1. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §7.3 2Var Optimization

2. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §7.2 → Partial Derivatives  Any QUESTIONS About HomeWork §7.2 → HW-05 7.2

3. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 3 Bruce Mayer, PE Chabot College Mathematics §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

4. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 4 Bruce Mayer, PE Chabot College Mathematics 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.

5. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 5 Bruce Mayer, PE Chabot College Mathematics 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)

6. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 6 Bruce Mayer, PE Chabot College Mathematics Quick Example  The function of x,y below has a maximum of about 0.5 at approximately (0.6, 0) Relative Maximum

7. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 7 Bruce Mayer, PE Chabot College Mathematics 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

8. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 8 Bruce Mayer, PE Chabot College Mathematics 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 Critical Points (CPs), NOT all CPs are Extrema Points A Surface that contains a CP that is NOT an Extremum is called a Saddle Surface

9. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 9 Bruce Mayer, PE Chabot College Mathematics Example  Find Critical Points  Find All Critical Points for  Critical points occur for a function of two variables wherever both 1 st Partials = 0 For the given 2-Variable function  Setting BOTH partials to Zero Generates 2-Eqns in 2-Unknwns

10. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 10 Bruce Mayer, PE Chabot College Mathematics Example  Find Critical Points  Using the x-Partial:  Now SubStitute into the y-Partial  Or  Then  BackSubbing:

11. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 11 Bruce Mayer, PE Chabot College Mathematics 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:

12. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 12 Bruce Mayer, PE Chabot College Mathematics Saddle Surface  At the “Saddle Point” (0,0,0)  But, the Curve is a MINIMUM in the xz plane MAXIMUM in the yz plane

13. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 13 Bruce Mayer, PE Chabot College Mathematics 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 Partials exist then let

14. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 14 Bruce Mayer, PE Chabot College Mathematics 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

15. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 15 Bruce Mayer, PE Chabot College Mathematics 2nd Partials Test Procedure  If D(a,b) is NEGATIVE, then (a,b) is a SADDLE POINT  If D(a,b) POSITIVE, Then Calculate → If f xx (a,b) is POSITIVE, then (a,b) is a MIN If f xx (a,b) is NEGATIVE, then (a,b) is a MAX 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

16. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 16 Bruce Mayer, PE Chabot College Mathematics Quick Example  For the Previous Example Calc  Then D  And  Now D>0 & f xx >0 so (2,1) is a MAX

17. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 17 Bruce Mayer, PE Chabot College Mathematics 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 in kCans  ; q R in kBags x ≡ Cashew Price in $ per Can y ≡ RaisinNut Price in $ per Bag

18. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 18 Bruce Mayer, PE Chabot College Mathematics 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.

19. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 19 Bruce Mayer, PE Chabot College Mathematics 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

20. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 20 Bruce Mayer, PE Chabot College Mathematics 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)

21. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 21 Bruce Mayer, PE Chabot College Mathematics Discriminant Prep → 2nd Partials  From Last Slide  Thus

22. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 22 Bruce Mayer, PE Chabot College Mathematics Example  Find Max Revenue  Next Find the Discriminant Function  The above calculation along with the fact that ∂ 2 R/∂x 2 = −1 (<0) shows that the critical point is a maximum  Then the Revenue at max

23. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 23 Bruce Mayer, PE Chabot College Mathematics 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 the Domain Region, R, at a CRITICAL POINT of the 2Var Function

24. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 24 Bruce Mayer, PE Chabot College Mathematics 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. For this example BOTH the Max & Min are on the EDGES of R

25. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 25 Bruce Mayer, PE Chabot College Mathematics 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]

26. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 26 Bruce Mayer, PE Chabot College Mathematics 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

27. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 27 Bruce Mayer, PE Chabot College Mathematics 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

28. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 28 Bruce Mayer, PE Chabot College Mathematics Example  Production Constraints  Then the THREE Constraints:  The Constrained Domain Region Graphically Practical Price Region in Dark Blue

29. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 29 Bruce Mayer, PE Chabot College Mathematics 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

30. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 30 Bruce Mayer, PE Chabot College Mathematics 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  Then

31. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 31 Bruce Mayer, PE Chabot College Mathematics 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

32. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 32 Bruce Mayer, PE Chabot College Mathematics Example  Production Constraints  With  Taking dR(x,4x−33.33)/dx = 0 produces a Maximum at

33. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 33 Bruce Mayer, PE Chabot College Mathematics 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 TrailMix as Cashews, so it is omitted. We then compare revenue for each of those three boundary points and identify the largest revenue.

34. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 34 Bruce Mayer, PE Chabot College Mathematics Example  Production Constraints  SUMMARY: We then compare revenue for each of those three boundary points and identify the largest revenue Tabulating the Results: (Cashews, T-Mix)  Pricing the Jumbos at $13.93 and Trial- by-Trail at $23.71 provides maximum revenue, given the constraints. (x,y)(0,15)(8,0)(13.93, 23.71) R45 32126.76

35. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 35 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §7.3 P13 → Critical Point Practice P51 → Box Design Optimization

36. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 36 Bruce Mayer, PE Chabot College Mathematics All Done for Today Saddle Point City

37. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 37 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

38. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 38 Bruce Mayer, PE Chabot College Mathematics P7.3.-13 → Find Critical Points

39. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 39 Bruce Mayer, PE Chabot College Mathematics P7.3-13

40. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 40 Bruce Mayer, PE Chabot College Mathematics Bottom View

41. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 41 Bruce Mayer, PE Chabot College Mathematics % Bruce Mayer, PE % MTH-15 16Feb14 % MTH15_Quick_3Var_3D_Plot_BlueGreenBkGnd_140113.m % clear; clc; clf; % clf clears figure window % % The Domain Limits xmin = -2; xmax = 2; % BASE max & min2 ymin = -2; ymax = 2; NumPts = 51 % The GRID ************************************** xx = linspace(xmin,xmax,NumPts); yy = linspace(ymin,ymax,NumPts); [x,y]= meshgrid(xx,yy); % The FUNCTION*********************************** zp = (x.^2 + 2*y.^2).*exp(1 - x.^2 - y.^2); % z =f(x,y) % the Plotting Range = 1.05*FcnRange zmin = min(min(zp)); zmax = max(max(zp)); % the Range Limits R = zmax - zmin; zmid = (zmax + zmin)/2; zpmin = zmid - 1.025*R/2; zpmax = zmid + 1.025*R/2; % % the Domain Plot axes; set(gca,'FontSize',12); mesh(x,y,zp, 'LineWidth', 2),grid, axis([xmin xmax ymin ymax zpmin zpmax]), grid, box,... xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y'), zlabel('\fontsize{14}z = f(x,y)'),... title(['\fontsize{16}MTH16 Bruce Mayer, PE',]),... annotation('textbox',[.73.05.0.1 ], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', 'MTH15 3Var 3D Plot.m','FontSize',7) %

42. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 42 Bruce Mayer, PE Chabot College Mathematics

43. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 43 Bruce Mayer, PE Chabot College Mathematics

44. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 44 Bruce Mayer, PE Chabot College Mathematics

45. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 45 Bruce Mayer, PE Chabot College Mathematics

46. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 46 Bruce Mayer, PE Chabot College Mathematics

47. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 47 Bruce Mayer, PE Chabot College Mathematics

48. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 48 Bruce Mayer, PE Chabot College Mathematics

49. BMayer@ChabotCollege.edu MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 49 Bruce Mayer, PE Chabot College Mathematics