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Chabot Mathematics
§7.3 2Var
Optimization
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
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Review §
7.2
 Any QUESTIONS About
• §7.2 → Partial Derivatives
 Any QUESTIONS
About
HomeWork
• §7.2 → HW-05
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§7.3 Learning Goals
 Locate and classify relative extrema for
a function of two variables using the
second partials test
 Examine applied problems
involving optimization of
functions of two variables
 Discuss and apply the extreme value
property for functions of two variables to
find absolute extrema on a closed,
bounded region
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Local Maximum & Minimum
 DEFINITION
 A function of two variables has a local
maximum at (a,b) if f(x y) ≤ f(a,b) when
(x,y) is near (a,b). [This means that
f(x,y) ≤ f(a,b) for all points (x,y) in some
DISK with center (a,b).] The number
f(a,b) is called a local maximum value.
 If f(x,y) ≥ f(a,b) when (x,y) is near (a,b),
then f(a,b) is a local minimum value.
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Local MaxMin Illustrated
 The DISK centered
at (x,y)=(a,b)
produces a “Hill”
with peak at f(a,b)
 Local (and Absolute)
max/min for z = f(x,y)
f a , b 
z  f x, y 
a , b 
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Quick Example
 The function of x,y below has a
maximum of about 0.5 at approximately
(0.6, 0)
Relative Maximum
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CriticalPoints and Extrema
 DEFINITION: A point (a,b) in the
domain of f(x,y), for which the first order
partial derivatives of f(x,y) exist, the
point (a,b) is a CRITICAL POINT if:
f x a , b  
f
x
xa
y b
0
&
f
y
xa
 0  f y a , b 
y b
 That is, BOTH Partials must equal Zero
at the Same Time
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CriticalPoints and Extrema
 THEOREM: If f has a local maximum or
minimum at (a,b) then point (a,b) MUST
be a Critical Point at which both Partials
Simultaneously equal Zero.
 While ALL max/min (Extrema) occur at
Critcal Points (CPs), NOT all CPs are
Extrema Points
• A Surface that contains a CP that is NOT
an Extremum is called a Saddle Surface
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Example  Find Critical Points
4
2
 Find All Critical Points for f ( x , y )  xy  
x y
 Critical points occur for a function of two
variables wherever both 1st Partials = 0
f
4
• For the given
 y 2
2-Variable function  x
x
&
f
y
x
 Setting BOTH partials to Zero
Generates 2-Eqns in 2-Unknwns
f
x
0 y
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4
x
2
&
f
y
0x
2
y
2
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2
y
2
Example  Find Critical Points
4
4
 Using the x-Partial: 0  y  x 2  y  x 2
 Now SubStitute into the y-Partial
0 x
2

 y 2
 Or x 
2
4
2
x
4
2
0 x

4

x 
 BackSubbing:
4
 Then
x 8
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
2 2
3
y
x
3
2
1
 83

0 x
4
2 41

x 
x 
16

x
1
3
x

2

1
1
x
2
x
2 2
3
8

x 2
y
4
2 
2

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y 1
Example  Find Critical Points
 SOLUTION
 Thus the only relative extremum of the
function occurs at (2,1)
MTH16 • Bruce Mayer, PE
 Whether this
extremum is a
maximum,
minimum, or
neither is not yet
known. Its graph
suggests a minimum:
12
z = f(x,y)
11
10
9
8
7
6
0.5
1
1.5
1
2
2.5
3
3.5
0.5
MTH15 3Var 3D Plot.m
x
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y
1.5
Saddle Surface
 At the “Saddle
Point” (0,0,0)
z
x
0
0
0
z
y
0
0
 But, the Curve is a
• MINIMUM in the
xz plane
• MAXIMUM in the
yz plane
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Critical Point Condition
 A Critical Point
ALWAYS
marks the
location of
a “Flat” Tangent
Plane, and can be
one of
• A MAXimum
• A MIMimum
• NEITHER
– i.e.; a SADDLE point
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 The Nature of a CP
can (usually) Be
determined by the
Second Partials
Test
 Assume for f(x,y) that
all needed Partial
exist then let
D x, y  
 f
2
x
2
 f
2

y
2
 f 



x

y


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2
2nd Partials Test Procedure
 Find a Critical Point (a,b) such that
f x a , b  
f
x
xa
y b
0
&
f
y
xa
y b
 0  f y a , b 
 Evaluate the “Discriminant” fcn, D(x,y)
from last slide, at the CP. That is, find
 2
 f
D a , b  
  x 2

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xa
y b
  2
  f

   y 2
 
xa
y b





 2
 f

 xy



xa 
y b 
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2
2nd Partials Test Procedure
 If D(a,b) is NEGATIVE, then (a,b) is a
SADDLE POINT
2
 f
 For D(a,b)
f xx  a , b  
2
x x a
POSITIVE calc
y b
• If fxx(a,b) is POSITIVE, then (a,b) is a MAX
• If fxx(a,b) is NEGATIVE, then (a,b) is a MIN
 If D(a,b) = 0 then the test is Inconclusive
• The pt (a,b) can be of ANY of the Three
forms; max, min, saddle
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Quick Example
 For the Previous
Example Calc
f xx
 
4

y 2
x 
x
f yy
 
2
 x  2

y 
y
f xy
 
2
 x  2

x 
y
 Then D
 f 
D x, y  



2
2
x
y
8

 xy 
 3
8 4
2
 x
D 1, 2   3  3  1
1 2

4
4
  3
D 1, 2   8   1  4  1  3
y

8
8
 And f xx  2 ,1   3  1

  1
2
 Now D>0 & fxx>0 so

 f
2
 f
2
(2,1) is a MAX
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2
2
Example  Find Max Revenue
 The Gladiator Goodies Company sells
Jumbo Cashews and the popular “Trial
by Trail” RaisinNut mix. Then GG
Industrial Engineering develops
Regression models for the products
 q C  8  0 .5 x  0 .05 y ; qC in kCans
 q R  6  0 . 2 x  0 . 2 y ; qR in kBags
• x ≡ Cashew Price in $ per Can
• y ≡ RaisinNut Price in $ per Bag
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Example  Find Max Revenue
 For this Financial Model
• Find a revenue function,
• determine at what prices revenue is
maximized, and
• find the maximum revenue from the sale of
these two products.
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Example  Find Max Revenue
 SOLUTION
 The Company Revenue is the sum of
revenue from the Two products, so
R ( x, y )  pC qC  p R q R
= x (8- 0.5x + 0.05y) + y ( 6 + 0.2x - 0.2y)
 8 x  6 y  0 . 25 xy  0 . 5 x  0 . 2 y
2
2
 To Maximize R, take the Partials and
set them to Zero
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Example  Find Max Revenue
 The Partial Derivatives
 R  x  8  0 . 25 y  x
&
 R  y  6  0 . 25 x  0 . 4 y
0  8  0 . 25 y  x
&
0  6  0 . 25 x  0 . 4 y
x  8  0 . 25 y
&
1 . 6 y  24  x
 Combine the Two Equations
x  1 . 6 y  24  8  0 . 25 y  x
1 . 6 y  0 . 25 y  8  24

y  32 / 1 . 35  23 . 70
 BackSub to find: x  8  1 .25 ( 23 .70 )  13 .93
 So have ONE Critical Point at About
(13.93 $/can, 23.70 $/bag)
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Example  Find Max Revenue
 Next Find the Discriminant Function
2
2

 f  f
 f 
D x, y  

 

