Chabot Mathematics
§3.5 Added
Optimization
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Review §
3.4
 Any QUESTIONS
About
• §3.4 → Optimization
& Elasticity
 Any QUESTIONS
About HomeWork
• §3.4 → HW-16
Chabot College Mathematics
2
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
§3.5 Learning Goals
 List and explore
guidelines for solving
optimization problems
 Model and analyze a
variety of optimization
problems
 Examine inventory
control
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Chabot Mathematics
TransLate:
Words → Math
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Applications Tips
 The Most Important Part of Solving
REAL WORLD (Applied Math) Problems
 The Two Keys to the Translation
• Use the LET Statement to ASSIGN
VARIABLES (Letters) to Unknown Quantities
• Analyze the RELATIONSHIP Among the
Variables and Constraints (Constants)
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Basic Terminology
 A LETTER that can be any one of
various numbers is called a VARIABLE.
 If a LETTER always represents a
particular number that NEVER
CHANGES, it is called a CONSTANT
Chabot College Mathematics
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A & B are
CONSTANTS
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Algebraic Expressions
 An ALGEBRAIC EXPRESSION
consists of variables, numbers, and
operation signs.
• Some
y

t
,
2

l

2

w
,
m

x

b
.
Examples 4
 When an EQUAL SIGN is placed
between two expressions, an equation
is formed →
y
2
y  m x b
E  mc
 t  37
4
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Translate: English → Algebra
 “Word Problems” must be stated in
ALGEBRAIC form using Key Words
Addition
add
sum of
plus
Subtraction
Multiplication
subtract
multiply
difference of
product of
minus
times
increased by decreased by
more than
Chabot College Mathematics
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less than
Division
divide
divided by
quotient of
twice
ratio
of
per
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Translation
 Translate this Expression:
Eight more than twice
the product of 5 and a number
 SOLUTION
• LET n ≡ the UNknown Number
8  2  5  n
Eight more than twice the product of 5 and a number.
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Mathematical Model
 A mathematical model is an
equation or inequality that
describes a real situation.
 Models for many applied (or
“Word”) problems already exist and
are called FORMULAS
 A FORMULA is a mathematical
equation in which variables are
used to describe a relationship
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Formula Describes Relationship
Relationship
Mathematical Formula
Perimeter of a triangle:
a
c
h
Area of a triangle:
b
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Volume of Cone
Relationship
Mathematical Formulae
Volume of a cone:
h
Surface area of a cone:
r
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  °F ↔ °C
Relationship
Celsius
Fahrenheit
Mathematical Formulae
Celsius to Fahrenheit:
Fahrenheit to Celsius:
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Mixtures
Relationship
Mathematical Formula
Percent Acid, P:
Acid
Base
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Solving Application Problems
1. Read the problem as many times as
needed to understand it thoroughly.
Pay close attention to the questions
asked to help identify the quantity the
variable(s) should represent. In other
Words, FAMILIARIZE yourself with the
intent of the problem
•
Often times performing a GUESS &
CHECK operation facilitates this
Familiarization step
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Solving Application Problems
2. Assign a variable or variables to
represent the quantity you are looking
for, and, when necessary, express all
other unknown quantities in terms of
this variable. That is, Use at LET
statement to clearly state the
MEANING of all variables
•
Frequently, it is helpful to draw a
diagram to illustrate the problem or to set
up a table to organize the information
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Solving Application Problems
3. Write an equation or equations that
describe(s) the situation. That is,
TRANSLATE the words into
mathematical Equations
4. Solve the equation; i.e., CARRY OUT
the mathematical operations to solve
for the assigned Variables
5. CHECK the answer against the
description of the original problem (not
just the equation solved in step 4)
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Solving Application Problems
6. Answer the question
asked in the
problem. That is,
make at
STATEMENT in
words that clearly
addressed the
original question
posed in the
problem description
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Mixture Problem
 A coffee shop is considering a new
mixture of coffee beans. It will be
created with Italian Roast beans costing
$9.95 per pound and the Venezuelan
Blend beans costing $11.25 per pound.
The types will be mixed to form a 60-lb
batch that sells for $10.50 per pound.
 How many pounds of each type
of bean should go into the blend?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Coffee Beans cont.2
1. Familiarize – This problem is similar
to our previous examples.
•
•
•

