Chabot Mathematics
§3.4 Elasticity
& Optimization
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Review §
3.3
 Any QUESTIONS
About
• §3.3 → Graph
Sketching
 Any QUESTIONS
About
HomeWork
• §3.3 → HW-15
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
§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
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Absolute Extrema
A function f has an absolute maximum of
f (c) if for every x in an open interval
containing c,
f (c )  f ( x )
A function f has an absolute minimum of
f (c) if for every x in an open interval
containing c,
f (c )  f ( x )
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Absolute Extrema
c , f c 
 Consider the
Function Graph
Shown at Right
 The function
c , f c 
appears to
have an
absolute
maximum of 7 at x = 1, and an absolute
minimum of 2 at x = 6 (and at x = 12).
max
max
min
Chabot College Mathematics
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min
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Extreme Value Explained
 The Absolute-Max or Absolute-Min over
some x-span of any function occurs
EITHER at
Abs-MAX
Abs-MAX
• The ENDS
• SomeWhere
in the Middle
• (Duh!!!)
Chabot College Mathematics
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Abs-MIN
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Find Abs Extrema
 Find the absolute extrema
2
x
for the fcn at Right
f x 
on the interval [−3,1]
x2
 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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Find Abs Extrema
 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
Chabot College Mathematics
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d  x2 
0  f ' x   

dx  x  2 
( x  2)  2 x  x 2 1
0
x  22
Quotient
Rule
2x2  4x  x2 x2  4x
0

2
2
x  2
x  2
0 = x 2 - 4x
0 = x ( x - 4)
Zero
x = 0 or x - 4 = 0 Products
x  0 or x  4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Find Abs Extrema
x2
 The endpoints are −3 and 1. f  x  
x2
So need compare the value for  3  x  1
of f(x) at x=−3 & x=1, and
also at the critical point,
x=0 on the interval.
• Making a T-Table:
x
-3
0
1
y=f (x )
-1.80
0.00
-1.00
 The Table shows that the absolute max
is 0 (attained at x = 0) and absolute min
is −1.8 (attained at x = −3).
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Find Abs Extrema
MTH15 • Bruce Mayer, PE
0
y = f(x) = x2/(x-2)
 The
fcn
Graph
0.5
2
x
f x  
x2
for  3  x  1
Chabot College Mathematics
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-0.5
-1
-1.5
-2
-3
XY f cnGraph6x6BlueGreenBkGndTemplate1306.m
-2.5
-2
-1.5
-1
-0.5
0
0.5
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
1
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:
Chabot College Mathematics
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dRq 
dC q 

dq qmax
dq qmax
 and
d 2 Rq 
dq 2 q
max
d 2C q 

dq 2 qmax
in
$
UNIT
in
$
UNIT 2
 That is
• where marginal revenue
equals marginal cost
• The CHANGE in the RSlope is less than the
CHANGE in the
C-Slope Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Finding Max Profit
 The Math Model for Pricing
2


p
q

50

100
q
of “StillStomping” Staplers:
• Where
– p ≡ Stapler Selling Price in $/Unit
– q ≡ Qty of Staplers Sold in kUnit
 The Total Cost model for the
4
2
StillStomping Staplers: C q   10q  q 
10
• Where
– C ≡ Stapler Production Cost in $k
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Finding Max Profit
 Use marginal analysis to find the
production level at which profit is
maximized, as well as the amount of
the maximum profit.
 SOLUTION:
 The marginal analysis criterion for
maximum Profit specifies that we should
examine where marginal revenue
equals marginal cost
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Finding Max Profit
 Now Total Revenue = [price]·[Qty-Sold]
2
R  q   pq   q  50  100q


• R in (kUnit)·($/Unit) = k$
 The Marginal Analysis requires
Derivatives for R & C
dR d


50q  100q 3   50  300q 2
dq dq
dC d 
4
2

10q  q    20q  1
dq dq 
10 
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Finding Max Profit
 Now set equal the two marginal
functions and solve using the quadratic
formula or a computer algebra system
such as MuPAD (c.f., MTH25)
dR
dC
2
 50  300q  20q  1 
dq
dq
2
0  300q  20q  49
 1  2 37
q
 q  0.3722 or q  0.4389
30
Qty, q, canNOT be Negative
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Finding Max Profit
 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
Pmax  R0.372  C 0.372


Pmax  500.372   1000.372   100.372   0.372  0.4
3
2
Pmax  $11.296k
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx

Ex:  Find Max Profit
MTH15 • Bruce Mayer, PE
Revenue
Cost
MaxProfit,
12
R & C ($)
10
8
6
4
2
0
0
0.1
Chabot College Mathematics
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0.2
0.3
0.4
Staplers (kUnit)
0.5
0.6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 01Aug13 • Rev 11Sep13
% MTH15_Quick_Plot_BlueGreenBkGnd_130911.m
%
clear; clc; clf; % clf clears figure window
%
% The Domain Limits
xmin = 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*FcnRange
ymin = min(min(y1),min(y2)); ymax = max(max(y1),max(y2)); % the Range Limits
R = ymax - ymin; ymid = (ymax + ymin)/2;
ypmin = ymid - 1.025*R/2; ypmax = ymid + 1.025*R/2
%
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ypmin*1.05 ypmax*1.05];
%
% mark max Profit
qm = 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 Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
plot(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')
%
hold
plot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)
hold off
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
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:
dC q 
Aqmin  
dq min
 In other words, Average Cost is
Minimized when Average Cost equals
Marginal Cost.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Find Min Avg Cost
 Recall from the previous example that
to produce k-Staplers it costs
StillStomping this amount in $k:
4
2
C q   10q  q 
10
 Use marginal analysis to find the
production level at which average cost
is minimized, as well as the minimum
average cost amount.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Find Min Avg Cost
 SOLUTION:
 The marginal analysis criterion for
minimum average cost specifies
determination of where average cost
equals marginal
 Recall AvgCost  TotalCost
QtyProduce d
that
 In this
C q  10q 2  q  0.4
0.4
Aq  

 10q  1 
Case
q
q
q
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Find Min Avg Cost
 ReCall also: dC dq  20q  1
 Now equate these functions and solve
0.4
dC
Aq   10q  1 
 20q  1 
q
dq
 0.4
 q
0.4
 10q  
 10q  
 0.04  q 2
q
 q
 10
q   0.04  0.2
 Again a Production Qty must be
positive, so q = 0.2 kUnits at min cost
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Find Min Avg Cost
Amin
 When producing 200 staplers, average
cost is minimized at a value of
C qmin 
0.4
0.4

 10q  1 
 100.2  1 
 2 1 2  5
qmin
min
qmin
0.2
 The units for Amin are $k/kUnit or $/Unit
 Thus the factory incurs a minimum
average cost of $5 per Stapler when
producing 200 units
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
MTH15 • Bruce Mayer, PE
9
Ac & dC/dq
Ac & dC/dq ($/Unit)
8
7
6
5
4
3
0.1
AvgCost
Marginal Cost
0.15
Chabot College Mathematics
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0.2
0.25
0.3
Staplers (kUnit)
0.35
0.4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
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
p dq
dq q
%
the commodity
E p  


is given by
q dp
dp p
%
 This Expression has the interpretation:
“the percentage rate of decrease in
demand q produced by a 1% increase
in price p.”
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Price Elasticity
 The Weekly demand for a pair of highend headphones follows the model
2
D p   0.01 p  20 p  10000
• 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.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Price Elasticity
 SOLUTION:
p dq
 ReCall The price elasticity
E p  
of demand Formula
q dp
 Use the Quantity-Demanded Eqn:
q = 0.01p2 - 20 p +10000,
 Then dq d
2
= ( 0.01p - 20 p +10000)
dp dp
 0.02 p  20
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Price Elasticity
p dq
 Then: E  p   
q dp
p
=( 0.02 p - 20)
2
0.01p - 20 p +10000
 so
500
0.02(500)  20
E 500  
2
0.01(500)  20(500)  10000
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Example  Price Elasticity
 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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Hi Levels of Elasticity: E p  1
 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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Lo Levels of Elasticity: E p  1
 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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Neutral Elasticity: E p  1
 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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
WhiteBoard Work
 Problems From §3.4
• P52 → Bird
Flight Power
• P58 → Radio Ratings
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
All Done for Today
HiElastic
vs.
LoElastic
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_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
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
ConCavity Sign Chart
ConCavity
Form
d2f/dx2 Sign
++++++
Critical (Break)
Points
Chabot College Mathematics
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−−−−−−
a
Inflection
−−−−−−
b
NO
Inflection
++++++
c
Inflection
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
x
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
P3.4-58 Plot by MuPAD
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-16_sec_3-4_Optimization.pptx