Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Chabot Mathematics §3.5 Added Optimization Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. 3.4 Review § • Any QUESTIONS About • §3.4 → Optimization & Elasticity • Any QUESTIONS About HomeWork • §3.4 → HW-16

3. §3.5 Learning Goals • List and explore guidelines for solving optimization problems • Model and analyze a variety of optimization problems • Examine inventory control

4. Chabot Mathematics TransLate: Words → Math Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

5. Applications Tips • The Most Important Part of Solving REAL WORLD (Applied Math) Problems Translating • The Two Keys to the Translation • Use the LET Statement to ASSIGN VARIABLES (Letters) to Unknown Quantities • Analyze the RELATIONSHIP Among the Variables and Constraints (Constants)

6. Basic Terminology • A LETTER that can be any one of various numbers is called a VARIABLE. • If a LETTER always represents a particular number that NEVER CHANGES, it is called a CONSTANT A & B are CONSTANTS

7. Algebraic Expressions • An ALGEBRAIC EXPRESSION consists of variables, numbers, and operation signs. • Some Examples • When an EQUAL SIGN is placed between two expressions, an equation is formed →

8. Addition Subtraction Multiplication Division add subtract multiply divide sum of difference of product of divided by plus minus times quotient of increased by decreased by twice ratio more than less than of per Translate: English → Algebra • “Word Problems” must be stated in ALGEBRAIC form using Key Words

9. Example  Translation • Translate this Expression: Eight more than twice the product of 5 and a number • SOLUTION • LET n ≡ the UNknown Number Eight more than twice the product of 5 and a number.

10. Mathematical Model • A mathematical model is an equation or inequality that describes a real situation. • Models for many applied (or “Word”) problems already exist and are called FORMULAS • A FORMULA is a mathematical equation in which variables are used to describe a relationship

11. c a h b Formula Describes Relationship Relationship Mathematical Formula Perimeter of a triangle: Area of a triangle:

12. h r Example  Volume of Cone Relationship Mathematical Formulae Volume of a cone: Surface area of a cone:

13. Celsius Fahrenheit Example  °F ↔ °C Relationship Mathematical Formulae Celsius to Fahrenheit: Fahrenheit to Celsius:

14. Acid Base Example  Mixtures Relationship Mathematical Formula Percent Acid, P:

15. Solving Application Problems • Read the problem as many times as needed to understand it thoroughly. Pay close attention to the questions asked to help identify the quantity the variable(s) should represent. In other Words, FAMILIARIZE yourself with the intent of the problem • Often times performing a GUESS & CHECK operation facilitates this Familiarization step

16. Solving Application Problems • Assign a variable or variables to represent the quantity you are looking for, and, when necessary, express all other unknown quantities in terms of this variable. That is, Use at LET statement to clearly state the MEANING of all variables • Frequently, it is helpful to draw a diagram to illustrate the problem or to set up a table to organize the information

17. Solving Application Problems • Write an equation or equations that describe(s) the situation. That is, TRANSLATE the words into mathematical Equations • Solve the equation; i.e., CARRY OUT the mathematical operations to solve for the assigned Variables • CHECK the answer against the description of the original problem (not just the equation solved in step 4)

18. Solving Application Problems • Answer the question asked in the problem. That is, make at STATEMENT in words that clearly addressed the original question posed in the problem description

19. Example  Mixture Problem • A coffee shop is considering a new mixture of coffee beans. It will be created with Italian Roast beans costing $9.95 per pound and the Venezuelan Blend beans costing $11.25 per pound. The types will be mixed to form a 60-lb batch that sells for $10.50 per pound. • How many pounds of each type of bean should go into the blend?

20. Example  Coffee Beans cont.2 • Familiarize –This problem is similar to our previous examples. • Instead of pizza stones we have coffee beans • We have two different prices per pound. • Instead of knowing the total amount paid, we know the weight and price per pound of the new blend being made. • LET: • i≡ no. lbs of Italian roast and • v≡ no. lbs of Venezuelan blend

21. i + v = 60 9.95i + 11.25v = 630 Example  Coffee Beans cont.3 • Translate –Since a 60-lb batch is being made, we have i + v = 60. • Present the information in a table.

22. Example  Coffee Beans cont.4 • Translate –We have translated to a system of equations • Carry Out –When equation (1) is solved for v, we have: v = 60  i. • We then substitute for v in equation (2).

23. Example  Coffee Beans cont.5 • Carry Out –Find v using v = 60  i. • Check –If 34.6 lb of Italian Roast and 25.4 lb of Venezuelan Blend are mixed, a 60-lb blend will result. • The value of 34.6 lb of Italian beans is 34.6•($9.95), or $344.27. • The value of 25.4 lb of Venezuelan Blend is 25.4•($11.25), or $285.75,

24. Example  Coffee Beans cont.6 • Check –cont. • so the value of the blend is [$344.27 + $285.75] = $630.02. • A 60-lb blend priced at $10.50 a pound is also worth $630, so our answer checks • State–The blend should be made from • 34.6 pounds of Italian Roast beans • 25.4 pounds of Venezuelan Blend beans

25. Example  Max Enclosed Area • A rancher wants to build rectangular enclosures for her cows and horses. She divides the rectangular space in half vertically, using fencing to separate the groups of animals and surround the space. • If she has purchased 864 yards of fencing, what dimensions give the maximum area of the total space and what is the area of each enclosure?

26. Example  Max Enclosed Area • SOLUTION: • First Draw Diagram, Letting • w ≡ EnclosureWidth in yards • l ≡ Enclosure Length in yards • Then the total Enclose Area for the large Rectangle

27. Example  Max Enclosed Area • The fencing required for the enclosure is the perimeter of the rectangle plus the length of the vertical fencing between enclosures

28. Example  Max Enclosed Area • Need to Maximize This Fcn: • However the fcn includes TWO UNknowns: length and width. • Need to eliminate one variable (either one) in order to Product a function of one variable to maximize. • Use the equation for total fencing and isolate length l: Solving for l

29. Example  Max Enclosed Area • Now we substitute the value for l into the area equation: • Maximize this function first by finding critical points by setting the first Derivative equal to Zero

30. Example  Max Enclosed Area • Set dA/dw to zero, then solve • Since There is only one critical point, the Extrema at w = 144 is Absolute • Thus apply the second derivative test (ConCavity) to determine max or min

31. Example  Max Enclosed Area • Since d 2A/dw 2 is ALWAYS Negative, then the A(w) curve is ConCave DOWN EveryWhere • Thus a MAX exists at w = 144 • Now find the length of the total space using our perimeter equation when solved for length

32. Example  Max Enclosed Area • Then The total space should be a 144yd by 216yd Rectangle. Each enclosure then is 144 yards wide and 216/2 - 108 yards long, and the area of each is 144yd·108yd = 15 552 sq-yd ← 216yd → ↑144yd↓ 15 552 yd2 15 552 yd2 ← 108yd → ← 108yd →

33. Example  Find Minimum Cost • The daily production cost associated with a company’s principal product, the ChabotPad (or cPad), is inversely proportional to the length of time, in weeks, since the cPad’s release. • Also, maintenance costs are linear and increasing. • At what time is total cost minimized? • The answer may contain constants

34. Example  Find Minimum Cost • SOLUTION: • Translate: at any given time, t, • Or • Now for CP → production cost associated with the cPadis inversely proportional to the length of time • Formulaically TotalCost = ProductionCost + MaintenanceCost

35. Example  Find Minimum Cost • t ≡ time in Weeks • K ≡ The Constant of ProPortionality in k$·weeks • Now for CM → maintenance costs are linear and increasing • Translated to a Eqn • m ≡ Slope Constant (positive) in $k/week • b ≡ Intercept Constant (positive) in $k • Then, the Total Cost

36. Example  Find Minimum Cost • find potential extrema by solving the derivative function set equal to zero: • Since tMUST be POSITIVE →

37. Example  Find Minimum Cost • Now use the second derivative test for absolute extrema to verify that this value of t produces a positiveConCavity (UP) which confirm a minimum value for cost: • At theZeroValue

38. Example  Find Minimum Cost • Since K and m are BOTH Positive then Is also Positive • The 2nd Derivative Test Confirms that the function is ConCave UP at the zero point, which confirms the MINIMUM • The Min Cost:

39. Example  Find Minimum Cost • STATE: for the cPad • Minimum Total Cost will occur at this many weeks • The Total Cost at this time in $k

40. Example  Minimize Travel Time • Gonzalo walks west on a sidewalk along the edge of the grass in front of the Education complex of the University of Oregon. • The grassy area is 200 feet East-West and 300 feet North-South. Gonzalo strolls at • 4 ft/sec on sidewalk • 2 ft/sec on grass. • From the NE corner how long should he walk on the sidewalk before cutting diagonally across the grass to reach the SW corner of the field in the shortest time?

41. Example  Minimize Travel Time • SOLUTION: • Need to TransLateWords to MathRelations • First DRAW DIAGRAM Letting: • x ≡ The SideWalkDistance • d ≡ The DiaGonal Grass Distance

42. Example  Minimize Travel Time • The total distance traveled is x+d, and need to minimize the time spent traveling, so use the physical relationship [Distance] = [Speed]·[Time] • Solving the above “Rate” Eqn for Time: • So the time spent traveling on the SideWalk at 4 ft/s

43. Example  Minimize Travel Time • Next, the time on Grass → • Writing in terms of x requires the use of the Pythagorean Theorem: • Then

44. Example  Minimize Travel Time • And the Total Travel Time, t, is the SideWalk-Time Plus the Grass-Time • Now Set the 1st Derivative to Zero to find tmin

45. Example  Minimize Travel Time • Continuing with the Reduction • OR

46. Example  Minimize Travel Time • UsingMoreAlgebra

47. Example  Minimize Travel Time • WithYetMoreAlgebra • But the Diagram shows that x can NOT be more than 200ft, thus 26.79ft is the only relevant location of a critical point

48. Example  Minimize Travel Time • Alternatives to check for max OR min: • 2nd Derivative Test • We could check using the second derivative test for absolute extrema to see if 26.79 corresponds to an absolute minimum, but that involves even more messy calculations beyond what we’ve already accomplished. • Slope Value-Diagram and Direction-Diagram (Sign Charts) • Instead, check the critical point against the two endpoints on either side of x = 26.79; say x=0 & x=200

49. Example  Minimize Travel Time • Findt-Valueat x=0,x=26.79 andx=200

50. Example  Minimize Travel Time • Finddt/dxSlopeat x=0 andx=200