Instead of pizza stones we have coffee beans
We have two different prices per pound.
Instead of knowing the total amount paid, we
know the weight and price per pound of the
new blend being made.
LET:
•
•
i ≡ no. lbs of Italian roast and
v ≡ no. lbs of Venezuelan blend
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Coffee Beans cont.3
2. Translate – Since a 60-lb batch is
being made, we have i + v = 60.
•
Present the information in a table.
Italian
Number of
pounds
v
Price per
pound
$9.95 $11.25
Value of
beans
9.95i
Chabot College Mathematics
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i
Venezuelan New Blend
11.25v
60
i + v = 60
$10.50
630
9.95i + 11.25v = 630
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Coffee Beans cont.4
2. Translate – We have translated
1
i  v  60
to a system
of equations 9.95i  11.25v  630 2
3. Carry Out – When equation (1) is
solved for v, we have: v = 60  i.
•
We then substitute for v in equation (2).
9.95i  11.2560  i  630
9.95i  675  11.25i  630
 1.30i  45  i  45 1.3  i  34.61
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Coffee Beans cont.5
3. Carry Out – Find v using v = 60  i.
v  60  i  60  34.61  v  25.39
4. Check – If 34.6 lb of Italian Roast and
25.4 lb of Venezuelan Blend are
mixed, a 60-lb blend will result.
•
The value of 34.6 lb of Italian beans is
34.6•($9.95), or $344.27.
•
The value of 25.4 lb of Venezuelan Blend
is 25.4•($11.25), or $285.75,
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Coffee Beans cont.6
4. Check – cont.
•
so the value of the blend is [$344.27 +
$285.75] = $630.02.
•
A 60-lb blend priced at $10.50 a pound is
also worth $630, so our answer checks
5. State – The blend should be made from
•
•
34.6 pounds of Italian Roast beans
25.4 pounds of Venezuelan Blend beans
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Max Enclosed Area
 A rancher wants to build rectangular
enclosures for her cows and horses.
She divides the rectangular space in
half vertically, using fencing to separate
the groups of animals and surround the
space.
 If she has purchased 864 yards of
fencing, what dimensions give the
maximum area of the total space and
what is the area of each enclosure?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Max Enclosed Area
 SOLUTION:
 First Draw Diagram,
Letting
• w ≡ Enclosure
Width in yards
• l ≡ Enclosure Length in yards
 Then the total Enclose Area for the
large Rectangle
A  lw
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Max Enclosed Area
 The fencing required
for the enclosure is
the perimeter of the
rectangle plus the
length of the vertical
fencing between
enclosures
P  2l  2 w  w
P  2l  3w
P  864
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Max Enclosed Area
 Need to Maximize This Fcn: A  l  w
 However the fcn includes TWO
UNknowns: length and width.
• Need to eliminate one variable (either one)
in order to Product a function of one
variable to maximize.
2l + 3w = 864
 Use the equation
Solving for l
for total fencing
864  3w
and isolate length l:
l
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Max Enclosed Area
 Now we substitute
the value for l into
the area equation:
A =l×w
æ 864 - 3w ö
=ç
÷× w
è
ø
2
 Maximize this function first by finding
critical points by setting the first
Derivative equal to Zero
Aw  432 w  1.5w
Chabot College Mathematics
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2

dA
 432  3w
dw
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Max Enclosed Area
 Set dA/dw to zero, 0 = 432 -3w
432
then solve
w=
=144.
3
 Since There is only
one critical point, the Extrema at
w = 144 is Absolute
 Thus apply the second derivative test
(ConCavity) to determine max or min
d  dA  d 2 A d
432  3w  3

 
2
dw  dw  dw
dw
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Max Enclosed Area
 Since d 2A/dw 2 is ALWAYS Negative,
then the A(w) curve is ConCave DOWN
EveryWhere
• Thus a MAX exists at w = 144
 Now find the length of the total space
using our perimeter equation when
solved for length
864  3w 864  3(144)
l

 216
2
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Max Enclosed Area
 Then The total space should be a 144yd
by 216yd Rectangle. Each enclosure
then is 144 yards wide and 216/2 - 108
yards long, and the area of each is
144yd·108yd = 15 552 sq-yd
← 216yd →
↑
144yd
↓
Chabot College Mathematics
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15 552 yd2
15 552 yd2
← 108yd →
← 108yd →
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Find Minimum Cost
 The daily production cost associated
with a company’s principal product, the
ChabotPad (or cPad), is inversely
proportional to the length of time, in
weeks, since the cPad’s release.
• Also, maintenance costs are linear and
increasing.
 At what time is total cost minimized?
• The answer may contain constants
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Find Minimum Cost
 SOLUTION:
 Translate: at any given time, t,
TotalCost = ProductionCost + MaintenanceCost
 Or CT t   CP t   CM t 
 Now for CP → production cost
associated with the cPad is inversely
proportional to the length of time
• Formulaically
Chabot College Mathematics
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K
C P t  
t
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Find Minimum Cost
– t ≡ time in Weeks
– K ≡ The Constant of
ProPortionality in k$·weeks
K
C P t  
t
 Now for CM → maintenance costs are
linear and increasing
• Translated to a Eqn
CM t   mt  b
– m ≡ Slope Constant (positive) in $k/week
– b ≡ Intercept Constant (positive) in $k
 Then, the Total Cost
K
CT t   C P t   CM t    mt  b 
t
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Find Minimum Cost
 find potential extrema by solving the
derivative function set equal to zero:
d
d K
K

CT t     mt  b    2  m
dt
dt  t
t

0   Kt 2  m
m
1 m
K
2
2
t   2  t 
K
t
K
m
 Since t MUST
K
be POSITIVE → tmin  m
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Find Minimum Cost
 Now use the second derivative test for
absolute extrema to verify that this
value of t produces a positive
ConCavity (UP) which confirm a
minimum value for cost:
d  dCT t   d  K
K
 d
2

    2  m    Kt  m   2 3
dt  dt  dt  t
t
 dt
 At the
Zero
Value
Chabot College Mathematics
37
d 2CT
dt 2
2
K m

K
K m

3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Find Minimum Cost
 Since K and m are BOTH Positive then
d 2CT
K
Is also Positive
2
2
3
dt 
K m

K m

 The 2nd Derivative Test Confirms that
the function is ConCave UP at the zero
point, which confirms the MINIMUM
 The Min Cost:
K
K
CT min tmin  
m
 b  Km  Km  b
m
K m
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Find Minimum Cost
 STATE: for the cPad
• Minimum Total Cost
will occur at this
many weeks
tmin
K

m
• The Total Cost at this time in $k
CT min t min   2 Km  b
Chabot College Mathematics
39
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
 Gonzalo walks west on a sidewalk along the
edge of the grass in front of the Education
complex of the University of Oregon.
 The grassy area is 200 feet East-West and
300 feet North-South. Gonzalo strolls at
• 4 ft/sec on sidewalk
• 2 ft/sec on grass.
 From the NE corner how long should he walk
on the sidewalk before cutting diagonally
across the grass to reach the SW corner of
the field in the shortest time?
Chabot College Mathematics
40
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
 SOLUTION:
 Need to TransLate
Words to Math
Relations
 First DRAW
DIAGRAM Letting:
• x ≡ The SideWalk
Distance
• d ≡ The DiaGonal Grass Distance
Chabot College Mathematics
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•
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
 The total distance traveled is x+d,
and need to minimize the time spent
traveling, so use the physical
relationship [Distance] = [Speed]·[Time]
 Solving the above
distance
time 
“Rate” Eqn for Time:
rate
 So the time spent
x
t1 
traveling on the
4 ft sec
SideWalk at 4 ft/s
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
 Next, the time
d
t
=
2
on Grass →
2
 Writing in terms of x
requires the use of the
Pythagorean Theorem:
d=
( 200 - x)
2
+ 300
2
 Then
d 1
t2  
2 2
Chabot College Mathematics
43
200  x 2  3002
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
 And the Total Travel Time, t, is the
SideWalk-Time Plus the Grass-Time
x
t  t1  t 2  
4
200  x 
2
 300 2
2
 Now Set the 1st Derivative to Zero to
find tmin
2
é
2 ù
d êx
0=
+
dx ê 4
ë
Chabot College Mathematics
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( 200 - x )
+ 300 ú
ú
2
û
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
 Continuing with the Reduction
d x
 
