130
2.7
CHAPTER 2
Functions and Their Graphs
Mathematical Models: Constructing Functions
OBJECTIVE
1
Construct and Analyze Functions
1 Construct and Analyze Functions
✓
Real-world problems often result in mathematical models that involve functions.These
functions need to be constructed or built based on the information given. In constructing functions, we must be able to translate the verbal description into the language of
mathematics. We do this by assigning symbols to represent the independent and dependent variables and then finding the function or rule that relates these variables.
EXAMPLE 1
Area of a Rectangle with Fixed Perimeter
The perimeter of a rectangle is 50 feet. Express its area A as a function of the length
l of a side.
Solution
Figure 70
w
l
A
Consult Figure 70. If the length of the rectangle is l and if w is its width, then the sum
of the lengths of the sides is the perimeter, 50.
l + w + l + w
2l + 2w
l + w
w
l
=
=
=
=
50
50
25
25 - l
The area A is length times width, so
w
A = lw = l125 - l2
The area A as a function of l is
A1l2 = l125 - l2
Note that we use the symbol A as the dependent variable and also as the name
of the function that relates the length l to the area A. This double usage is common
in applications and should cause no difficulties.
EXAMPLE 2
Economics: Demand Equations
In economics, revenue R, in dollars, is defined as the amount of money received
from the sale of a product and is equal to the unit selling price p, in dollars, of the
product times the number x of units actually sold. That is,
R = xp
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SECTION 2.7
Mathematical Models: Constructing Functions
131
In economics, the Law of Demand states that p and x are related: As one increases, the other decreases. Suppose that p and x are related by the following
demand equation:
1
p = - x + 20,
0 … x … 200
10
Express the revenue R as a function of the number x of units sold.
Solution
Since R = xp and p = -
1
x + 20, it follows that
10
R1x2 = xp = xa -
1
1
x + 20b = - x2 + 20x
10
10
NOW WORK PROBLEM
EXAMPLE 3
3.
Finding the Distance from the Origin to a Point on a Graph
Let P = 1x, y2 be a point on the graph of y = x2 - 1.
(a) Express the distance d from P to the origin O as a function of x.
(b) What is d if x = 0?
(c) What is d if x = 1?
22
?
2
(e) Use a graphing utility to graph the function d = d1x2, x Ú 0. Rounded to two
decimal places, find the value(s) of x at which d has a local minimum. [This gives
the point(s) on the graph of y = x2 - 1 closest to the origin.]
(d) What is d if x =
Solution
Figure 71
(a) Figure 71 illustrates the graph. The distance d from P to O is
d = 41x - 022 + 1y - 022 = 3x2 + y2
y
2
Since P is a point on the graph of y = x2 - 1, we substitute x2 - 1 for y. Then
y x2 1
1
(x, y)
(0, 0) d
1
1
2
d1x2 = 4x2 + 1x2 - 122 = 3x4 - x2 + 1
x
We have expressed the distance d as a function of x.
1
(b) If x = 0, the distance d is
d102 = 21 = 1
(c) If x = 1, the distance d is
d112 = 21 - 1 + 1 = 1
(d) If x =
Figure 72
2
da
2
0
0
22
, the distance d is
2
12 4
12 2
1
1
13
12
b =
a
b - a
b + 1 =
- + 1 =
2
2
A4 2
2
B 2
(e) Figure 72 shows the graph of Y1 = 3x4 - x2 + 1. Using the MINIMUM feature
on a graphing utility, we find that when x L 0.71 the value of d is smallest.The local
minimum is d L 0.87 rounded to two decimal places. [By symmetry, it follows that
when x L -0.71 the value of d is also a local minimum.]
NOW WORK PROBLEM
9.
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132
CHAPTER 2
Functions and Their Graphs
EXAMPLE 4
Filling a Swimming Pool
A rectangular swimming pool 20 meters long and 10 meters wide is 4 meters deep
at one end and 1 meter deep at the other. Figure 73 illustrates a cross-sectional
view of the pool. Water is being pumped into the pool to a height of 3 meters at the
deep end.
Figure 73
20 m
10 m
1m
1m
L
x
3m
(a) Find a function that expresses the volume V of water in the pool as a function of
the height x of the water at the deep end.
(b) Find the volume when the height is 1 meter.
(c) Find the volume when the height is 2 meters.
(d) At what height is the volume 20 cubic meters? 100 cubic meters?
Solution
(a) Let L denote the distance (in meters) measured at water level from the deep
end to the short end. Notice that L and x form the sides of a triangle that is similar to the triangle whose sides are 20 meters by 3 meters. This means L and x
are related by the equation
L
20
=
x
3
or L =
20x
,
3
0 … x … 3
The volume V of water in the pool at any time is
V = ¢
Since L =
1
cross-sectional
≤ 1width2 = a Lxb1102 cubic meters
triangular area
2
20x
, we have
3
V1x2 = a
1 # 20x #
100 2
xb1102 =
x
2 3
3
cubic meters
(b) When the height x of the water is 1 meter, the volume V = V1x2 is
V112 =
100 # 2
1
1 = 33
3
3
cubic meters
(c) When the height x of the water is 2 meters, the volume V = V1x2 is
V122 =
100 # 2
400
1
2 =
= 133
3
3
3
cubic meters
100 2
x = 20, we find that when x L 0.77 meter, the vol3
100 2
x = 100, we find that when
ume is 20 cubic meters. By solving the equation
3
x L 1.73 meters, the volume is 100 cubic meters.
(d) By solving the equation
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 2.7
EXAMPLE 5
Figure 74
1
(a) Express the area A of the rectangle as a function of x.
(x, y )
y 25 x 2
10
1 2 3 4 5
(0,0)
Area of a Rectangle
A rectangle has one corner on the graph of y = 25 - x2, another at the origin, a
third on the positive y-axis, and the fourth on the positive x-axis. See Figure 74.
y
30
20
133
Mathematical Models: Constructing Functions
(b) What is the domain of A?
(c) Graph A = A1x2.
(d) For what value of x is the area largest?
x
Solution
(a) The area A of the rectangle is A = xy, where y = 25 - x2. Substituting this expression for y, we obtain A1x2 = x125 - x22 = 25x - x3.
(b) Since x represents a side of the rectangle, we have x 7 0. In addition, the area
must be positive, so y = 25 - x2 7 0, which implies that x2 6 25, so
-5 6 x 6 5. Combining these restrictions, we have the domain of A as
5x ƒ 0 6 x 6 56 or 10, 52 using interval notation.
(c) See Figure 75 for the graph of A = A1x2.
(d) Using MAXIMUM, we find that the area is a maximum of 48.11 at x = 2.89,
each rounded to two decimal places. See Figure 76.
Figure 75
Figure 76
50
50
0
0
5
NOW WORK PROBLEM
EXAMPLE 6
5
0
0
15.
Making a Playpen*
A manufacturer of children’s playpens makes a square model that can be opened at
one corner and attached at right angles to a wall or, perhaps, the side of a house. If
each side is 3 feet in length, the open configuration doubles the available area in
which the child can play from 9 square feet to 18 square feet. See Figure 77.
Now suppose that we place hinges at the outer corners to allow for a configuration like the one shown in Figure 78.
Figure 77
Figure 78
3
3
3
3
3
x
3
h
3
3
18 ft2
*
Adapted from Proceedings, Summer Conference for College Teachers on Applied Mathematics (University of Missouri, Rolla), 1971.
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134
CHAPTER 2
Functions and Their Graphs
(a) Express the area A of this configuration as a function of the distance x between
the two parallel sides.
(b) Find the domain of A.
(c) Find A if x = 5.
(d) Graph A = A1x2. For what value of x is the area largest? What is the maximum
area?
Solution
(a) Refer to Figure 78. The area A that we seek consists of the area of a rectangle
(with width 3 and length x) and the area of an isosceles triangle (with base x and
two equal sides of length 3). The height h of the triangle may be found using the
Pythagorean Theorem.
x 2
h2 + a b = 32
2
x 2
x2
36 - x2
h2 = 32 - a b = 9 =
2
4
4
1
336 - x2
2
h =
The area A enclosed by the playpen is
A = area of rectangle + area of triangle = 3x +
A1x2 = 3x +
1 1
xa 336 - x2 b
2 2
x336 - x2
4
Now the area A is expressed as a function of x.
(b) To find the domain of A, we note first that x 7 0, since x is a length. Also, the
expression under the square root must be positive, so
36 - x2 7 0
x2 6 36
-6 6 x 6 6
Figure 79
Combining these restrictions, we find that the domain of A is 0 6 x 6 6, or
10, 62 using interval notation.
24
(c) If x = 5, the area is
A152 = 3152 +
0
6
0
5
36 - 1522 L 19.15 square feet
44
If the width of the playpen is 5 feet, its area is 19.15 square feet.
(d) See Figure 79. The maximum area is about 19.82 square feet, obtained when x is
about 5.58 feet.
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