T. James 1
Notebook Resource
Portfolio of Problems
Towanda James
Problem Solving, EMAT 6600
Dr. Jim Wilson
August 2, 2006
T. James 2
Portfolio of Problems
1. Solve 2 x  x10 by iteration. I used an excel spreadsheet in order to do the iteration process.
The items marked in red are where I noticed the answers were very close. x  - .9371 or x
1.07755.
-2
-1
0
1
2
-0.9
-0.8
-0.95
-0.94
-0.93
-0.92
-0.935
-0.939
-0.937
-0.9371
1.07
1.075
1.076
1.077
1.0775
1.07755
=2^A1
0.25
0.5
1
2
4
0.535887
0.574349
0.517632
0.521233
0.524858
0.528509
0.523042
0.521594
0.522318
0.522282
2.099433
2.106722
2.108183
2.109645
2.110376
2.110449
-2
-1
0
1
2
-0.9
-0.8
-0.988
-0.97
-0.96
-0.95
-0.94
-0.93
-0.935
-0.939
-0.937
-0.9371
1.1
1.05
1.06
1.07
1.075
1.076
1.077
1.0775
1.07755
Look at the graphs of the functions.
=D1^10
1024
1
0
1
1024
0.348678
0.107374
0.886277
0.737424
0.664833
0.598737
0.538615
0.483982
0.510642
0.532913
0.52167
0.522227
2.593742
1.628895
1.790848
1.967151
2.061032
2.080284
2.099699
2.109467
2.110446
T. James 3
2. Volume of a Frustum of a Cone
A frustum may be formed from a cone with a circular base by cutting off the tip of the cone with
a cut perpendicular to the height, forming a lower base and an upper base that are circular and
parallel.
Let h be the height, R the radius of the lower base, and r the radius of the upper base. One
picture of the frustum is the following.
Given R, r, and h, find the volume of the frustum.
The Formula
V
h
3
( R 2  Rr  r 2 )
Extensions:
1. Derive the formula.
2. What is the area of the curved surface of the frustum of a right circular cone? What is
the total surface area of the frustum?
3. What is the volume of a frustum formed from square based pyramid of height h, lower
base having a side length of S, and an upper base of side length s?
4. What is the total surface area of the square based frustum?
Solution:
Consider the cone before it was cut. Let the height of the cut part = H.
T. James 4
The volume of a cone is V =
1
3
r h.
2
The volume of the original pre-cut cone is Vo  13  R 2 ( H  h) .
The volume of the cut part is Vc  13  r 2 H .
Now we have a subtraction problem in order to get the volume of the frustum. Consider the
difference of the two cones. Subtracting the cut-off part from the original pre-cut cone would
leave the bottom part of the cone called the frustum.
VF  Vo  Vc  13  R 2 ( H  h) - 13  r 2 H
= 13  R 2 H  13  R 2 h  13  r 2 H
= 13  ( R 2 H  R 2 h  r 2 H )
= 13  ( R 2 h  R 2 H  r 2 H )
 

= 13  ( R 2 h  H R 2  r 2 )
= 13  ( R 2 h  H  R  r )( R  r )
= 13   R 2 h  H  R  r )( R  r  
Now, recall the original pre-cut cone. Consider the triangles formed by the height and the bases.
The triangles are similar by AA similarity and the sides are proportional.
H H h

r
R
H
r
h
R
Cross Multiplying yields,
HR  r ( H  h)
HR  Hr  hr
HR  Hr  hr
H ( R  r )  hr
Substitute H ( R  r ) for hr in the last step for VF .
Recall:
2
1
VF = 3   R h  H  R  r )( R  r  
T. James 5
=   R h  hr  R  r  
= 13  h  R 2  r  R  r  
2
1
3
=
h
3
= 13  h  R 2  Rr  r 2 
( R 2  Rr  r 2 ) , which is the formula.
The formula for the lateral area of a right circular cone is  r .
The slant height of the original pre-cut cone can be found by using the Pythagorean theorem
( H  h) 2  R 2  2 .