0
dx  4

200  x 2  3002 
2
 OR

(
1 1 1
2
0 = + × ( 200 - x ) + 300 2
4 2 2

1
 1
2
2





200

x

300

2
 4
Chabot College Mathematics
45

1
d x 1
2
2 2

  200  x   300 
 dx  4 2


)
-1/2
1 / 2
× [ 2(200 - x)× (-1)]
 2 
 200  x  
 1 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
(
æ1
ç
è2
Chabot College Mathematics
46
)
-1/2
1
2
2
= ( 200 - x ) + 300
× [ 200 - x ]
2
1
1

 200  x
2
2
200  x   3002
 Using
More
Algebra
1
2
( 200 - x )
2
+ 300 = 200 - x
2
2
2
2ö
( 200 - x ) + 300 ÷ø = ( 200 - x )
1
2
1
2
2
( 200 - x) + × 300 = ( 200 - x )
4
4
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
3
2
1
 With
2
- ( 200 - x ) = - × 300
4
4
Yet
2
( 200 - x) = 30000
More
Algebra
200 - x = ± 30000
x = 200 ± 30000
x  26.79 OR x  373.21
 But the Diagram shows that x can NOT
be more than 200ft, thus 26.79ft is the
only relevant location of a critical point
Chabot College Mathematics
47
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
 Alternatives to check for max OR min:
• 2nd Derivative Test
– We could check using the second derivative
test for absolute extrema to see if 26.79
corresponds to an absolute minimum, but that
involves even more messy calculations beyond
what we’ve already accomplished.
• Slope Value-Diagram and DirectionDiagram (Sign Charts)
– Instead, check the critical point against the two
endpoints on either side of x = 26.79;
say x=0 & x=200
Chabot College Mathematics
48
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
2
 Find
t x   x 4  200  x   3002 2
t-Value
2
2




t
0

0
4

200

0

300
2
at x=0,
x=26.79
t 0  180.3
and
2
2
x=200 t 26.79  26.79 4  200  26.79  300 2
t 26.79  179.9
t 200  200 4  200  200  3002 2
t 200   200
2
Chabot College Mathematics
49
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
 Find
dt/dx
Slope
at x=0
and
x=200
dt 1 1
 
dx 4 2
dt
dx
x 0
dt
dt
dx
50
200  x 2  3002
200  0
200  0
2
 300 2
dxx0  0.0273 sec ft
1 1
 
4 2
x  200
dt
Chabot College Mathematics
1 1
 
4 2
200  x
200  200
200  200
2
 300 2
dxx 200  0.250 sec ft
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
Value Summary Bar Chart
Slope Summary Bar Chart
MTH15 • t(x) Value-Chart
MTH15 • dt/dx Slope-Chart
181
Bruce May er, PE • 15Jul13
180.8
0.2
180.6
dt/dx (sec/ft)
t(x) (sec)
180.4
180.2
180
179.8
179.6
179.4
0.15
0.1
0.05
0
179.2
179
1
2
3
x = [0, 26.79, 200]
 t is smallest at x = 26.79
Chabot College Mathematics
51
-0.05
1
2
3
x = [0, 26.79, 200]
 Slopes have different
SIGNS on either side
of x = 26.79 Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
% Bruce Mayer, PE
% MTH-15 • 15Jul13
%
% The Bar Values for
x = [0 26.79 200]
t = [-0.0273 0 .25]
%
% the Bar Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
bar(t),axis([0 4 -.05 .25]),...
grid, xlabel('\fontsize{14}x = [0, 26.79, 200]'),
ylabel('\fontsize{14}dt/dx (sec/ft)'),...
title(['\fontsize{16}MTH15 • dt/dx Slope-Chart',]),...
annotation('textbox',[.15 .8 .0 .1], 'FitBoxToText', 'on', 'EdgeColor',
'none', 'String', 'Bruce Mayer, PE • 15Jul13 ','FontSize',7)
Chabot College Mathematics
52
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 15Jul13
%
% The Bar Values for
x = [0 26.79 200]
t = [180.3 179.9 200]
%
% the Bar Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
bar(t),axis([0 4 179 181]),...
grid, xlabel('\fontsize{14}x = [0, 26.79, 200]'),
ylabel('\fontsize{14}t(x) (sec)'),...
title(['\fontsize{16}MTH15 • t(x) Value-Chart',]),...
annotation('textbox',[.15 .8 .0 .1], 'FitBoxToText', 'on', 'EdgeColor',
'none', 'String', 'Bruce Mayer, PE • 15Jul13 ','FontSize',7)
Example  Minimize Travel Time
 The T-Tables for the
Value and Slope
Diagrams
x
0
26.79
200
t (x )
180.3
179.9
200.0
x
0
26.79
200
dt /dx
-0.0273
0.0000
0.2500
 Both the Value &
Slope Analyses
confirm that
x ≈ 26.79 is an
absolute minimum
Chabot College Mathematics
53
 In other words, if
Gonzalo walks on
the sidewalk for
about 26.79 feet and
then walks directly
to the southwest
corner through the
grass, he will spend
the minimum time of
about 179.90
seconds walking.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Example  Minimize Travel Time
 Plot by MuPAD
Chabot College Mathematics
54
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Bruce Mayer, PE
MTH15 • 15Jul13
MTH15_Minimize_Travel_Time_1307.mn
t := x/4 + sqrt((200-x)^2+300^2)/2
t0 := subs(t, x = 0)
float(t0)
MuPAD Code
t200 := subs(t, x=200)
dtdx := Simplify(diff(t,x))
u := solve(dtdx=0, x)
float(u)
dtdx0 := subs(dtdx, x = 0)
float(dtdx0)
tmin := subs(t, x = u)
float(tmin)
dtdx200 := subs(dtdx, x = 200)
float(dtdx200)
plot(t, x =0..200, GridVisible = TRUE,
LineWidth = 0.04*unit::inch,
plot::Scene2d::BackgroundColor = RGB::colorName([.8, 1, 1]),
XAxisTitle = " x (ft) ", YAxisTitle = " t (s) " )
Chabot College Mathematics
55
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
MuPAD Code
Chabot College Mathematics
56
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
WhiteBoard Work
 Problems From §3.5
• P9 → GeoMetry
+ Calculus
• Special Prob →
Enclosure Cost
– Total Enclosed
Area = 1600 ft2
– Fence Costs in
$/Lineal-Ft
 Straight = 30
 Curved = 40
Chabot College Mathematics
57
See File →
MTH15_Lec-17a_Fa13_sec_35_Round_End_Fence_Enclosu
re.pptx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
All Done for Today
TravelTime
or
TimeTravel?
Chabot College Mathematics
58
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
59
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
ConCavity Sign Chart
ConCavity
Form
d2f/dx2 Sign
++++++
Critical (Break)
Points
Chabot College Mathematics
60
−−−−−−
a
Inflection
−−−−−−
b
NO
Inflection
++++++
c
Inflection
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
x
Chabot College Mathematics
61
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Chabot College Mathematics
62
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Chabot College Mathematics
63
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx
Chabot College Mathematics
64
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx