4.7 FORMING FUNCTIONS
FROM VERBAL DESCRIPTIONS
Objectives:
1) Form a function in one variable from a
verbal description
2) Use the graphing calculator to
approximate a value for a graphic solution
to a verbal problem
Question 1
A pile of sand is in the shape of a cone with a diameter
that is twice its height. Sketch the diagram. Express the
volume V as a function of the height of the pile.
Since the volume of a cone is given by
V = 1/3r2h
and d = 2h, therefore 2r = 2h or h = r, we have
V = 1/3h3
In this last equation, the volume is a function of the
height of the sand pile (cone).
Strategy
1) Identify given information.
2) Assign variables to the related quantities.
3) Use formula(s) to link all variables together
(Primary Equation).
4) Find out the relationship among other variables
to one variable (Secondary Equation(s)).
5) Rewrite the formula(s) again in terms of one
variable.
Question 2 A winch at the tip of a 12meter building pulls pipe of the same
length to a vertical position. The
length of a rope at top of the building
holds the pipe. Find the length of
the rope in terms of the height.
s
12
[Solution] The relationships among
the three variables are:
x 2  y 2  122
x2  144  y2
and
s 2  x2  (12  y)2
s2  144  y 2  (12  y)2
s2  144  y 2  144  24 y  y 2  288  24 y
s  288  24 y
12
(x, y)
Question 3 An open top box with a square base is to be constructed
from sheet metal in such a way that the completed box is made
from 2 square yards of sheet metal.
1) Express the volume of the box as a multivariable function of the
base and width of the square base dimensions.
2) Express the volume of the box as a single function of the width of
the square base.
3) Find the feasible domain of the side of the base.
[Solution] (1)
Let the length of the base be x inch,
and the height be h inch.
Then the volume of the box in terms
of x and h inch is
V ( x, h)  x h
2
Primary Equation
h
x
(2) Notice that “made from a 2 yd2 sheet metal” provides
S ( x, h)  x2  4xh  2
Secondary Equation
From the secondary equation, we have
h  (2  x2 ) / 4 x
Then the volume of the box in terms of x in the primary
equation is
V ( x)  x2 (2  x2 ) / 4x  x(2  x2 ) / 4
(3) Since the height h > 0, then we need to solve the
rational inequality
2 x
0
4x
2
This results in
0 x 2
4 x(2  x 2 )  0

x  0
Question 4 A manufacturer wants to design an open box
having a square base and a surface area of 108 in2. What
dimension will produce a box with maximum volume?
Solution
Let the length of the base be x inch,
and the height be h inch.
Then the volume of the box in terms
of x and h inch is
V ( x, h)  x h
2
h
x
Primary Equation
Notice that “surface area of 108 in2 ”
provides
S
( x, h)  x 2  4xh  108
Secondary Equation
Question 4 A manufacturer wants to design an open box
having a square base and a surface area of 108 in2. What
dimension will produce a box with maximum volume?
Solution
From the secondary equation, we have
h
h  (108 x 2 ) / 4 x
Then the volume of the box in terms
of x in the primary equation is
V ( x)  x 2 (108 x 2 ) / 4x  27x  x3 / 4
x
Now, before we aim on V(x) and try to find its maximum
volume while x has certain length, we should find the feasible
domain of x.
Notice that
S ( x, h)  108  x 2  4 xh  x 2
and
x0
This means
0  x  108
Feasible Domain
Therefore, V(6, 3) is maximum when x = 6 and h = 3.
The box dimension is 6 x 6 x 3.
Question 5 Which points on the graph of y = 4 – x2 are
closest to the point (0, 2)?

Solution

The distance between the point (0, 2)
to any point (x, y) on the graph is

d  ( x  0)  ( y  2)  x  ( y  2)
2
2
Notice that y = 4 –
2

2

then
x2,
2
2
4
(x, y)
d

d ( x)  x  (4  x  2)  x  3 x  4
2
y
2

Also notice that the x
value makes the d(x)
reaches the minimum is the
same as makes the d2(x)
reaches the minimum.
x










Question 5 Which points on the graph of y = 4 – x2 are
closest to the point (0, 2)?
Solution
Therefore,
2
 2 3 7
d ( x)  x  3 x  4   x   
2 4

2
When
4
3
x 
2
2
2
3
6
x

2
2
or
and
3 5
y  4 
2 2
The d2(x) reaches the minimum. The minimum value is
7
2
d min ( x) 
4