2
2
x
y
 xy 
2
2
   1     0 . 4   0 . 25   0 . 3375  0
2
 The above calculation along with the
fact that ∂2R/∂x2 = −1 (<0) shows that
the critical point is a maximum
 Then the Revenue at max
R max 13 . 93 , 23 . 7   8 13 . 93   6  23 . 7   0 . 25 13 . 93  23 . 7   0 . 5 13 . 93   0 . 2  23 . 7 
2
R max 13 . 93 , 23 . 7   $ 126 815
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2
Constrained Domain Extrema
 DOMAIN of many RealWorld 2Var Math
Models are Constrained For Various
Practical Reasons; Call This Domain, R
 In a finite Constrained-Domain an
ABSOLUTE Max or Min is Present
• The Absolute Extrema exists at ONE of
– The EDGES, or BOUNDARY, of the Domain
Region, R
– The INTERIOR of R, at a Critical Point of the
2Var Function
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Constrained Domain Illustrated
 Consider a 2Var MathModel, z = f(x,y)
with x & y constrained in the XY-Plane
by the 2D function g(x,y)=k (k a const)
as Illustrated below.
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Example  Production Constraints
 Gladiator Goodies Company (GGC)
Factory does NOT have Unlimited
Production Capacity.
 The Factory has been Engineered to
this Design Constraint:
[No. RaisinNut Bags] ≥ 2·[ No. Cashew Cans]
• e.g., if the factory produces 2300 Cashew
Cans per week, then at least 4600
RaisinNut Bags also come off the Line
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Example  Production Constraints
 Recall the GGC Revenue MathModel as
developed by the Industrial Engineer
R ( x , y )  8 x  6 y  0 . 25 xy  0 . 5 x  0 . 2 y
2
2
 Given the Factory-Production
Constraint, Find the Maximum Revenue
that may be realized using the Cashew
and RaisinNut Production Line
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Example  Production Constraints
 The Goal is the as in the Previous
example; to Maximize Revenue. Stated
Mathematically →
q R  2 qC
(6 + 0.2x - 0.2y) ³ 2(8- 0.5x + 0.05y)
-10 +1.2x ³ 0.3y
y  4 x  100 3
 A further constraint is that GGC will
NOT Give Away their products; thus
x0
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&
y 0
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Example  Production Constraints
 Then the THREE Constraints:
y 0
&
 The
Constrained
Domain
Region
Graphically
27
y  4 x  100 3
45
40
35
30
25
20
15
10
• Practical
Price Region 5
0
in Dark Blue 0
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&
MTH16 • Bruce Mayer, PE
y = RaisinNut Price ($/Bag)
x0
MTH15 Quick Plot BlueGreenBkGnd 130911.m
5
10
15
20
x = Cashew Price ($/Can)
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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 = 20;
% The FUNCTION **************************************
x = linspace(xmin,xmax,10000); y1 = 4*x-100/3; yFilter = (y1>0);
y=y1.*yFilter;
% ***************************************************
% the Plotting Range = 1.05*FcnRange
ymin = min(y); ymax = max(y); % 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];
%
vxR = [xmax,xmax]; vyR = [0,ymax]; % close the constraint line at Right
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
area(x,y, 'LineWidth', 4),grid, axis([xmin 1.05*xmax ypmin ypmax]),...
xlabel('\fontsize{14}x = Cashew Price ($/Can)'), ylabel('\fontsize{14}y
= RaisinNut Price ($/Bag)'),...
title(['\fontsize{16}MTH16 • Bruce Mayer, PE',]),...
annotation('textbox',[.53 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor',
'none', 'String', 'MTH15 Quick Plot BlueGreenBkGnd 130911.m','FontSize',7)
hold on
plot(zxv,zyv, 'k', zxh,zyh, 'k', vxR,vyR, 'k', 'LineWidth', 2)
hold off
Example  Production Constraints
 Now consider all boundary points at
which critical values can occur. First,
consider the vertical line at x = 0:
R  x , y   8 x  6 y  0 . 25 xy  0 . 5 x  0 . 2 y
2
R 0 , y   6 y  0 . 2 y
2
2
 Taking dR(0,y)/dy = 0 produces a
Maximum at
6
y
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 0 .4
 15
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Example  Production Constraints
 Now consider all boundary points at
which critical values can occur. First,
consider the vertical line at x = 0:
R(x, y) = 8x + 6y + 0.25xy - 0.5x - 0.2y
2
2
R(0, y) = 6y - 0.2y2
 Taking dR(0,y)/dy = 0 produces a
Maximum at
6
y
Chabot College Mathematics
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 0 .4
 15
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Example  Production Constraints
 Next consider the horizontal line y = 0:
R  x , y   8 x  6 y  0 . 25 xy  0 . 5 x  0 . 2 y
2
R  x ,0   8 x  0 .5 x
2
2
8
 Taking dR(x,0)/dx = 0
x
8
1
Results in an x-maximum
100
 Lastly examine the Slanted Line: y  4 x 
3
 Sub the Slanted Line Constraint into the
Revenue Function
R ( x , y )  8 x  6 y  0 . 25 xy  0 . 5 x  0 . 2 y
2
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2
Example  Production Constraints
 With
y  4x 
100
3
100 

R  x ,4 x 
  R  x , 4 x  33 . 33 
3 

100 
100 
100 



2
 8 x  6 4 x 
  0 . 25 x  4 x 
  0 .5 x  0 .2  4 x 

3 
3 
3 



= -2.7x + 77x - 422.22
2
 Taking dR(x,4x−33.33)/dx = 0 produces
a Maximum at
x
Chabot College Mathematics
32
 77
 5 .4
 14 . 26
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2
Example  Production Constraints
 SUMMARY:
• Have 3 boundary critical points to consider:
– (0,15), (8 0), and (14.26, 23.71)
• We would normally also consider the only critical
point on the interior (found in the previous
Example 2). However, this point does not satisfy
the condition that there be at least twice as many
health fusion nuts as cashews, so it is omitted.
• We then compare revenue for each of those
three boundary points and identify the largest
revenue.
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Example  Production Constraints
 SUMMARY:
• We then compare revenue for each of
those three boundary points and identify
the largest revenue
• Tabulating the Results
(x,y)
(0,15)
(8,0)
(14.26, 23.71)
R
45
32
126.76
 Pricing the cashews at $14.26 and Trialby-Trail at $23.71 provides maximum
revenue, given the constraints.
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[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
WhiteBoard Work
 Problems From §7.3
• P51 → Box Design Optimization
Chabot College Mathematics
35
Bruce Mayer, PE
[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
All Done for Today
Saddle
Point
City
Chabot College Mathematics
36
Bruce Mayer, PE
[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
2a
–
Chabot College Mathematics
37
[email protected]
2b
Bruce Mayer, PE
[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
38
Bruce Mayer, PE
[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
39
Bruce Mayer, PE
[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
40
Bruce Mayer, PE
[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
41
Bruce Mayer, PE
[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
42
Bruce Mayer, PE
[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx