAXIMUM & MINIMUM VALUES Definition 1
A function f has an absolute maximum
(or global maximum) at c if f(c) ≥ f(x) for
all x in D, where D is the domain of f.
The number f(c) is called the maximum value
of f on D.
11. MAXIMUM & MINIMUM VALUES Definition 1
Similarly, f has an absolute minimum at c
if f(c) ≤ f(x) for all x in D and the number f(c)
is called the minimum value of f on D.
The maximum and minimum values of f
are called the extreme values of f.
12. MAXIMUM & MINIMUM VALUES
The figure shows the graph of a function f
with absolute maximum at d and absolute
minimum at a.
 Note that (d, f(d)) is
the highest point on
the graph and (a, f(a))
is the lowest point.
13. LOCAL MAXIMUM VALUE
If we consider only values of x near b—for
instance, if we restrict our attention to the
interval (a, c)—then f(b) is the largest of those
values of f(x).
 It is called a local
maximum value of f.
14. LOCAL MINIMUM VALUE
Likewise, f(c) is called a local minimum value
of f because f(c) ≤ f(x) for x near c—for
instance, in the interval (b, d).
 The function f also has
a local minimum at e.
15. MAXIMUM & MINIMUM VALUES Definition 2
In general, we have the following definition.
A function f has a local maximum (or relative
maximum) at c if f(c) ≥ f(x) when x is near c.
 This means that f(c) ≥ f(x) for all x in some
open interval containing c.
 Similarly, f has a local minimum at c if f(c) ≤ f(x)
when x is near c.
16. MAXIMUM & MINIMUM VALUES Example 1
The function f(x) = cos x takes on its (local
and absolute) maximum value of 1 infinitely
many times—since cos 2nπ = 1 for any
integer n and -1 ≤ cos x ≤ 1 for all x.
Likewise, cos (2n + 1)π = -1 is its minimum
value—where n is any integer.
17. MAXIMUM & MINIMUM VALUES Example 2
If f(x) = x2, then f(x) ≥ f(0) because
x2 ≥ 0 for all x.
 Therefore, f(0) = 0 is the absolute (and local)
minimum value of f.
18. MAXIMUM & MINIMUM VALUES Example 2
This corresponds to the fact that the origin
is the lowest point on the parabola y = x2.
 However, there is no highest point on the parabola.
 So, this function has no maximum value.
19. MAXIMUM & MINIMUM VALUES Example 3
From the graph of the function f(x) = x3,
we see that this function has neither
an absolute maximum value nor an absolute
minimum value.
 In fact, it has no local
extreme values either.
20. MAXIMUM & MINIMUM VALUES Example 4
The graph of the function
f(x) = 3x4 – 16x3 + 18x2 -1 ≤ x ≤ 4
is shown here.
21. MAXIMUM & MINIMUM VALUES Example 4
You can see that f(1) = 5 is a local
maximum, whereas the absolute maximum
is f(-1) = 37.
 This absolute maximum
is not a local maximum
because it occurs at
an endpoint.
22. MAXIMUM & MINIMUM VALUES Example 4
Also, f(0) = 0 is a local minimum and
f(3) = -27 is both a local and an absolute
 Note that f has
neither a local nor
an absolute maximum
at x = 4.
23. MAXIMUM & MINIMUM VALUES
We have seen that some functions have
extreme values, whereas others do not.
The following theorem gives conditions
under which a function is guaranteed to
possess extreme values.
24. EXTREME VALUE THEOREM Theorem 3
If f is continuous on a closed interval [a, b],
then f attains an absolute maximum value f(c)
and an absolute minimum value f(d) at some
numbers c and d in [a, b].
25. EXTREME VALUE THEOREM
The theorem is illustrated
in the figures.
 Note that an extreme value
can be taken on more than once.
26. EXTREME VALUE THEOREM
Although the theorem is intuitively
very plausible, it is difficult to prove
and so we omit the proof.
27. EXTREME VALUE THEOREM
The figures show that a function need not
possess extreme values if either hypothesis
(continuity or closed interval) is omitted from
the theorem.
28. EXTREME VALUE THEOREM
The function f whose graph is shown is
defined on the closed interval [0, 2] but has
no maximum value.
 Notice that the range of f
is [0, 3).
 The function takes on values
arbitrarily close to 3, but never
actually attains the value 3.
29. EXTREME VALUE THEOREM
This does not contradict the theorem
because f is not continuous.
 Nonetheless, a discontinuous
function could have maximum
and minimum values.
30. EXTREME VALUE THEOREM
The function g shown here is continuous
on the open interval (0, 2) but has neither
a maximum nor a minimum value.
 The range of g is (1, ∞).
 The function takes on
arbitrarily large values.
the theorem because
the interval (0, 2) is not closed.
31. EXTREME VALUE THEOREM
The theorem says that a continuous function
on a closed interval has a maximum value
and a minimum value.
However, it does not tell us how to find these
extreme values.
 We start by looking for local extreme values.
32. LOCAL EXTREME VALUES
The figure shows the graph of a function f
with a local maximum at c and a local
minimum at d.
33. LOCAL EXTREME VALUES
It appears that, at the maximum and
minimum points, the tangent lines are
horizontal and therefore each has slope 0.
34. LOCAL EXTREME VALUES
We know that the derivative is the slope
of the tangent line.
 So, it appears that f ’(c) = 0 and f ’(d) = 0.
35. LOCAL EXTREME VALUES
The following theorem says that
this is always true for differentiable
36. FERMAT’S THEOREM Theorem 4
If f has a local maximum or
minimum at c, and if f ’(c) exists, then
f ’(c) = 0.
37. FERMAT’S THEOREM Proof
Suppose, for the sake of definiteness, that f
has a local maximum at c.
Then, according to Definition 2, f(c) ≥ f(x)
if x is sufficiently close to c.
38. FERMAT’S THEOREM Proof (Equation 5)
This implies that, if h is sufficiently close to 0,
with h being positive or negative, then
f(c) ≥ f(c + h)
and therefore
f(c + h) – f(c) ≤ 0
39. FERMAT’S THEOREM Proof
We can divide both sides of an inequality
by a positive number.
 Thus, if h > 0 and h is sufficiently small,
we have:
f (c  h )  f (c )
0
h
40. FERMAT’S THEOREM Proof
Taking the right-hand limit of both sides
of this inequality (using Theorem 2 in
Section 2.3), we get:
f (c  h )  f ( c )
lim  lim 0 0
h 0 h h 0
41. FERMAT’S THEOREM Proof
However, since f ’(c) exists,
we have: f ( c  h )  f (c )
f '(c) lim
h 0 h
f (c  h )  f (c )
 lim
h 0 h
 So, we have shown that f ’(c) ≤ 0.
42. FERMAT’S THEOREM Proof
If h < 0, then the direction of the inequality
in Equation 5 is reversed when we divide by
f (c  h )  f (c )
0 h0
h
43. FERMAT’S THEOREM Proof
So, taking the left-hand limit,
we have: f ( c  h )  f (c )
f '(c) lim
h 0 h
f (c  h )  f (c )
 lim 0
h 0 h
44. FERMAT’S THEOREM Proof
We have shown that f ’(c) ≥ 0 and also
that f ’(c) ≤ 0.
Since both these inequalities must be true,
the only possibility is that f ’(c) = 0.
45. FERMAT’S THEOREM Proof
We have proved the theorem for
the case of a local maximum.
 The case of a local minimum can be proved
in a similar manner.
 Alternatively, we could use Exercise 76
to deduce it from the case we have just proved.
46. FERMAT’S THEOREM
The following examples caution us
against reading too much into the theorem.
 We can’t expect to locate extreme values
simply by setting f ’(x) = 0 and solving for x.
47. FERMAT’S THEOREM Example 5
If f(x) = x3, then f ’(x) = 3x2, so f ’(0) = 0.
 However, f has no maximum or minimum at 0—as you
can see from the graph.
 Alternatively, observe
that x3 > 0 for x > 0
but x3 < 0 for x < 0.
48. FERMAT’S THEOREM Example 5
The fact that f ’(0) = 0 simply means that
the curve y = x3 has a horizontal tangent
at (0, 0).
a maximum or minimum
at (0, 0), the curve crosses
its horizontal tangent there.
49. FERMAT’S THEOREM Example 6
The function f(x) = |x| has its (local and
absolute) minimum value at 0.
 However, that value can’t be found by setting f ’(x) = 0.
 This is because—as
shown in Example 5
in Section 2.8—f ’(0)
does not exist.
50. Examples 5 and 6 show that we must
be careful when using the theorem.
 Example 5 demonstrates that, even when f ’(c) = 0,
there need not be a maximum or minimum at c.
 In other words, the converse of the theorem
is false in general.
 Furthermore, there may be an extreme value
even when f ’(c) does not exist (as in Example 6).
51. FERMAT’S THEOREM
The theorem does suggest that we should
at least start looking for extreme values of f
at the numbers c where either:
 f ’(c) = 0
 f ’(c) does not exist
52. FERMAT’S THEOREM
Such numbers are given a
special name—critical numbers.
53. CRITICAL NUMBERS Definition 6
A critical number of a function f is
a number c in the domain of f such that
either f ’(c) = 0 or f ’(c) does not exist.
54. CRITICAL NUMBERS Example 7
Find the critical numbers of
f(x) = x3/5(4 - x).
 The Product Rule gives:
3/ 5  2/5
f '( x)  x ( 1)  (4  x)( x
3
5 )
3/ 5 3(4  x)
 x  2/5
5x
 5 x  3(4  x) 12  8 x
 2/5
 2/5
5x 5x
55. CRITICAL NUMBERS Example 7
 The same result could be obtained by
first writing f(x) = 4x3/5 – x8/5.
 Therefore, f ’(x) = 0 if 12 – 8x = 0.
 That is, x = 32 , and f ’(x) does not exist
when x = 0.
 Thus, the critical numbers are 3
2
and 0.
56. CRITICAL NUMBERS
In terms of critical numbers, Fermat’s
Theorem can be rephrased as follows
(compare Definition 6 with Theorem 4).
57. CRITICAL NUMBERS Theorem 7
If f has a local maximum or
minimum at c, then c is a critical
number of f.
58. CLOSED INTERVALS
To find an absolute maximum or minimum
of a continuous function on a closed interval,
we note that either:
 It is local (in which case, it occurs at
a critical number by Theorem 7).
 It occurs at an endpoint of the interval.
59. CLOSED INTERVALS
Therefore, the following
three-step procedure always
60. CLOSED INTERVAL METHOD
To find the absolute maximum and minimum
values of a continuous function f on a closed
interval [a, b]:
1. Find the values of f at the critical numbers of f
in (a, b).
2. Find the values of f at the endpoints of the interval.
3. The largest value from 1 and 2 is the absolute
maximum value. The smallest is the absolute minimum
value.
61. CLOSED INTERVAL METHOD Example 8
Find the absolute maximum
and minimum values of the function
f(x) = x3 – 3x2 + 1 -½ ≤ x ≤ 4
62. CLOSED INTERVAL METHOD Example 8
As f is continuous on [-½, 4], we
can use the Closed Interval Method:
f(x) = x3 – 3x2 + 1
f ’(x) = 3x2 – 6x = 3x(x – 2)
63. CLOSED INTERVAL METHOD Example 8
As f ’(x) exists for all x, the only critical
numbers of f occur when f ’(x) = 0, that is,
x = 0 or x = 2.
Notice that each of these numbers lies in
the interval (-½, 4).
64. CLOSED INTERVAL METHOD Example 8
The values of f at these critical numbers
are: f(0) = 1 f(2) = -3
The values of f at the endpoints of the interval
are: f(-½) = 1/8 f(4) = 17
 Comparing these four numbers, we see that
the absolute maximum value is f(4) = 17 and
the absolute minimum value is f(2) = -3.
65. CLOSED INTERVAL METHOD Example 8
Note that the absolute maximum occurs
at an endpoint, whereas the absolute
minimum occurs at a critical number.
66. CLOSED INTERVAL METHOD Example 8
The graph of f is sketched here.
67. EXACT VALUES
If you have a graphing calculator or
a computer with graphing software, it is
possible to estimate maximum and minimum
values very easily.
 However, as the next example shows,
calculus is needed to find the exact values.
68. EXACT VALUES Example 9
a.Use a graphing device to estimate
the absolute minimum and maximum values
of the function f(x) = x – 2 sin x, 0 ≤ x ≤ 2π.
b.Use calculus to find the exact minimum
and maximum values.
69. EXACT VALUES Example 9 a
The figure shows a graph of f in
the viewing rectangle [0, 2π] by [-1, 8].
70. EXACT VALUES Example 9 a
By moving the cursor close to the maximum
point, we see the y-coordinates don’t change
very much in the vicinity of the maximum.
 The absolute
maximum value
 It occurs when
x ≈ 5.2
71. EXACT VALUES Example 9 a
Similarly, by moving the cursor close to
the minimum point, we see the absolute
minimum value is about –0.68 and it occurs
when x ≈ 1.0
72. EXACT VALUES Example 9 a
It is possible to get more accurate
estimates by zooming in toward
the maximum and minimum points.
73. EXACT VALUES Example 9 b
The function f(x) = x – 2 sin x is continuous
on [0, 2π].
As f ’(x) = 1 – 2 cos x, we have f ’(x) = 0
when cos x = ½.
 This occurs when x = π/3 or 5π/3.
74. EXACT VALUES Example 9 b
The values of f at these critical points
  
f ( / 3)   2sin   3  0.684853
3 3 3
5 5 5
f (5 / 3)   2sin   3 6.968039
3 3 3
75. EXACT VALUES Example 9 b
The values of f at the endpoints
f(0) = 0
f(2π) = 2π ≈ 6.28
76. EXACT VALUES Example 9 b
Comparing these four numbers and
using the Closed Interval Method, we see
the absolute minimum value is
f(π/3) = π/3 - 3
and the absolute maximum value is
f(5π/3) = 5π/3 + 3
 The values from (a) serve as a check on our work.
77. MAXIMUM & MINIMUM VALUES Example 10
The Hubble Space Telescope was
deployed on April 24, 1990, by the space
shuttle Discovery.
78. MAXIMUM & MINIMUM VALUES Example 10
A model for the velocity of the shuttle
during this mission—from liftoff at t = 0
until the solid rocket boosters were jettisoned
at t = 126 s—is given by:
v(t) = 0.001302t3 – 0.09029t2 + 23.61t – 3.083
(in feet per second)
79. MAXIMUM & MINIMUM VALUES Example 10
Using this model, estimate the absolute
maximum and minimum values of
the acceleration of the shuttle between
liftoff and the jettisoning of the boosters.
80. MAXIMUM & MINIMUM VALUES Example 10
We are asked for the extreme values
not of the given velocity function,
but rather of the acceleration function.
81. MAXIMUM & MINIMUM VALUES Example 10
So, we first need to differentiate to find
the acceleration:
a (t ) v '(t )
d 3 2
 (0.001302t  0.09029t
dt
 23.61t  3.083)
2
0.003906t  0.18058t  23.61
82. MAXIMUM & MINIMUM VALUES Example 10
We now apply the Closed Interval Method
to the continuous function a on the interval
0 ≤ t ≤ 126.
Its derivative is:
a’(t) = 0.007812t – 0.18058
83. MAXIMUM & MINIMUM VALUES Example 10
The only critical number occurs
when a’(t) = 0:
0.18058
t1  23.12
0.007812
84. MAXIMUM & MINIMUM VALUES Example 10
Evaluating a(t) at the critical number
and at the endpoints, we have:
a(0) = 23.61 a(t1) ≈ 21.52 a(126) ≈ 62.87
 The maximum acceleration is about 62.87 ft/s2.
 The minimum acceleration is about 21.52 ft/s2.