[email protected] mth15_lec-17_sec_3-5_added_optimization.pptx 1 bruce mayer, pe chabot...
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[email protected] • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx 1
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
§3.5 Added
Optimization
[email protected] • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx 2
Bruce Mayer, PE Chabot College Mathematics
Review §
Any QUESTIONS About• §3.4 → Optimization
& Elasticity
Any QUESTIONS About HomeWork• §3.4 → HW-16
3.4
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Bruce Mayer, PE Chabot College Mathematics
§3.5 Learning Goals
List and explore guidelines for solving optimization problems
Model and analyze a variety of optimization problems
Examine inventory control
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Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
TransLate: Words →
Math
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Bruce Mayer, PE Chabot College Mathematics
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)
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Bruce Mayer, PE Chabot College Mathematics
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
A & B are CONSTANTS
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Bruce Mayer, PE Chabot College Mathematics
Algebraic Expressions
An ALGEBRAIC EXPRESSION consists of variables, numbers, and operation signs.• Some
Examples , 2 2 , .4
yt l w m x b
When an EQUAL SIGN is placed between two expressions, an equation is formed →
374
2 ty
cmEbxmy
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Bruce Mayer, PE Chabot College Mathematics
Translate: English → Algebra
“Word Problems” must be stated in ALGEBRAIC form using Key Words
per of less than more than
ratio twicedecreased byincreased by
quotient of times minus plus
divided byproduct ofdifference of sum of
divide multiply subtract add
DivisionMultiplicationSubtractionAddition
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Bruce Mayer, PE Chabot College Mathematics
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.
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Bruce Mayer, PE Chabot College Mathematics
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
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Bruce Mayer, PE Chabot College Mathematics
Formula Describes Relationship
Relationship Mathematical Formula
Perimeter of a triangle:
a
b
ch
Area of a triangle:
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Bruce Mayer, PE Chabot College Mathematics
Example Volume of Cone
Relationship Mathematical Formulae
h
r
Volume of a cone:
Surface area of a cone:
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Bruce Mayer, PE Chabot College Mathematics
Example °F ↔ °C
Relationship Mathematical Formulae
Celsius to Fahrenheit:
Fahrenheit to Celsius:
Celsius Fahrenheit
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Bruce Mayer, PE Chabot College Mathematics
Example Mixtures
Relationship Mathematical Formula
Percent Acid, P:
Base
Acid
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Bruce Mayer, PE Chabot College Mathematics
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
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Bruce Mayer, PE Chabot College Mathematics
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
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Bruce Mayer, PE Chabot College Mathematics
Solving Application Problems3. 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)
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Bruce Mayer, PE Chabot College Mathematics
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
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Bruce Mayer, PE Chabot College Mathematics
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?
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Bruce Mayer, PE Chabot College Mathematics
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
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Bruce Mayer, PE Chabot College Mathematics
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 Venezuelan New Blend
Number of pounds i v 60
Price per pound $9.95 $11.25 $10.50
Value of beans 9.95i 11.25v 630
i + v = 60
9.95i + 11.25v = 630
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Bruce Mayer, PE Chabot College Mathematics
Example Coffee Beans cont.4
2. Translate – We have translated to a system of equations
263025.1195.9
160
vi
vi
3. Carry Out – When equation (1) is solved for v, we have: v = 60 i. • We then substitute for v in equation (2).
6306025.1195.9 ii
63025.1167595.9 ii
61.343.1454530.1 iii
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Bruce Mayer, PE Chabot College Mathematics
Example Coffee Beans cont.5
3. Carry Out – Find v using v = 60 i. 39.2561.346060 viv
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,
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Bruce Mayer, PE Chabot College Mathematics
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
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Bruce Mayer, PE Chabot College Mathematics
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?
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Bruce Mayer, PE Chabot College Mathematics
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
wlA
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Bruce Mayer, PE Chabot College Mathematics
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
wlP
wwlP
32
22
864P
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Bruce Mayer, PE Chabot College Mathematics
Example Max Enclosed Area
Need to Maximize This Fcn: 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.
Use the equation for total fencing and isolate length l:
wlA
2
3864 wl
Solving for l
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Bruce Mayer, PE Chabot College Mathematics
Example Max Enclosed Area
Now we substitute the value for l into the area equation:
Maximize this function first by finding critical points by setting the first Derivative equal to Zero
wdw
dAwwwA 34325.1432 2
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Bruce Mayer, PE Chabot College Mathematics
Example Max Enclosed Area
Set dA/dw to zero, then solve
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
334322
2
wdw
d
dw
Ad
dw
dA
dw
d
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Bruce Mayer, PE Chabot College Mathematics
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
2
3864 wl
2
)144(3864 216
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Bruce Mayer, PE Chabot College Mathematics
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
↑144yd
↓
← 216yd →
← 108yd → ← 108yd →
15 552 yd2 15 552 yd2
[email protected] • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx 33
Bruce Mayer, PE Chabot College Mathematics
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
[email protected] • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx 34
Bruce Mayer, PE Chabot College Mathematics
Example Find Minimum Cost
SOLUTION: Translate: at any given time, t,
Or Now for CP → production cost
associated with the cPad is inversely proportional to the length of time• Formulaically
TotalCost = ProductionCost + MaintenanceCost
tCtCtC MPT
t
KtCP
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Bruce Mayer, PE Chabot College Mathematics
Example Find Minimum Cost– t ≡ time in Weeks– K ≡ The Constant of
ProPortionality in k$·weeks
Now for CM → maintenance costs are linear and increasing• Translated to a Eqn
– m ≡ Slope Constant (positive) in $k/week– b ≡ Intercept Constant (positive) in $k
Then, the Total Cost
t
KtCP
bmttCM
bmtt
KtCtCtC MPT
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Bruce Mayer, PE Chabot College Mathematics
Example Find Minimum Cost
find potential extrema by solving the derivative function set equal to zero:
Since t MUST be POSITIVE →
mt
Kbmt
t
K
dt
dtC
dt
dT
2
mKt 20
m
Kt
K
m
tK
mt 2
22 1
m
Kt min
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Bruce Mayer, PE Chabot College Mathematics
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:
At theZeroValue
3
22
2t
KmKt
dt
dm
t
K
dt
d
dt
tdC
dt
d T
32
2
2mK
K
dt
Cd
mK
T
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Bruce Mayer, PE Chabot College Mathematics
Example Find Minimum Cost
Since K and m are BOTH Positive then
Is also Positive
The 2nd Derivative Test Confirms that the function is ConCave UP at the zero point, which confirms the MINIMUM
The Min Cost:
32
2
2mK
K
dt
Cd
mK
T
bKmKmbm
Km
mK
KtCT minmin
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Bruce Mayer, PE Chabot College Mathematics
Example Find Minimum Cost
STATE: for the cPad• Minimum Total Cost
will occur at this many weeks
• The Total Cost at this time in $k
m
Kt min
bKmtCT 2minmin
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Bruce Mayer, PE Chabot College Mathematics
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?
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time
SOLUTION: Need to TransLate
Words to MathRelations
First DRAW DIAGRAM Letting:• x ≡ The SideWalk
Distance• d ≡ The DiaGonal Grass Distance
•
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Bruce Mayer, PE Chabot College Mathematics
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 “Rate” Eqn for Time:
So the time spent traveling on the SideWalk at 4 ft/s
rate
distancetime
secft 41
xt
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time
Next, the time on Grass →
Writing in terms of x requires the use of the Pythagorean Theorem:
Then
222 300200
2
1
2 x
dt
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time
And the Total Travel Time, t, is the SideWalk-Time Plus the Grass-Time
Now Set the 1st Derivative to Zero to find tmin
2
300200
4
22
21
xxttt
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time
Continuing with the Reduction
OR
2
122
22
3002002
1
42
300200
40 x
x
dx
dxx
dx
d
1
2200300200
2
1
4
1 2/122 xx
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time
UsingMoreAlgebra
x
x
200
300200
1
2
122
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time
WithYetMoreAlgebra
But the Diagram shows that x can NOT be more than 200ft, thus 26.79ft is the only relevant location of a critical point
21.373 OR 79.26 xx
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Bruce Mayer, PE Chabot College Mathematics
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 Direction-Diagram (Sign Charts)– Instead, check the critical point against the two
endpoints on either side of x = 26.79; say x=0 & x=200
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time
Findt-Valueat x=0,x=26.79 andx=200
23002004 22 xxxt
23000200400 22 t
3.1800 t
230079.26200479.2679.26 22 t
9.17979.26 t
23002002004200200 22 t
200200 t
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time
Finddt/dxSlopeat x=0 andx=200
22 300200
200
2
1
4
1
x
x
dx
dt
220 3000200
0200
2
1
4
1
xdx
dt
ftsec 0273.00 xdxdt
22200 300200200
200200
2
1
4
1
xdx
dt
ftsec 250.0200 xdxdt
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel TimeValue Summary Bar Chart
t is smallest at x = 26.79
Slope Summary Bar Chart
Slopes have different SIGNS on either sideof x = 26.79
1 2 3179
179.2
179.4
179.6
179.8
180
180.2
180.4
180.6
180.8
181
x = [0, 26.79, 200]
t(x)
(se
c)
MTH15 • t(x) Value-Chart
1 2 3-0.05
0
0.05
0.1
0.15
0.2
x = [0, 26.79, 200]
dt/d
x (s
ec/
ft)
MTH15 • dt/dx Slope-Chart
Bruce May er, PE • 15Jul13
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Bruce Mayer, PE Chabot College Mathematics
MA
TL
AB
Co
de
% Bruce Mayer, PE% MTH-15 • 15Jul13%% The Bar Values for x = [0 26.79 200]t = [180.3 179.9 200]% % the Bar Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenbar(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)
% Bruce Mayer, PE% MTH-15 • 15Jul13%% The Bar Values for x = [0 26.79 200]t = [-0.0273 0 .25]% % the Bar Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenbar(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)
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time The T-Tables for the
Value and Slope Diagrams
Both the Value & Slope Analyses confirm that x ≈ 26.79 is an absolute minimum
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.
x t (x ) x dt /dx0 180.3 0 -0.0273
26.79 179.9 26.79 0.0000200 200.0 200 0.2500
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Bruce Mayer, PE Chabot College Mathematics
Example Minimize Travel Time Plot by MuPAD
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Bruce Mayer, PE Chabot College Mathematics
Mu
PA
D C
od
e
Bruce Mayer, PEMTH15 • 15Jul13MTH15_Minimize_Travel_Time_1307.mn
t := x/4 + sqrt((200-x)^2+300^2)/2
t0 := subs(t, x = 0)
float(t0)
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) " )
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Bruce Mayer, PE Chabot College Mathematics
MuPAD Code
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Bruce Mayer, PE Chabot College Mathematics
WhiteBoard Work
Problems From §3.5• P9 → GeoMetry
+ Calculus• Special Prob →
Enclosure Cost– Total Enclosed
Area = 1600 ft2
– Fence Costs in $/Lineal-FtStraight = 30Curved = 40
See File →MTH15_Lec-17a_Fa13_sec_3-5_Round_End_Fence_Enclosu
re.pptx
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Bruce Mayer, PE Chabot College Mathematics
All Done for Today
TravelTimeor
TimeTravel?
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Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
Appendix
–
srsrsr 22
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Bruce Mayer, PE Chabot College Mathematics
ConCavity Sign Chart
a b c
−−−−−−++++++ −−−−−− ++++++
x
ConCavityForm
d2f/dx2 Sign
Critical (Break)Points Inflection NO
InflectionInflection
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Bruce Mayer, PE Chabot College Mathematics
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Bruce Mayer, PE Chabot College Mathematics
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Bruce Mayer, PE Chabot College Mathematics
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Bruce Mayer, PE Chabot College Mathematics