Copyright
c
Oklahoma State University. All rights reserved.
Full file at
258
Solution Guide for Chapter 2
(c) The cannonball reaches its maximum height where the graph reaches a peak. In
the righthand picture below, we see that this occurs when
x
= 1
.
61
and
y
=
1
.
39
. Thus the cannonball reaches a maximum height of 1.39 miles at 1.61 miles
downrange.
2.
Profit
:
(a) The exercise suggests a horizontal span of
0
to
7
. The table of values on the left
below led us to choose a vertical span of

3
to
6
. The graph is on the right below.
(b) From the table of values we see that
P
(0) =

2
. This means that, if no widgets
are sold, the weekly profit will be

2
thousand dollars, so the producer will have
a loss of
$2000
.
(c) To find where the largest profit occurs, we locate the peak of the graph. In the
figure below, we see that the peak occurs when
n
= 4
.
83
and
P
= 5
.
01
. Thus the
profit is as large as possible at a sales level of
4
.
83
thousand widgets.
Copyright
c
Oklahoma State University. All rights reserved.
Full file at
SECTION 2.5
Optimization
259
3.
Marine fishery
:
(a) The exercise suggests a horizontal span of
0
to
1
.
5
. The table of values on the left
below led us to choose a vertical span of

0
.
1
to
0
.
1
. The graph is on the right
below.
(b) The growth rate if the population size is
0
.
24
million tons is represented by
G
(0
.
24)
in functional notation. From a table of values or the graph we see that the value is
0
.
04
million tons per year.
(c) From a table of values or the graph we see that
G
(1
.
42) =

0
.
02
. This means that
if the population size is
1
.
42
million tons then the growth rate is

0
.
02
million tons
per year, so the population is decreasing at a rate of
0
.
02
million tons per year.
(d) To find where the largest growth rate occurs, we locate the peak of the graph. In
the figure below, we see that the peak occurs when
n
= 0
.
67
and
G
= 0
.
07
. Thus
the growth rate is the largest at a population size of
0
.
67
million tons.
Copyright
c
Oklahoma State University. All rights reserved.
Full file at
260
Solution Guide for Chapter 2
4.
Enclosing a field
:
(a) The picture should be a rectangle with width marked 3 miles and length marked
5 miles. Since the perimeter of the rectangle is two lengths plus two widths, the
enclosure uses
2
×
3 + 2
×
5 = 16
miles of fence, as required.
The area of the
rectangle is
5
×
3 = 15
square miles.
(b) The picture should be a rectangle with width marked 2 miles and length marked
6 miles. Since the perimeter of the rectangle is two lengths plus two widths, the
enclosure uses
2
×
2 + 2
×
6 = 16
miles of fence, as required.
The area of the
rectangle is
6
×
2 = 12
square miles.
(c)
i. Since there are 16 miles of fence, the width
w
is somewhere between 0 and
8 miles, and so we use that for a horizontal span. The table below suggests
a vertical span of 0 to 20 square miles. In the graph below, width is on the
horizontal axis, and area is on the vertical axis.
The graph shows the area
enclosed by a rectangle of perimeter 16 as a function of the width. From the
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 Spring '13
 YOYL
 Calculus, Limit, Oklahoma State University