 ( H  h) 2  R 2
The lateral area of the original pre-cut cone is Lo   r ( H  h) 2  R 2 .
The lateral area of the cut part is Lc   r H 2  r 2 .
LF  Lo  Lc   r ( H  h) 2  R 2 -  r H 2  r 2
In order to get the total surface area of the frustum, add on the area of the base, which is a circle.
The formula in order to find the area of a circle is  r 2 .
Therefore, the total surface area of a frustum is  r ( H  h) 2  R 2 -  r H 2  r 2 +  r 2 .
I think this is a good activity to share with students. I wonder if the formula for the area of a
circle makes any since to them. Consider the following activity. This is problem solving on a
middle school or lower high school level.
10
8
6
4
2
-15
-10
-5
5
-2
-4
-6
-8
10
15
T. James 6
A portion of this circle is covered by a shaded square. The length of a side of the shaded square
is the same length as the radius of the circle. We call such a square a “radius square.”
1. How many radius squares does it take to cover the circle? (You can cut out the radius
squares, cover the circle and see how many it takes to cover.) Record your data in a table
with columns for the radius, area of the radius square, number of radius squares needed to
cover the circle, and estimated area of the circle.
2.
Do you think the same number of radius squares is needed to fill all circles? Construct a
different size circle in GSP. Construct a quadrilateral interior for the radius square in the
circle like the picture above. Copy your circle in word. Copy the grid and radius squares in
Word. Crop pictures where needed. Record your new data in the second row in the table
you made for the 1st circle. Repeat with another different circle. Do all the circles take the
same number of radius squares to cover them? How many is that again?
3. Did you get a little over 3 squares to fill the circle? What number have we been studying
that’s a little over 3?
4. Describe any patterns and relationships you see in your table that will allow you to predict
the area of the circle from its radius square. Test you ideas on some other circular objects.
5. How can you find the area of a circle if you know the radius?
6. How can you find the radius of a circle if you know the area?
7. Calculate the expression (5) 2  (3.14) .How does this expression relate to the area of the 1st
circle you worked with? Now calculate the expression using the  symbol on the calculator.
Which expression is more accurate?
T. James 7
3. PROBLEM: Average Rate
Danielsville is 20 miles from Jim's home in Athens. He Drives 60 mph going to a meeting (he is
almost late) but coming home the weather is bad and drives 30 mph. What is his average speed
for the time he is on the road?
A. Discuss why 45 mph is not correct.
Going to D'ville took 1/3 of an hour. Coming home took 2/3 of an hour. So the total 40 miles
took one hour. The speed is 40 mph.
B. Use
d=rt
to verify and develop an understanding of how to find average rates.
Solution:
The average rate is over the same distance each way. The average rate is not the arithmetic
mean of the two rates. That is, the rate is not 60230  45mph.
Distance is rate times time. Since Jim drove two different rates, we need to calculate going and
returning.
Going
d  rg t g
Returning
d  rr tr
20  60t g
20  30tr
t g  13
tr 
2
3
It took 13 hr traveling at 60 mph and 23 hr traveling at 30 mph for the entire trip. The total
distance was 40 miles that took one hour. The speed, therefore, is 40 mph.
d g  rg t g and d r  rr tr
For this problem, r g = 60 and r r = 30
tg 
r=
2( rg rr )
rg  rr
r=
2(6030)
60  30
and tr 
dg
rg
d = rt
2d = r ( t g  tr )
2d = r(
dg
rg
+
dr
rr
)
dr
rr
r = 40
The average is 40 mph.
2 = r( r1g  r1r )
1
r
 12 ( r1g  r1r ) Note: This is the harmonic mean of r g and r r because it is the arithmetic mean of the inverse.
r=
2( rg rr )
rg  rr
T. James 8
4. Volume of a land Fill;
Curved Path, Trapezoid Cross-Section
The following e-mail was sent to me:
I am an engineering geologist. I am involved in a case where a landowner filled in a stream with
a road, several bridges, and an island. The local regulatory agency would like to estimate the
volume of fill placed in the stream. That's my job. I have been able to calculate most of the
volume by slicing the fill into cross sections and calculating the area of cross section and then
multiplying the area for whatever length I decide to evaluate. My problem is where some cross
sections are attached at a pivot point and I have a angular "length" to evaluate. For example how
would you calculate the volume where two trapezoid cross sections of different areas that have a
common pivot point and are separated by 50 degrees ? Or more easily, what is the volume of a
trapezoid that sweeps across an angle of 50 degrees?
No wonder I am a Geologist, couldn't handle the math ! Thanks for your
help.
--------------------Interpret what he is asking for and devise an easily understood strategy for estimating the volume
he is looking for.
There is a pivot point P and a distance R from the pivot point to the center of the trench. We have
a trapezoid cross section with bases a and b and height h. The pivot angle is 50 degrees.
Devise a spreadsheet that can be used to estimate the volume given input of the dimensions
of the trapezoid, the distance from the pivot point to the center of the trench, and the pivot
angle.
Possible Solution:
h = height of the trapezoid
a = base 1 of the trapezoid
b = base 2 of the trapezoid
R = distance from the pivot point to the center of the trench
 = pivot angle
V = volume of a land fill
Area of trapezoid =
Arc Length = R  (
Volume =
1
h( a  b)
2

180
)
1


h( a  b)  R  (
h(a  b)  R
)=
2
180
360
Spreadsheet on the next page.

h
=A1+2
a
=B1+4
b
=C1+3
R
D2-5
E1-2
12
14
16
18
20
22
28
30
32
34
36
10
12
14
16
18
20
22
24
26
20
24
28
32
36
40
44
48
52
56
60
64
68
72
76
80
84
88
92
96
10
13
16
19
22
25
28
31
34
12
15
18
21
24
27
30
33
36
39
42
100
97
92
87
82
77
72
67
62
57
52
47
42
37
32
27
22
17
12
7
50
48
46
44
42
40
38
36
14
12
10
8
6
24
22
20
18
16
14
12
T. James 9
V
PI()/360*A1*(B1+C1)*D1*E1
15707.96
21047
25999.54
30666.34
34863.3
38435.64
48134.23
49885.35
20845.65
13800.39
12252.21
2690.6
2348.655
10415.01
10124.59
9330.53
8086.459
6475.311
4609.345
2630.141
T. James 10
5. HERON'S FORMULA
See Also: Problem Solving with Heron's Formula
Introduction
Heron's formula for the area of a triangle with sides of length a, b, c is
where
Problem:
Develop a proof of Heron's Formula for the area of a triangle.
Algebraic Proof:
h is perpendicular to the base. So using Pythagorean theorem we
know that
Also, c = p+q, q = c-p and p = c-q.
Since q = c-p, then q 2 =(c-p)
2
and q 2 =c 2 -2cp +p 2 . Adding h 2 to each side gives
h 2 +q 2 =h 2 =c 2 -2cp +p 2 . Substituting, b 2 =a 2 -2cp+c 2 solving for p we
get,
Now since
. Substitute and get an expression in terms of a, b and c.
T. James 11
T. James 12
6. Square Root of 2
Prove that the Square Root of 2 is an irrational number.
Discussion: That is, prove that this number can not be expressed as a rational fraction in lowest
terms, where the greatest common divisor of a and b is 1.
For what positive integers N does your proof generalize to show that
is irrational?
Why? or Why not?
Show
irrational.
Solution:
1. Assume that √2 is a rational number. This would mean that there exist integers a and b
such that a / b = √2.
2. Then √2 can be written as an irreducible fraction a / b such that a and b are coprime
integers and (a / b)2 = 2.
3. It follows that a2 / b2 = 2 and a2 = 2 b2.
4. Therefore a2 is even because it is equal to 2 b2 which is obviously even.
5. It follows that a must be even (odd square numbers have odd square roots and even
square numbers have even square roots).
6. Because a is even, there exists a k that fulfills: a = 2k.
7. We insert the last equation of (3) in (6): (2k)2 = 2b2 is equivalent to 4k2 = 2b2 is
equivalent to 2k2 = b2.
8. Because 2k2 is even it follows that b2 is also even which means that b is even because
only even numbers have even squares.
9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated
in (2).
Since we have found a contradiction the assumption (1) that √2 is a rational number must be
false. The opposite is proven. √2 is irrational.
This proof can be generalized to show that any root of any natural number is either a natural
number or irrational.
T. James 13
7. PROBLEM: The Box Problem

Use the Arithmetic Mean -- Geometric Mean Inequality to find the maximum volume of
a box made from a 25 by 25 square sheet of cardboard by removing a small square from
each corner and folding up the sides to form a lidless box.

Determine what shape boxes could be created by this method from the 25 by 25 square
sheet to hold a volume of 200 cu. units. 400 cu units? 800 cu units?

Generalize. Use the AM-GM Inequality to discuss the maximum volume of a box formed
from an n X n square sheet of cardboard.

Why will the AM-GM Inequality not be a useful tool when the sheet of cardboard is 20
by 25?
Solution:

Let’s name the length of the corners that were cut out x. With the corners (x) cut out, each
side has length 25-2x. The height of the box is x. Thus a box formed would have volume
2
V  x  25  2 x  = 14 (4 x)(25  2 x)(25  2 x) . Using the AM-GM inequality [ a 2 b  ab or for
this problem
1
4
a b  c
3
 3 abc ]
(4 x)(25  2 x)(25  2 x) 
1
4
 4 x  25  2 x  25  2 x 


3


3
= 14  503   1157.407407
3
So, 14 (4 x)(25  2 x)(25  2 x)  1157.407407 . Therefore, the maximum volume would be
1157.407407 cubic units. It would be equal to 1157.407407 cubic units when 4x=25-2x, 6x=25
and x = 256 = 4. 16 .
 To get V=200 cubic units
a. 0.338 x 24.324 x 24.324
b. 10.296 x 4.408 x 4.408
T. James 14
2
obtained by solving x(25-2x) = 200
To get V=400 cubic units
a. .7207 x 6.59 x 23.5586
b. 9.204 x 6.592 x 6.592
obtained by solving x(25-2x) 2 =400
To get V = 800 cubic units
a. 1.7215 x 21.557 x 21.557
b. 7.2464 x 10.5072 x 10.5072
obtained by solving x(25-2x) 2 =800
These are rounded so it won’t be exactly 200, 400 or 800.

To generalize, use AM-GM inequality to discuss the maximum volume of a box formed from
an n by n square sheet.
V=x(n-2x)(n-2x)
 4 x  n  2 x  n  2 x  1 2n 3

 =4 3 
3


3
= 14 (4x)(n-2x)(n-2x) 
1
4
So, the max volume would be
1
4
 23n 
3
cubic units, with equality when 4x=n-2x, 6x=n and x= n6

The AM-GM Inequality
 ab or for this problem a3bc  3 abc states that equality is reached if and only if a=b or in
this case a=b=c. If we have a 20 by 25 sheet of cardboard, the volume would be V=x(20-2x)(252x) = 14 (4x) (20-2x)(25-2x)
Let a=4x and b=20-2x and c=25-2x
a b
2
Can we satisfy the equality condition of a=b=c?
4x=20-2x=25-2x
4x=20-2x 4x=25-2x
6x=20
6x=25
20
x= 6
x= 256
For a 20 x 25 sheet of cardboard, the AM-GM inequality would not be a useful tool because 4x =
20-2x = 25-2x will not hold.
T. James 15
8. PROBLEM: Maximum area -- triangle.
Use the Arithmetic Mean -- Geometric Mean Inequality to show that the maximum area of a
triangular region with a given perimeter is attained when the triangle is equilateral.
Solution:
abc
2
2S=a+b+c
Semi-perimeter =S=
The area using Heron’s Formula is A= s(s  a)(s  b)(s  c)
Using the AM-GM Inequality

(s-a)(s-b)(s-c)  

s  a    s b    s  c 
3
3 s  a  b  c 

= 
3




3
3
3
=  3 s 32 s  substituting 2s for a + b + c
s3
s 3
=  3  =
27
3
(s-a)(s-b)(s-c)   3s  with equality when s-a = s-b = s-c, since a+b+c is a constant, then
S=
abc
is also a constant so a=b=c which mean an equilateral triangle. A=s
2
 3s 
3

s4
27
T. James 16
9. PROBLEM: Minimum of x + 1/x.
Use the Arithmetic Mean -- Geometric Mean Inequality to show that
Solution:
Using graphing calculator, we see that the minimum value for x, y>0, is y = f(x)=2.
Using AM-GM inequality, we know that
ab
 ab
2
1
1
x  x

*x
2
x
1
x  x
 1
2
1
x  x
1
2
1
x  x 2
Thus, the minimum value for f(x) =
1
 x is 2.
x
T. James 17
10. PROBLEM: Find AM-GM Problems
Find maximization or minimization problems for which the A.M.-G.M. inequality can be used as
a tool. Many of the traditional max/min problems from calculus are candidates (except that they
might not be interesting . . .) Create a file of good problems.
1. Two nonnegative numbers sum is 10. How large can their product be?
2. A piece of wire 100cm long is going to be cut into several pieces and used to
construct a skeleton of a rectangular box with a square base. What are the
dimensions of the box with the largest volume?
3. Federalist Express limits the size of a mailable parcel. The longest side plus the
girth (the perimeter of a cross-section perpendicular to the longest side) may not
exceed 108 inches. Which mailable cube has the largest volume?
11. Problem: Comparison of areas in square of sides a+b
Interpret the comparison of the areas of the total square and the five shown parts.
Solution:
The total area of the square = (a+b) 2 . The area of each unshaded rectangular part = ab. The
total area of the unshaded part = 4ab. We also have a shaded square part which we will denote
as I. Thus, (a+b) 2 =4ab + I. (a+b) 2 = 4ab if the area of I = 0.
So, (a+b) 2  4ab
(a+b)  2 ab
a b
ab AM-GM Inequality!
2 
T. James 18
(a+b)
2
 4ab with equality iff a=b.
12. PROBLEM: Maximum of f(x) = (1-x)(1+x)(1+x)
Find the maximum of
in the interval [0, 1].
A. Graph with TI-81, TI-82, or a computer graphing tool.
B. Use AM-GM Inequality.
C. Use calculus
AM-GM Inequality
Find the maximum of f(x) = (1-x)(1+x)(1+x)
= (2-2x)(1+x)(1+x) 
1
2
1
2
 2  2x 1 x 1 x 


3


=
x0
1
2
 43 
3
= 12
3
 6427  = 3227  1.1851852
T. James 19
with equality when
1 + x = 2 – 2x, 3x =1, x =
Calculus
1
3
 .33333333
F(x) = (1-x)(1+x) 2
=(1-x)(1+2x+x 2 )
=1+2x+ x 2 -x-2 x 2 -x 3
=1 + x - x 2 - x 3
f’(x) = 0 gives the maximum value of the function
f’(x) =1-2x-3 x 2
solving f’(x) =0,
1-2x-3 x 2 = 0
x =-1 or x= 13
Since we are restricted to [0,1], x=-1 is not allowed. So the maximum value of the function F(x)
= (1-x)(1+x) 2 is when x= 13 ( 13 , 1.185185)
13.
Given segment AB with parallel intersecting segments AE and BC on the same side of AB.
Construct the segments AC and BE, intersecting at F and construct the segment FG parallel to
AE and BC with G on AB.
Prove:
Given:
GB AB
AG AB


and
FG AE
FG BC
Adding the two equations together yields,
GB AG AB AB



FG FG AE BC
T. James 20
GB  AG AB * BC  AB * AE

FG
AE * BC
Substituting AB for GB+AG, since AB=GB+AG, and factoring yields,
AB AB ( BC  AE )

FG
AE * BC
Dividing both sides by AB yields,
1
BC  AE

FG AE * BC
Simplifying the right-hand side yields,
1
BC
AE


FG AE * BC AE * BC
=
Thus:
1
1

AE BC
1
1
1

=
FG AE BC
14. PROBLEM: Arithmetic Mean -- Geometric Mean Inequality
Find 5 different demonstrations (proofs) of the Arithmetic Mean -- Geometric Mean inequality
for two non-negative variables

diameter
A stands for Arithmetic Mean, H stands for Harmonic
Mean and G stands for Geometric Mean. a + b =
A =radius.
A
H
a
G
b
T. James 21
It’s given that segments a+b form the diameter of the circle and that A is the radius. Since it
takes two radii to equal a diameter, 2A = a+b. Solving for A we get, A = a 2 b . A is an arithmetic
mean of a and b because it is an average of a+b. Now look at the big right triangle. H + the
dashed line (the hypotenuse of the right triangle) = r because it is a segment from the center of
the circle to the edge of the circle. The short leg of the right triangle is r – b. By using the
Pythagorean Theorem, we can solve for G, r 2  (r  b)2  G 2 . This simplifies to 2rb  b 2  G 2 ,
but 2r = a+b because it is give that a+b is a diameter and 2 radii = a diameter. Now we have
(a  b)b  b2  G 2 . Simplifying that we get ab  b 2  b 2  G 2 . So, ab  G . This shows the G
is the geometric mean of a and b because it is the square root of the product a and b. The
geometric mean between two positive numbers a and b is the positive number x where ax  bx .
Solving for x the geometric mean is ab , which is what G equals. The harmonic mean between
a and b is x if 1x is the arithmetic mean of 1a and b1 . Let’s take the arithmetic mean of 1a and b1 .
 b1
. Simplifying that we have b2aba . Then we must invert which gives b2ab
. Now that
a
2
we now that b2ab
is the harmonic means of a and b, how do we know that it is the length of the
a
line segment H? Now for the inequalities, H  G  A. From a geometric point of view, the
radius will always be the biggest segment in a circle coming from the center. G is parallel to A
in a different spot of a semi-circle it has to be smaller. H is a leg where G is the hypotenuse, H
must be smaller than G.
We have

1
a
Geometric Proof:
y
x
xy
y
xy
x
I
x
xy
xy
y
y
x
A =(x+y) 2
We see that the area of each of the 4 rectangles within each square is xy.
Thus, (x+y) 2  4xy
(x+y)  2 xy
x y
2
 xy
T. James 22
AM  GM
(x+y) 2 =4xy if the area of I = 0. For the area of I to equal 0, x must equal y.
 Algebraic Proof
(x-y) 2  0 for all x,y > 0, with equality iff x =y.
x 2 -2xy+y 2  0
x 2 +4xy-2xy+y 2  4xy
x 2 +2xy+y 2  4xy
(x-y) 2  4xy
x+y  2 xy
x y
 xy
2
AM  GM

a-b
x
x
Let’s name the blue circle P and the red circle O. Given two circles, O & P, with radii b & a
respectively.
(a-b) 2 + x 2 = (a+b) 2
x 2 = (a+b) 2 - (a-b) 2
x 2 = a 2 + 2ab + b 2 - [a 2 - 2ab + b 2 ]
x 2 = 4ab
x = 4ab
x = 2 ab
a+b  x (a+b is the hypotenuse and the longest side of the triangle) and x = 2 ab substituting
a+b  2 ab
T. James 23
ab
 ab
2
AM  GM
15. Triangle Mid-Segment Theorem
Take any triangle ABC. Find the mid-point D of AB and the mid-point E or AC.
Click here to open a GSP file. Explore with measures of DE and BC as any vertex is moved.
One of the most useful theorems in geometry of plane triangles is the Triangle Mid-Segment
Theorem:
In any triangle, a segment joining the midpoints of any two sides will be parallel to
the third side and half its length.
Prove the Triangle Mid-Segment Theorem.
Solution:
I extended DE and constructed
EF  DE and CF AD.
AD  DB and AE  EC by definition of midpoint.
 AED   CEF since they are vertical angles
EF  DE by construction
AED  CEF by SAS
 A   ECF by CPCTC
AB CF as constructed
AD  FC CPCTC
DB  FC transitive property
BCFE is a parallelogram by definition
DE BC since opposite sides of a parallelogram are parallel
DF  BC since opposite sides of a parallelogram are 
DE+EF=DF segment addition postulate
DE+DE=DF substitution
2DE=DF substitution
DE= 12 DF division property of equality
T. James 24
Recall DF=BC, so ED= 12 BC
Thus ED BC and ED= 12 BC
16. Problem: Square roots relation
Characterize the set of solutions of the equation
for positive integers x, y, and z.
The set of solutions of the equation
squares.
for x,y,z > 0 is the set of perfect
Any of the perfect squares can be substituted for x, y and z. Irrational numbers wil not work,
because they can not be expressed as ba . where b can not equal 0.
17. Problem: Sum of Unit Fractions
Consider all fractions of the form
where n is a positive integer. Which of these fractions may be expressed as the sum of two
different unit fractions (reciprocals) of integers?
I only found a couple of fractions of the form
different unit fractions.
n
n1
T. James 25
, n>0, that can be expressed as the sum of two
N=3 331  43  12  14
N =5 551  65  12  13
n
n 1
 1x  1y
 yxy x
nxy=(n+1)(y+x)
nxy=ny+nx+y+x
nxy-ny-nx=x+y
n(xy-y-x)=x+y
n(xy-(x+y))=x+y
n= xy x(xy y )
n
n 1
18. Problem: Million Drops of Water
If you had a million drops of water, would you be more likely to
 drink it?

take a bath in it?

swim in it?
I would drink the water if it were clean, because as hot as it is this summer I am always thirsty.
19. PROBLEM: Volume of a Can
What is the volume of a 12 ounce can?
In other words, how may cubic inches in a 12 oz can?
Make a guess.

I don’t know 40 cubic inches.
Measure the radius and height of a soda can and compute an estimate.
T. James 26
2
V=  r 2 h =  (2in) 5in  63 cubic inches in an 8 oz can. Approximately 95 cubic inches in a
12 oz can.
How many cubic centimeters are in a 12 ounce can?

2
V=  r 2 h =  (5.5cm) 11.5cm  1092 cubic centimeters
20. PROBLEM: Big Tires
My Ford Bronco was fitted at the factory with 30 inch diameter tires. That means its
speedometer is calibrated for 30 inch diameter tires. I "enhanced" the vehicle with All Terrain
tires that have a 31 inch diameter. How will this change the speedometer readings? Specifically,
assuming the speedometer was accurate in the first place, what should I make the speedometer
read as I drive with my 31 inch tires so that the actual speed is 55 mph?
Forego the discussion of whether it is safe to drive a Bronco that fast . . .
One would expect some use of ratios here . . .

30in
31in
, x = 56.83 mph

55mph
x
21. Problem: Ladder and Box
A ladder 5 meters long leans against a wall, reaching over the top of a box that is 1
meter on each side. The box is against the wall. What is the maximum height on the
wall that the ladder can reach? The side view is:
Assume the wall is perpendicular to the floor. Use your calculator to find the
maximum height to the nearest .01 meter.
Possible Solution:
T. James 27