1. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §3.4 Elasticity & Optimization

2. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §3.3 → Graph Sketching  Any QUESTIONS About HomeWork §3.3 → HW-15 3.3

3. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 3 Bruce Mayer, PE Chabot College Mathematics §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

4. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 4 Bruce Mayer, PE Chabot College Mathematics Absolute Extrema A function f has an absolute maximum of if for every x in an open interval containing c, A function f has an absolute minimum of if for every x in an open interval containing c,

5. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 5 Bruce Mayer, PE Chabot College Mathematics Example  Absolute Extrema  Consider the Function Graph Shown at Right  The function appears to have an absolute maximum of 7 at x = 1, and an absolute minimum of 2 at x = 6 (and at x = 12).

6. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 6 Bruce Mayer, PE Chabot College Mathematics 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

7. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 7 Bruce Mayer, PE Chabot College Mathematics Extreme Value Explained  The Absolute-Max or Absolute-Min over some x-span of any function occurs EITHER at The ENDS SomeWhere in the Middle (Duh!!!) Abs-MAX Abs-MIN

8. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 8 Bruce Mayer, PE Chabot College Mathematics Example  Find Abs Extrema  Find the absolute extrema for the fcn at Right on the interval [−3,1]  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

9. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 9 Bruce Mayer, PE Chabot College Mathematics 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 Quotient Rule Zero Products

10. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 10 Bruce Mayer, PE Chabot College Mathematics Example  Find Abs Extrema  The endpoints are −3 and 1. So need compare the value of f(x) at x=−3 & x=1, and also at the critical point, x=0 on the interval. Making a T-Table:  The Table shows that the absolute max is 0 (attained at x = 0) and absolute min is −1.8 (attained at x = −3).

11. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 11 Bruce Mayer, PE Chabot College Mathematics Example  Find Abs Extrema  The fcn Graph

12. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 12 Bruce Mayer, PE Chabot College Mathematics 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:  and  That is where marginal revenue equals marginal cost The CHANGE in the R- Slope is less than the CHANGE in the C-Slope

13. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 13 Bruce Mayer, PE Chabot College Mathematics Example  Finding Max Profit  The Math Model for Pricing of “StillStomping” Staplers: Where –p ≡ Stapler Selling Price in $/Unit –q ≡ Qty of Staplers Sold in kUnit  The Total Cost model for the StillStomping Staplers: Where –C ≡ Stapler Production Cost in $k

14. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 14 Bruce Mayer, PE Chabot College Mathematics 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

15. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 15 Bruce Mayer, PE Chabot College Mathematics Example  Finding Max Profit  Now Total Revenue = [price]·[Qty-Sold] R in (kUnit)·($/Unit) = k$  The Marginal Analysis requires Derivatives for R & C xx

16. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 16 Bruce Mayer, PE Chabot College Mathematics 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) Qty, q, canNOT be Negative

17. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 17 Bruce Mayer, PE Chabot College Mathematics 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

18. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 18 Bruce Mayer, PE Chabot College Mathematics Ex:  Find Max Profit

19. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 19 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 =.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

20. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 20 Bruce Mayer, PE Chabot College Mathematics 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:  In other words, Average Cost is Minimized when Average Cost equals Marginal Cost.

21. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 21 Bruce Mayer, PE Chabot College Mathematics Example  Find Min Avg Cost  Recall from the previous example that to produce k-Staplers it costs StillStomping this amount in $k:  Use marginal analysis to find the production level at which average cost is minimized, as well as the minimum average cost amount.

22. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 22 Bruce Mayer, PE Chabot College Mathematics Example  Find Min Avg Cost  SOLUTION:  The marginal analysis criterion for minimum average cost specifies determination of where average cost equals marginal  Recall that  In this Case

23. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 23 Bruce Mayer, PE Chabot College Mathematics Example  Find Min Avg Cost  ReCall also:  Now equate these functions and solve  Again a Production Qty must be positive, so q = 0.2 kUnits at min cost

24. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 24 Bruce Mayer, PE Chabot College Mathematics Example  Find Min Avg Cost  When producing 200 staplers, average cost is minimized at a value of  The units for A min are $k/kUnit or $/Unit  Thus the factory incurs a minimum average cost of $5 per Stapler when producing 200 units

25. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 25 Bruce Mayer, PE Chabot College Mathematics Ac & dC/dq

26. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 26 Bruce Mayer, PE Chabot College Mathematics 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 the commodity is given by  This Expression has the interpretation: “the percentage rate of decrease in demand q produced by a 1% increase in price p.”

27. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 27 Bruce Mayer, PE Chabot College Mathematics Example  Price Elasticity  The Weekly demand for a pair of high- end headphones follows the model 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.

28. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 28 Bruce Mayer, PE Chabot College Mathematics Example  Price Elasticity  SOLUTION:  ReCall The price elasticity of demand Formula  Use the Quantity-Demanded Eqn:  Then

29. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 29 Bruce Mayer, PE Chabot College Mathematics Example  Price Elasticity  Then:  so

30. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 30 Bruce Mayer, PE Chabot College Mathematics 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

31. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 31 Bruce Mayer, PE Chabot College Mathematics Hi Levels of Elasticity:  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

32. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 32 Bruce Mayer, PE Chabot College Mathematics Lo Levels of Elasticity:  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

33. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 33 Bruce Mayer, PE Chabot College Mathematics Neutral Elasticity:  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

34. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 34 Bruce Mayer, PE Chabot College Mathematics 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

35. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 35 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §3.4 P52 → Bird Flight Power P58 → Radio Ratings

36. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 36 Bruce Mayer, PE Chabot College Mathematics All Done for Today HiElastic vs. LoElastic

37. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.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 MTH15_Lec-16_sec_3-4_Optimization.pptx 38 Bruce Mayer, PE Chabot College Mathematics ConCavity Sign Chart abc −−−−−−++++++−−−−−−++++++ x ConCavity Form d 2 f/dx 2 Sign Critical (Break) Points InflectionNO Inflection Inflection

39. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 39 Bruce Mayer, PE Chabot College Mathematics

40. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 40 Bruce Mayer, PE Chabot College Mathematics

41. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 41 Bruce Mayer, PE Chabot College Mathematics

42. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 42 Bruce Mayer, PE Chabot College Mathematics

43. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 43 Bruce Mayer, PE Chabot College Mathematics

44. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 44 Bruce Mayer, PE Chabot College Mathematics

45. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 45 Bruce Mayer, PE Chabot College Mathematics P3.4-58 Plot by MuPAD

46. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 46 Bruce Mayer, PE Chabot College Mathematics

47. BMayer@ChabotCollege.edu MTH15_Lec-16_sec_3-4_Optimization.pptx 47 Bruce Mayer, PE Chabot College Mathematics