So the points   6 , 5  are closest to (0, 2) and the distance is

2
2
7
d min ( x) 
2
Question 6 From a raft 50 m offshore, a lifeguard wants to swim
to shore and run to a snack bar 100 m down the beach, as show.
a) If the lifeguard swims at 1 m/s and run at 3 m/s, express the
total swimming and running time t as a function of the distance
x show in the diagram?
b) Find the minimum time.
Solution
The swimming distance in terms of x
is
s( x)  x2  502
The running distance in
terms of x is
50
100
r ( x)  100  x
The total time in second in terms of x is
100  x
2
2
T ( x)  x  50 
3
x
Question 7 A man is in boat 2 miles from the nearest point
R on the coast. His is to go to a point Q, located 3 miles
down the coast and 1 mile in land. He can row at 2 miles per
hour and walk at 4 miles per hour. Toward what point on the
coast should he row in order to reach point Q in the least time?
Solution
Let the point on the coast
toward by the man be P, and
PR = x, then PS = 3 – x.
Then
MP  x 2  4
PQ  (3  x) 2  1
M
2
R
x
P 3 3–x
S
1
Q
Then the total time from
point M to Q will be
T ( x) 
x2  4
x 2  6 x  10

2
4
M
2
R
x
P 3 3–x
S
1
Q
The feasible domain for x is [0, 3]. The solution for x within
the domain is x = 1.
Question 8 Four feet wire is to be used to form a square
and a circle. How much of the wire should be used for the
square and how much should be used for the circle to enclose
the maximum area?
x
Solution
Let the side of the square be x and the
radius of the circle be r. Then
A( x, r )  x 2   r 2
x
Primary Equation
Since the total length of wire is 4 feet, then
Secondary Equation
4 x  2 r  4
So,
r
2(1  x)

r
r
2(1  x)
x

x
 2(1  x)   x 2  4(1  x)
2
A( x)  x   


  
1
 [(  4) x 2  8 x  4]

2
The feasible domain of x is restricted by
the square’s perimeter
0  x 1
We conclude that when x  0 yields
the maximum area. That means all
the wire is used to form circle.
2
r
Question 9 The Rectangle ABCD has vertices C and D on
the x-axis and vertices A and B on the part of the parabola
y = 4 – x2 that is above the x-axis.

y
a) Express the perimeter P of the rectangle as a function of
the x-coordinate of A.
b) What is the domain of the perimeter
function?
c) For what value of x is the perimeter
a maximum?




B
Solution
(a)
P( x, y)  2(2 x)  2 y  4 x  2 y
P( x)  4 x  2(4  x )
2
 2 x 2  4 x  8

A (x, y)


x


C



D




Question 9 The Rectangle ABCD has vertices C and D on
the x-axis and vertices A and B on the part of the parabola
y = 4 – x2 that is above the x-axis.

y
a) Express the perimeter P of the rectangle as a function of
the x-coordinate of A.
b) What is the domain of the perimeter
function?
c) For what value of x is the perimeter
a maximum?




Solution
B
(b)

Domain of the perimeter
is
0 x2
A (x, y)

x



C



D




Question 9 The Rectangle ABCD has vertices C and D on
the x-axis and vertices A and B on the part of the parabola
y = 4 – x2 that is above the x-axis.

y
a) Express the perimeter P of the rectangle as a function of
the x-coordinate of A.
b) What is the domain of the perimeter
function?
c) For what value of x is the perimeter
a maximum?




Solution
B
(c)

From the expression in (a),
P( x)  2 x2  4 x  8
when x = 1, P(x) has a
maximum.
Pmax (1)  10
A (x, y)

x



C



D




Question 10 The Rectangle ABCD has vertices C and D on
the x-axis and vertices A and B on the part of the parabola
y = 4 – x2 that is above the x-axis.

y
a) Express the area A of the rectangle as a function of the xcoordinate of A.
b) What is the domain of the area
function?
c) For what value of x is the area
a maximum?




B
Solution
A (x, y)

(a)

A( x, y)  2 xy
x
A( x)  2 x(4  x )
2



C



D




Question 10 The Rectangle ABCD has vertices C and D on
the x-axis and vertices A and B on the part of the parabola
y = 4 – x2 that is above the x-axis.

y
a) Express the area A of the rectangle as a function of the xcoordinate of A.
b) What is the domain of the area
function?
c) For what value of x is the area
a maximum?




Solution
B
(b)

Domain of the area is
0 x2
A (x, y)

x



C



D




Question 10 The Rectangle ABCD has vertices C and D on
the x-axis and vertices A and B on the part of the parabola
y = 4 – x2 that is above the x-axis.

y
a) Express the area A of the rectangle as a function of the xcoordinate of A.
b) What is the domain of the area
function?
c) For what value of x is the area
a maximum?




Solution
B
(c)

From the expression in (a),
A( x)  2 x(4  x2 )
when x  1.155, P(x) has
a maximum.
A (x, y)

x

Pmax (1.155)  6.158


C



D




Assignment
Day 1
P. 150 #16 – 24 (even),
P. 161 #2, 3, 5, 7, 8, 10, 11
Day 2
P. 162 #14, 15, 18, 19, 21, WS 4.3B