[email protected] MTH15_Lec-16_sec_3-4_Optimization.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical

[email protected] • MTH15_Lec-16_sec_3-4_Optimization.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§3.4 Elasticity

& Optimization



Review §

Any QUESTIONS About• §3.3 → Graph

Sketching

Any QUESTIONS About HomeWork• §3.3 → HW-15

3.3

http://www.google.com/url?sa=i&rct=j&q=sketch+graph+calculus&source=images&cd=&cad=rja&docid=hIPjKPr2xN1YgM&tbnid=J0Hmn4qU-RezXM:&ved=0CAUQjRw&url=http://www.popscreen.com/v/6koIm/AP-Calculus-Stillwater-Sketch-Second-Derivative-Graphs&ei=xMTiUcaXD6PeiALQo4CoDw&bvm=bv.48705608,d.cGE&psig=AFQjCNGsCuxZoz6GhUPKh1EGkQStHz7T0g&ust=1373902377250416



§3.4 Learning Goals

Use the extreme value property to find absolute extrema

Compute absolute extrema in applied problems

Study optimization principles in economics

Define and examine price elasticity of demand

http://www.google.com/imgres?imgurl&imgrefurl=http://economicseducationhs.blogspot.com/2012/10/elasticity-of-demand.html&h=0&w=0&sz=1&tbnid=ko0_FRvzH5-o9M&tbnh=191&tbnw=264&prev=/search?q=elasticity+of+demand&tbm=isch&tbo=u&zoom=1&q=elasticity%20of%20demand&docid=76KCkB4itEbGVM&hl=en&ei=fcbiUcX9EIObjAKdh4D4Dg&ved=0CAEQsCU

http://www.google.com/imgres?imgurl&imgrefurl=http://economicseducationhs.blogspot.com/2012/10/elasticity-of-demand.html&h=0&w=0&sz=1&tbnid=Avtao6ecqiJU5M&tbnh=190&tbnw=265&prev=/search?q=elasticity+of+demand&tbm=isch&tbo=u&zoom=1&q=elasticity%20of%20demand&docid=76KCkB4itEbGVM&hl=en&ei=fcbiUcX9EIObjAKdh4D4Dg&ved=0CAIQsCU



Absolute Extrema

A function f has an absolute maximum of

if for every x in an open interval containing c,

A function f has an absolute minimum of

if for every x in an open interval containing c,

)()( xfcf

)()( xfcf



Example Absolute Extrema

Consider the Function Graph Shown at Right

The function appears to have an absolute maximum of 7 at x = 1, and an absolute minimum of 2 at x = 6 (and at x = 12).

maxmax , cfc

minmin , cfc



Extreme Value Property

All absolute extrema of a continuous function on a closed interval are found at one of:• a CRITICAL point on the interval• an ENDPOINT of the interval

ReCall Critical Points: • Let c be an x-value in the domain of f• If [df/dx]c =0 OR [df/dx]c →±∞, then f has a

Critical Point at c



Extreme Value Explained

The Absolute-Max or Absolute-Min over some x-span of any function occurs EITHER at • The ENDS• SomeWhere

in the Middle

• (Duh!!!)

Abs-MAX Abs-MAX

Abs-MIN

http://www.google.com/url?sa=i&rct=j&q=calculus+critical+points&source=images&cd=&cad=rja&docid=37WK2DnIIdtMfM&tbnid=U6FyjJ0ZBuu73M:&ved=0CAUQjRw&url=http://tutorial.math.lamar.edu/problemsns/calci/criticalpoints.aspx&ei=HdDiUZfIJImWigLG-YAg&psig=AFQjCNEglluzgsm4oFiJfEmp7XUyZnDKBg&ust=1373904515226804



Example Find Abs Extrema

Find the absolute extrema for the fcn at Right on the interval [−3,1]

SOLUTION: As indicated by the Extreme Value

Property, we need to check the values of the function at:• Any CRITICAL points and Both ENDpoints

– The LARGEST of these values is the absolute MAX, the smallest is the absolute min

2

2

x

xxf




First, find critical points by setting the derivative equal to zero and solving:

Recall, however, that the interval of interest is [−3,1]• Thus x=4 is NOT

part of the Slon

2

'02

x

x

dx

dxf

2

2

2

12)2(0

x

xxx

2

2

2

22

2

4

2

420

x

xx

x

xxx

4or 0 xx

QuotientRule

ZeroProducts




The endpoints are −3 and 1. So need compare the value of f(x) at x=−3 & x=1, and also at the critical point, x=0 on the interval. • Making a T-Table:

The Table shows that the absolute max is 0 (attained at x = 0) and absolute min is −1.8 (attained at x = −3).

x y=f (x )-3 -1.800 0.001 -1.00

13for

2

2

xx

xxf




ThefcnGraph

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2

-1.5

-1

-0.5

0

0.5

x

y =

f(x)

= x

2 /(x-

2)

MTH15 • Bruce Mayer, PE

XYf cnGraph6x6BlueGreenBkGndTemplate1306.m

13for

2

2

xx

xxf



Marginal Analysis for Max Profit Given:

• R ≡ annual Revenue in $

• C ≡ annual Cost in $• P ≡ annual Profit in $• q ≡ annual quantity

sold in Units

Then the absolute maximum of P occurs at the production level for which:

and

That is• where marginal revenue

equals marginal cost• The CHANGE in the R-

Slope is less than the CHANGE in the C-Slope

UNIT

$in

maxmax qqdq

qdC

dq

qdR

22

2

2

2

UNIT

$in

maxmax qqdq

qCd

dq

qRd



Example Finding Max Profit

The Math Model for Pricingof “StillStomping” Staplers:• Where

– p ≡ Stapler Selling Price in $/Unit– q ≡ Qty of Staplers Sold in kUnit

The Total Cost model for the StillStomping Staplers:• Where

– C ≡ Stapler Production Cost in $k

210050 qqp

10

410 2 qqqC




Use marginal analysis to find the production level at which profit is maximized, as well as the amount of the maximum profit.

SOLUTION: The marginal analysis criterion for

maximum Profit specifies that we should examine where marginal revenue equals marginal cost




Now Total Revenue = [price]·[Qty-Sold]

• R in (kUnit)·($/Unit) = k$

The Marginal Analysis requires Derivatives for R & C

x

210050 qqqpqR

23 3005010050 qqqdq

d

dq

dR

12010

410 2

qqq

dq

d

dq

dC




Now set equal the two marginal functions and solve using the quadratic formula or a computer algebra system such as MuPAD (c.f., MTH25)

dq

dCqq

dq

dR 12030050 2

49203000 2 qq

4389.0or3722.030

3721

qqq

Qty, q, canNOT be Negative




The negative solution makes no sense in this context as production level is always non-negativve. Thus have one solution at q ≈0.372k, or 372 staplers.

The maximum profit is

372.0372.0max CRP

k296.11$max P

4.0372.0372.010372.0100372.050 23max P



Ex:

Fin

d M

ax Pro

fit

0 0.1 0.2 0.3 0.4 0.5 0.60

2

4

6

8

10

12

Staplers (kUnit)

R &

C (

$)


RevenueCostMaxProfit,



MA

TL

AB

Co

de

% Bruce Mayer, PE% MTH-15 • 01Aug13 • Rev 11Sep13% MTH15_Quick_Plot_BlueGreenBkGnd_130911.m%clear; clc; clf; % clf clears figure window%% The Domain Limitsxmin = 0; xmax = .6;% The FUNCTION **************************************x = linspace(xmin,xmax,10000); y1 = 50*x - 100*x.^3;y2 = 10*x.^2 + x + 4/10;% ***************************************************% the Plotting Range = 1.05*FcnRangeymin = min(min(y1),min(y2)); ymax = max(max(y1),max(y2)); % the Range LimitsR = ymax - ymin; ymid = (ymax + ymin)/2;ypmin = ymid - 1.025*R/2; ypmax = ymid + 1.025*R/2% % The ZERO Lineszxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ypmin*1.05 ypmax*1.05];%% mark max Profitqm = 0.372; Rm = 50*qm - 100*qm.^3; Cm = 10*qm.^2 + qm + 4/10;Q = [qm, qm]; P = [Cm,Rm] % make vertical line whose length is Max-Profit%% the 6x6 Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenplot(x,y1, x,y2, Q,P, '--md', 'LineWidth', 3),grid, axis([xmin xmax ypmin ypmax]),... xlabel('\fontsize{14} Staplers (kUnit)'), ylabel('\fontsize{14} R & C ($)'),... title('\fontsize{16}MTH15 • Bruce Mayer, PE'), legend('Revenue', 'Cost', 'MaxProfit,','Location','NorthWest')%holdplot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)hold off



Marginal Analysis for Min Avg Cost

Given cost C as a function of production level q, then the production level at which the minimum average cost occurs satisfies:

In other words, Average Cost is Minimized when Average Cost equals Marginal Cost.

min

min dq

qdCqA



Example Find Min Avg Cost

Recall from the previous example that to produce k-Staplers it costs StillStomping this amount in $k:

Use marginal analysis to find the production level at which average cost is minimized, as well as the minimum average cost amount.

10

410 2 qqqC




SOLUTION: The marginal analysis criterion for

minimum average cost specifies determination of where average cost equals marginal

Recall that

In thisCase

dQtyProduce

TotalCostAvgCost

q

qq

qq

q

qCqA

4.0110

4.010 2




ReCall also: Now equate these functions and solve

Again a Production Qty must be positive, so q = 0.2 kUnits at min cost

120 qdqdC

dq

dCq

qqqA 120

4.0110

204.010

104.0

104.0

qq

qq

qq

2.004.0 q




When producing 200 staplers, average cost is minimized at a value of

The units for Amin are $k/kUnit or $/Unit

Thus the factory incurs a minimum average cost of $5 per Stapler when producing 200 units

52122.0

4.012.010

4.0110

minmin

min

minmin

qq

q

qCA



Ac &

dC

/dq

0.1 0.15 0.2 0.25 0.3 0.35 0.43

4

5

6

7

8

9

Staplers (kUnit)

Ac

& d

C/d

q (

$/U

nit)


AvgCostMarginal Cost



Price Elasticity of Demand

If q = D(p) units of a commodity are demanded by the market at a unit price p, where D is a differentiable function, then the price elasticity of demand for the commodity is given by

This Expression has the interpretation: “the percentage rate of decrease in demand q produced by a 1% increase in price p.”

%

%

pdp

qdq

dp

dq

q

ppE



Example Price Elasticity

The Weekly demand for a pair of high-end headphones follows the model

• Where– D ≡ Demand in Units– p ≡ Price in $/Unit

What is the price elasticity of demand when the headphones sell at $500 per pair? Interpret the result.

100002001.0 2 pppD




SOLUTION: ReCall The price elasticity

of demand Formula Use the Quantity-Demanded Eqn:

Then

dp

dq

q

ppE

2002.0 p




Then:

so

dp

dq

q

ppE

20)500(02.010000)500(20)500(01.0

500500

2

E

2




A price elasticity of 2 means that we expect each 1% INcrease in price to cause an associated 2% DEcrease in demand for the product.

HiEnd Headphones are a luxury good, so it may make sense that demand would respond sharply to a change in price; i.e., the demand is very Elastic



Hi Levels of Elasticity:

E(p)>1 signifies Elastic demand. The percentage decrease in demand is

greater than the percentage increase in price that caused it. Thus, demand is relatively sensitive to changes in price.

A decrease in price, conversely, causes an associated increase in revenue when demand is elastic.• i.e., Lowering/Raising the price produces

large changes in demand much

1pE



Lo Levels of Elasticity:

E(p)<1 signifies INelastic demand. The percentage decrease in demand is

less than the percentage increase in price that caused it. Thus, demand is relatively insensitive to changes in price.

A decrease in price causes an associated decrease in revenue when demand is INelastic.• i.e., Lowering/Raising the price does NOT

change demand much

1pE



Neutral Elasticity:

E(p)=1 signifies Neutral demand. Since the Demand is of unit elasticity,

The percentage changes in price and demand are approximately equal.

It can be shown that Revenue is maximized at a price for which demand is of unit elasticity

1pE



Elasticity Illustrated The demand for headphones in the

previous example is Elastic at a unit price of $500 (because E = 2, which is greater than 1)• Change in price will cause a large change

in Demand

At a unit price of $200 per pair of headphones, E = 0.5, so the demand is INelastic at that price• A price change causes a small

Demand change



WhiteBoard Work

Problems From §3.4• P52 → Bird

Flight Power• P58 → Radio Ratings

http://www.google.com/url?sa=i&rct=j&q=Bird+Power&source=images&cd=&cad=rja&docid=ooOsOZfLZCb8VM&tbnid=kEmeyM9vXCecyM:&ved=0CAUQjRw&url=http://www.humanevents.com/2012/06/27/bird-group-sues-obama-administration-wind-power/&ei=KRrjUZLwE43siwLL44HwCw&bvm=bv.48705608,d.cGE&psig=AFQjCNG5fdh5kdLbvbhOXh5zX5zhoS9jsg&ust=1373924208659885

http://www.google.com/url?sa=i&rct=j&q=talk+radio+studio&source=images&cd=&cad=rja&docid=T6mMPBEME2bA_M&tbnid=AtbDE_YLEJgmVM:&ved=0CAUQjRw&url=http://www.1100kfnx.com/index.php?/maingal/image_med/6/&ei=BxvjUcLiOafjiALStICoCw&bvm=bv.48705608,d.cGE&psig=AFQjCNGTB9Wb_vanH5NMVLOKh2SfZxNbZg&ust=1373924416298289



All Done for Today

HiElasticvs.

LoElastic

http://www.google.com/url?sa=i&rct=j&q=demand+elasticity&source=images&cd=&cad=rja&docid=JAfvFo2amPDfeM&tbnid=UPHm6piLNX8a7M:&ved=0CAUQjRw&url=http://www.s-cool.co.uk/a-level/economics/elasticities/revise-it/price-elasticity-of-demand&ei=pRvjUZCvIYmiiQKRsICgCw&bvm=bv.48705608,d.cGE&psig=AFQjCNHiakPEsU4QUPy_vTX0zM1UG5kuKQ&ust=1373924562716758



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix

–

srsrsr 22



ConCavity Sign Chart

a b c

−−−−−−++++++ −−−−−− ++++++

x

ConCavityForm

d2f/dx2 Sign

Critical (Break)Points Inflection NO

InflectionInflection















P3.4-58 Plot by MuPAD





http://www.google.com/url?sa=i&rct=j&q=derivative+power+rule&source=images&cd=&cad=rja&docid=NjbfKj_JZBSzwM&tbnid=jJkxz7X095T-NM:&ved=0CAUQjRw&url=http://mathworld.wolfram.com/Derivative.html&ei=TfPWUe_nA6akiQKZ64DIDw&bvm=bv.48705608,d.cGE&psig=AFQjCNGw-AM0Kc2tARyUGs-zPM7iK-qH3g&ust=1373127702659906

http://www.google.com/url?sa=i&rct=j&q=related+rates+calculus&source=images&cd=&cad=rja&docid=wSGtOAIrRxIoIM&tbnid=l4Q1f5xFVNRraM:&ved=0CAUQjRw&url=http://www.cantonschools.org/~lforastiere/ap%20calculus?Templates=Printable&ei=2TLcUY-gOa2UjALp4IHgDw&bvm=bv.48705608,d.cGE&psig=AFQjCNH42r6LfHpjrPyJXK2DJrvdwHWZIQ&ust=1373471717102765

http://www.google.com/url?sa=i&rct=j&q=related+rates+calculus&source=images&cd=&cad=rja&docid=nuuUeFKHM5pqjM&tbnid=x7raEhpCGJ8edM:&ved=0CAUQjRw&url=http://designatedderiver.wikispaces.com/Games+and+fun+stuff&ei=IzPcUYPxBaOLiwLjs4GYAQ&bvm=bv.48705608,d.cGE&psig=AFQjCNH42r6LfHpjrPyJXK2DJrvdwHWZIQ&ust=1373471717102765

http://www.google.com/url?sa=i&rct=j&q=relative+extrema+of+a+function&source=images&cd=&cad=rja&docid=r50JwQWumQU79M&tbnid=cBBX7LSiOy-pcM:&ved=0CAUQjRw&url=http://www.ltcconline.net/greenl/courses/115/applications/frsttst.htm&ei=C87eUYSWEoWpiQLjwoCYAg&bvm=bv.48705608,d.cGE&psig=AFQjCNETiKTT033nRi_5KJkrhn3WjmAx4w&ust=1373642329390150

Documents

[email protected] MTH15_Lec-16_sec_3-4_Optimization.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical