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Optimization (4.3)
Global maxima and minima on closed intervals. The single greatest (or least)
value of a function f on a closed interval [a,b] is called the global maximum (or minimum)
of f on [a,b] . COMPARE to local maximum (or minimum). DRAW figure 4.31.
2
Let's examine the global maximum and global minimum for y = e!x on [!1,1] .
DRAW graph. This function should look familiar by now. It appears that the global
maximum occurs at x = 0 and the global minimum occurs at x = ±1 . These maximum
1
and minimum values are 1 and , respectively.
e
Theorem 4.2 (Extreme Value Theorem) states (p. 186): If f is continuous on a
closed interval [a,b] , then f has a global maximum and a global minimum on [a,b] .
We saw in our example above that we may have more than one global maximum
or minimum point on [a,b] . The Extreme Value Theorem guarantees that we will have at
least one global maximum and at least one global minimum on [a,b] .
To find the global maximum and global minimum points on [a,b] , we must
compare the values of both endpoints and all critical points on [a,b] ( f !(x) = 0 ).
SOLVE 4.3.6. f ( x ) = sin 2 x ! cos x, 0 " x " # . DRAW graph.
(a) f !( x ) = 2sin x cos x + sin x = (sin x)(2cos x + 1) . f !( x ) = 0 when sin x = 0 or
1
1
when 2cos x + 1 = 0 (or cos x = ! ). sin x = 0 when x = 0 or x = ! . cos x = ! when
2
2
2!
2!
. So the critical points are x = 0, x =
x=
or x = ! . Note that sin x > 0 if
3
3
2!
2!
0 < x < ! . And 2cos x + 1 > 0 if 0 < x <
and 2cos x + 1 < 0 if
< x < ! . Therefore,
3
3
2"
2"
and f !( x ) < 0 for
f !( x ) > 0 for 0 < x <
< x < " . Thus, f has a local maximum at
3
3
2!
by the FDT and local minima at x = 0 or x = ! (WHY?).
x=
3
(b) To determine the global maximum and global minimum, we need to find
2
" 2! %
# 2" & # 3 & # 1 & 3 1 5
f (0), f $ ' and f (! ) . We find f (0) = !1, f % ( = % ( ! % ! ( = + = and
# 3&
$ 3 ' $ 2 ' $ 2' 4 2 4
5
2!
, and the global
f (! ) = 1. Therefore, the global maximum is and it occurs at x =
4
3
minimum is -1 and it occurs at x = 0.
Global maxima and minima on open intervals.
The Extreme Value Theorem only guarantees at least one global maximum and
one global minimum on a CLOSED interval – not on an OPEN interval. DRAW y = x
on (!1,1) . Here, we have neither a global maximum nor a global minimum. If a global
maximum or global minimum does occur, then it will occur at a critical value. DRAW
y = x(x ! 2)(x + 2) on (!2,2) . Here, the global maximum and global minimum occur at
the critical points. However, if we looked at the same function on the interval (!3,3) ,
then we would have NO global maximum nor global minimum. Therefore, we need to
calculate the values at the critical points and compare these values to the behavior of the
graph near the endpoints on the interval.
SOLVE 4.3.10. To find the global maximum and minimum values of
f ( x ) = x ! ln x on the interval (0,!) , we first find the values of the critical points.
DRAW graph. To find the critical points of f, we must solve f !( x ) = 0 . Since
1
f !( x ) = 1" . the only critical point is where x = 1. To determine if the critical point at
x
x = 1 is a local minimum or a local maximum, we can use either the First or Second
1
Derivative Tests. Using the First Derivative Test, we see that f !( x ) = 1" < 0 for x < 1
x
1
and f !( x ) = 1" > 0 for x > 1. Therefore, the critical point at x = 1 is a local minimum.
x
Now, we examine the behavior of the graph near its endpoints. We find that
f ( x ) ! " as x ! 0+ and f ( x ) ! " as x ! " . Therefore, the local minimum is a global
minimum. The minimum value is f (1) = 1! ln1 = 1! 0 = 1 . DOES the graph have a
global maximum? No.
Upper and lower bounds.
Discuss Example 4.
SOLVE 4.3.16. f ( x ) = x + sin x on [0,2! ] . DRAW graph. The graph suggests
that f is increasing on [0,2! ] . Let's see if we can confirm this. f !( x ) = 1" cos x and
f ! " 0 on [0,2! ] , so f is increasing there. Therefore, the lower bound for y occurs at x =
0 or y (0) = 0 + sin0 = 0 , and the upper bound for y occurs at x = 2! or
y (2! ) = 2! + sin(2! ) = 2! .
Applications.
dT
CD2 D3
SOLVE 4.3.21. We have T ( D) =
and
= CD ! D2 = D(C ! D) .
!
dD
2
3
The critical points are when D = 0 or D = C . We can ignore the first critical point. Why?
To determine if the point D = C is a local maximum or a local minimum, we note that
dT
is negative when D > C and positive when 0 < D < C . Therefore, by the FDT the
dD
dT
temperature change is maximized when D = C . (b) The sensitivity is
= CD ! D2 . To
dD
d 2T
= C ! 2D . The only critical point occurs
maximize sensitivity, we find its derivative
dD2
d 2T
C
C
=
C
!
2D
when D = . We note that
is
negative
when
and positive when
D
>
2
2
dD2
C
C
D < . Therefore, by the FDT the sensitivity is maximized when D = .
2
2