BMayer@ChabotCollege.edu MTH15_Lec-17_sec_3-5_Added_Optimization.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

BMayer@ChabotCollege.edu • MTH15_Lec-17_sec_3-5_Added_Optimization.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot Mathematics

§3.5 Added

Optimization

Review §

Any QUESTIONS About• §3.4 → Optimization

& Elasticity

Any QUESTIONS About HomeWork• §3.4 → HW-16

§3.5 Learning Goals

List and explore guidelines for solving optimization problems

Model and analyze a variety of optimization problems

Examine inventory control

Chabot Mathematics

TransLate: Words →

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)

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

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 →

cmEbxmy

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

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.

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

Formula Describes Relationship

Relationship Mathematical Formula

Perimeter of a triangle:

Area of a triangle:

Example Volume of Cone

Relationship Mathematical Formulae

Volume of a cone:

Surface area of a cone:

Example °F ↔ °C

Relationship Mathematical Formulae

Celsius to Fahrenheit:

Fahrenheit to Celsius:

Celsius Fahrenheit

Example Mixtures

Relationship Mathematical Formula

Percent Acid, P:

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

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

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)

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

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?

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

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

2. Translate – We have translated to a system of equations

263025.1195.9

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

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,

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

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?

SOLUTION: First Draw Diagram,

Letting• w ≡ Enclosure

Width in yards• l ≡ Enclosure Length in yards

Then the total Enclose Area for the large Rectangle

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

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:

3864 wl

Solving for l

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

dAwwwA 34325.1432 2

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

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

3864 wl

)144(3864 216

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

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

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

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

bmttCM

KtCtCtC MPT

find potential extrema by solving the derivative function set equal to zero:

Since t MUST be POSITIVE →

mKt 20

Kt min

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

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:

bKmKmbm

KtCT minmin

STATE: for the cPad• Minimum Total Cost

will occur at this many weeks

• The Total Cost at this time in $k

Kt min

bKmtCT 2minmin

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?

Example Minimize Travel Time

SOLUTION: Need to TransLate

Words to MathRelations

First DRAW DIAGRAM Letting:• x ≡ The SideWalk

Distance• d ≡ The DiaGonal Grass Distance

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

distancetime

secft 41

Next, the time on Grass →

Writing in terms of x requires the use of the Pythagorean Theorem:

222 300200

And the Total Travel Time, t, is the SideWalk-Time Plus the Grass-Time

Now Set the 1st Derivative to Zero to find tmin

300200

Continuing with the Reduction

3002002

300200

2200300200

1 2/122 xx

UsingMoreAlgebra

300200

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

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

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

Finddt/dxSlopeat x=0 andx=200

22 300200

220 3000200

ftsec 0273.00 xdxdt

22200 300200200

200200

ftsec 250.0200 xdxdt

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

x = [0, 26.79, 200]

MTH15 • t(x) Value-Chart

1 2 3-0.05

x = [0, 26.79, 200]

MTH15 • dt/dx Slope-Chart

Bruce May er, PE • 15Jul13

% 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)

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

Example Minimize Travel Time Plot by MuPAD

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) " )

MuPAD Code

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

All Done for Today

TravelTimeor

TimeTravel?

Chabot Mathematics

Appendix

srsrsr 22

ConCavity Sign Chart

−−−−−−++++++ −−−−−− ++++++

ConCavityForm

d2f/dx2 Sign

Critical (Break)Points Inflection NO

InflectionInflection

BMayer@ChabotCollege.edu MTH15_Lec-17_sec_3-5_Added_Optimization.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

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