Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§7.3 2VarOptimizati

on



Review §

Any QUESTIONS About• §7.2 → Partial Derivatives

Any QUESTIONS About HomeWork• §7.2 → HW-05

7.2

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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



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.



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)

ba,

yxfz ,

baf ,



Quick Example The function of x,y below has a

maximum of about 0.5 at approximately (0.6, 0) Relative Maximum



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:

That is, BOTH Partials must equal Zero

at the Same Time

bafyf

xfbaf y

byax

byaxx ,0&0,



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



Example Find Critical Points Find All Critical Points for Critical points occur for a function of two

variables wherever both 1st Partials = 0• For the given

2-Variable function Setting BOTH partials to Zero

Generates 2-Eqns in 2-Unknwns

yxxyyxf 24),(

22

2&4y

xyf

xy

xf

22

20&40y

xyf

xy

xf



Example Find Critical Points Using the x-Partial: Now SubStitute into the y-Partial

Or

Then

BackSubbing:

22

440x

yx

y

22

2222 420

42020 xxx

xy

x

3442 8

111162

42 x

xxxxx

288 31

31

33 xxx

124422 yy

xy



Example Find Critical Points SOLUTION Thus the only relative extremum of the

function occurs at (2,1) Whether this

extremum is a maximum, minimum, or neither is not yet known. Its graph suggests a minimum:

0.5 1 1.5 2 2.5 3 3.5 0.5

1

1.56

7

8

9

10

11

12

y

MTH16 • Bruce Mayer, PE

x

z =

f(x,y

)

MTH15 3Var 3D Plot.m



Saddle Surface At the “Saddle

Point” (0,0,0)

But, the Curve is a• MINIMUM in the xz plane

• MAXIMUM in the yz plane

00

000

yz

xz



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

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

22

2

2

2

2

,

yxf

yf

xfyxD



2nd Partials Test Procedure Find a Critical Point (a,b) such that

Evaluate the “Discriminant” fcn, D(x,y) from last slide, at the CP. That is, find

bafyf

xfbaf y

byax

byax

x ,0&0,

2

2

2

2

2

2

,

byax

byax

byax yx

fyf

xfbaD



2nd Partials Test Procedure If D(a,b) is NEGATIVE, then (a,b) is a

SADDLE POINT For D(a,b)

POSITIVE calc• 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

byax

xx xfbaf

2

2

,



Quick Example For the Previous

Example Calc Then D

And

Now D>0 & fxx>0 so (2,1) is a MAX

32

84xx

yx

f xx

32

42yy

xy

f yy

122

y

xx

f xy

22

2

2

2

2

,

yxf

yf

xfyxD

233 124

182,1 D

31418482,1 D

1281,2 3 xxf



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

; qC in kCans ; qR in kBags

• x ≡ Cashew Price in $ per Can• y ≡ RaisinNut Price in $ per Bag

yxqC 05.05.08 yxqR 2.02.06



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

To Maximize R, take the Partials and set them to Zero

RRCC qpqpyxR ),(

22 2.05.025.068 yxxyyx



Example Find Max Revenue The Partial Derivatives

Combine the Two Equations

BackSub to find: So have ONE Critical Point at About

(13.93 $/can, 23.70 $/bag)

yxyRxyxR 4.025.06&25.08

yxxy 4.025.060&25.080

xyyx 246.1&25.08

xyyx 25.08246.170.2335.1/3224825.06.1 yyy93.13)70.23(25.18 x



Example Find Max Revenue Next Find the Discriminant Function

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

22

2

2

2

2

,

yxf

yf

xfyxD

03375.025.04.01 2

22max 7.232.093.135.07.2393.1325.07.23693.1387.23,93.13 R

815126$7.23,93.13max R



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



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.



Example Production Constraints Gladiator Goodies Company (GGC)

Factory does NOT have Unlimited Production Capacity.

The Factory has been Engineered to this Design Constraint:

• e.g., if the factory produces 2300 Cashew Cans per week, then at least 4600 RaisinNut Bags also come off the Line

[No. RaisinNut Bags] ≥ 2·[ No. Cashew Cans]



Example Production Constraints Recall the GGC Revenue MathModel

as developed by the Industrial Engineer

Given the Factory-Production Constraint, Find the Maximum Revenue that may be realized using the Cashew and RaisinNut Production Line

22 2.05.025.068),( yxxyyxyxR



Example Production Constraints The Goal is the as in the Previous

example; to Maximize Revenue. Stated Mathematically →

A further constraint is that GGC will NOT Give Away their products; thus

CR qq 2

31004 xy

0&0 yx



Example Production Constraints Then the THREE Constraints:

The Constrained Domain RegionGraphically• Practical

Price Regionin Dark Blue

31004&0&0 xyyx

0 5 10 15 200

5

10

15

20

25

30

35

40

45

x = Cashew Price ($/Can)

y =

Rai

sinN

ut P

rice

($/B

ag)

MTH16 • Bruce Mayer, PE

MTH15 Quick Plot BlueGreenBkGnd 130911.m



MATLA

B C

ode% 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 = 20;% The FUNCTION **************************************x = linspace(xmin,xmax,10000); y1 = 4*x-100/3; yFilter = (y1>0); y=y1.*yFilter;% ***************************************************% the Plotting Range = 1.05*FcnRangeymin = min(y); ymax = max(y); % 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];%vxR = [xmax,xmax]; vyR = [0,ymax]; % close the constraint line at Right% the 6x6 Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenarea(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 onplot(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:

Taking dR(0,y)/dy = 0 produces a Maximum at

22 2.05.025.068, yxxyyxyxR

22.06,0 yyyR

154.06

y



Example Production Constraints Now consider all boundary points at

which critical values can occur. First, consider the vertical line at x = 0:

Taking dR(0,y)/dy = 0 produces a Maximum at

154.06

y



Example Production Constraints Next consider the horizontal line y = 0:

Taking dR(x,0)/dx = 0 Results in an x-maximum

Lastly examine the Slanted Line: Sub the Slanted Line Constraint into the

Revenue Function

22 2.05.025.068, yxxyyxyxR

25.080, xxxR

818

x

31004 xy

22 2.05.025.068),( yxxyyxyxR



Example Production Constraints With

Taking dR(x,4x−33.33)/dx = 0 produces a Maximum at

31004 xy

2

2

310042.05.0

3100425.0

3100468

33.334,31004,

xxxxxx

xxRxxR

26.144.577

x



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.



Example Production Constraints SUMMARY:

• We then compare revenue for each of those three boundary points and identify the largest revenue

• Tabulating the Results

Pricing the cashews at $14.26 and Trial-by-Trail at $23.71 provides maximum revenue, given the constraints.

(x,y) (0,15) (8,0) (14.26, 23.71)R 45 32 126.76



WhiteBoard Work Problems From §7.3

• P51 → Box Design Optimization



All Done for Today

SaddlePointCity

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Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix

–

srsrsr 22

a2 b2

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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