=
5. ‘(&)
g(z)
=
=
—
—
4
2
y
1
x
4
—
2
z
+4
+
2
±
<
9,—7r/2
2
sin
4
z
<
0 <
-I- sins, 0
<
12x ± v
2
6r
sin 20,0
3
x
4
x < 2w
g(t)
=
— (t
—
3
4 — 2x
x
IT
=
2)2/3
=
20. g(0)
IsinOI,0 <9
< 2n-
cos 0
O<t9<2r
1 -f sin 0’
18. f(x)
14.
X
-
f(x) = (x
16. r(s) = 3s +-
2
s
2)
3
-x
2
4
12.g(x)=x
+
f(x)
g(x)
22.
23.
26. F(x)
25. F(x)
24. h(x)
f(x)
21.
=
=
=
=
+
+
X
-f
2x
6\/
x
x
—
—
4
on [0,00)
[0, oo)
4x on [0,
00)
4x on [0,4]
on [0, co)
32
4
2 2x on [0, 2]
sin
in Problems 21—30, find, if possible, the (global) mnaxinnun and
ndnirnuni values of the given flnction on the indicated interval.
=
19. A(0)
t0
1
17. f(t)=t — ,
15.
13. H(x)
3x
11.f(x)=x
—
3
5,
3x + 1
10. f(x) =
x- + 1
Jn Problemns 11—20, fInd the critical points and use the test of your
choice to decide which critical points give a local irmaxini urn value
and which give a local mninimwn vahie. What are these local mnaxi
mum and niinimnum values?
9. h(y)
8.
7.f(x)=
=
=
4. f(.r)
6. r(z)
=
=
3. f(9)
2. f(x)
1. f(.r)
In Problems 1—10, identify the critical points. Then use (a) the First
Derivative Test and (ifpossible) (6) the Second Derivative Test to
decide which of the critical points give a local niaxirnuni and
which give a local nminiirmurn.
Problem Set 3.3
1. Iffis continuous at c, f’(x) > 0 near to con the left. and
value
f’(.x) < 0 near to c on the right, then f(c) is a local
forf
2. If f’(x) = (x + 2)(x 1), then f(—2) is a local
value for f.
value forf and f(l) is a local
—
Chapter 3 Applications of the Derivative
Concepts Review
166
on (8,oo)
1 on [—2,21
—
sin t2 on [0, w]
—
(8
2
16x
—on(0. r/2)
cosx
nor a
=
—
—
—
—
, A
2
B)
2)2(5
B
r
3)
(
2
— 4)2
A)(.r— B).0<A< B
r
4)
(
2
1 )2(
-
—
—
x(x
(x
(x
(x
— 1)2(x - 2)2(x - 3)(x -4)
-(x - l)(x - 2)(z - 3)(x -4)
(1
3
x
minimum 3. maximum
1. maximum 2. maximum;
4. local maximum; local minimum;0
Answers to Concepts Review:
—
37. f is differentiable, has domain [0. 61, and has two local
maxima and two local minima on (0, 6).
38. f is differentiable, has domain [0, 6], and has three local
maxima and two local minima on (0, 6).
39. f is continuous, but not necessarily differentiable, has do
main [0,6], and has one local minimnuni and one local maximum
on (0,6).
40. f is continuous, but not necessarily differentiable, has do
main [0,6], and has one local minimum and no local maximum on
(0,6).
41. f has domain [0,6], but is not necessarily continuous, and
has three local maxima and no local minimum on (0,6).
42. f has domain [0,6], but is not necessarily continuous, and
has two local maxima and no local minimum on (0, 6).
43. Consider f(x) = Ax
2 -I- Bx + C with A > 0. Show that
f(s)
0 forallx if andonlyifB
2 — 4AC 0.
2 + Cx ± D with A > 0.
44. Consider f(s) = Ax
3 -t- Bx
Show that f has one local maxirnllm and one local minimum if
2 3AC > 0.
and only if B
45. What conclusions can you draw aboutffrom the informa
tion thatf’(c) = f”(c) = 0 andf”(c) > 0?
In Problems 37—42, sketch a graph of a function with the given
properties. If it is impossible to graph such afunction, then indicate
this and justify your answer
36. f(s)
=
=
34. f(s)
35. f(s)
=
=
=
33. f’(x)
32. f’(x)
31. f(s)
f’ is given. Find all values
of x that make the function f(a) a local minimum and (b) a local
maxilnuni.
=
2 +
x
slnx
—--—- ±
27
In Problems 31—36, the first derivative
30. h(t)
=
=
g(x)
28.
29. B(s)
=
f(x)
27.
64
4. If 1(x) = x
, then f(0) is neither a
3
even though f”(O) =
3. If f’(c)
0 and f”(c) <0, we expect to find a local
value for fat c.
r
Figure 3
24—2x
24
IE 1
Figure 2
Figure 1
I_____
+
±
r
xt
I
9—2x
3.4
Practical Problems
167
Q to be maximized or mini
=
x(9 — 2x)(24
—
2x)
=
216x
—
2 + 4x
66x
3
216
—
132x + 12x
2
12(18
—
)
2
lix + x
=
12(9
—
x)(2
—
x)
=
0
iiiLE} A farmer has 100 meters of wire fence with which he plans to
build two identical adjacent pens, as shown in Figure 3. What are the dimensions of
the enclosure that has maximum area?
It is often helpful to plot the objective function. Plotting functions can be done
easily with a graphing calculator or a CAS. Figure 2 shows a plot of the function
V(x) = 216x — 66x
2 + 4x
. When x = 0, V(x) is equal to zero. In the context of
3
folding the box, this means that when the width of the cut-out corner is zero there
is nothing to fold up, so the volume is zero. Also, when x = 4.5, the cardboard gets
folded in half, so there is no base to the box; this box will also have zero volume.
Thus, V(0)
0 and V(4.5)
0. The greatest volume must be attained for some
value of x between 0 and 4.5. The graph suggests that the maximum volume occurs
when xis about 2; by using calculus, we can determine that the exact value of x that
maximizes the volume of the box is x
2.
This gives x = 2 or x = 9, but 9 is not in the interval [0, 4.5].We see that there are
only three critical points, 0, 2, and 4.5. At the end points 0 and 4.5, V
0; at 2,
V = 200. We conclude that the box has a maximum volume of 200 cubic inches if
x = 2, that is, if the box is 20 inches long, 5 inches wide, and 2 inches deep.
I
=
Now x cannot be less than 0 or more than 4.5. Thus, our problem is to maximize V
on [0, 4.5]. The stationary points are found by setting dV/dx equal to 0 and solving
the resulting equation:
V
SOlUTION Let x be the width of the square to be cut out and V the volume of
the resulting box. Then
A rectangular box is to be made from a piece of cardboard 24
inches long and 9 inches wide by cutting out identical squares from the four cor
ners and turning up the sides, as in Figure 1. Find the dimensions of the box of max
iniumn volume. What is this volume?
Throughout, use your intuition to get some idea of what the solution of the
problem should be. For many physical problems you can get a “ballpark” estimate
of the optimal value before you begin to carry out the details.
Step 5: Either substitute the critical values into the objective function or use the
theory from the last section (i.e., the First and Second Derivative Tests) to deter
mine the maximum or minimum.
Step 4: Find the critical points (end points, stationary points, singular points).
Step 3: Use the conditions of the problem to eliminate all but one of these vari
ables, and thereby express Q as a function of a single variable.
Step 2: Write a formula for the objective function
mized in terms of the variables from step 1.
Step 1: Draw a picture for the problem and assign appropriate variables to the im
portant quantities.
Based on the examples and the theory developed in the first three sections of this
chapter, we suggest the following step-by-step method that can be applied to many
practical optimization problems. Do not follow it slavishly; common sense may
sometimes suggest an alternative approach or omission of some steps.
Section 3.4 Practical Problems