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

1. Chabot Mathematics §3.4 Elasticity& Optimization Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. 3.3 Review § • Any QUESTIONS About • §3.3 → Graph Sketching • Any QUESTIONS About HomeWork • §3.3 → HW-15

3. §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. 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. 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. 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. Extreme Value Explained • The Absolute-Max or Absolute-Min over some x-span of any function occurs EITHER at • The ENDS • SomeWherein the Middle • (Duh!!!) Abs-MAX Abs-MAX Abs-MIN

8. Example  Find Abs Extrema • Find the absolute extremafor 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. Example  Find Abs Extrema • First, find critical points by setting the derivative equal to zeroand solving: • Recall, however, that the interval of interest is [−3,1] • Thus x=4 is NOTpart of the Slon QuotientRule ZeroProducts

10. 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. Example  Find Abs Extrema • ThefcnGraph

12. 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. Example  Finding Max Profit • The Math Model for Pricingof “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. 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. Example  Finding Max Profit • Now Total Revenue = [price]·[Qty-Sold] • R in (kUnit)·($/Unit) = k$ • The Marginal Analysis requires Derivatives for R & C • x

16. 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. 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. Ex:  Find Max Profit

19. % 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 = [xminxmax]; 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([xminxmaxypminypmax]),... 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 MATLAB Code

20. 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. 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. 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 thisCase

23. 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. Example  Find Min Avg Cost • When producing 200 staplers, average cost is minimized at a value of • The units for Amin are $k/kUnit or $/Unit • Thus the factory incurs a minimum average cost of $5 per Stapler when producing 200 units

25. Ac & dC/dq

26. 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. 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. Example  Price Elasticity • SOLUTION: • ReCall The price elasticity of demand Formula • Use the Quantity-Demanded Eqn: • Then

29. Example  Price Elasticity • Then: • so

30. 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. 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. 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. 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. 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 smallDemand change

35. WhiteBoard Work • Problems From §3.4 • P52 → Bird Flight Power • P58 → Radio Ratings

36. All Done for Today HiElasticvs.LoElastic

37. Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu –

38. ConCavity Sign Chart ConCavityForm ++++++ −−−−−− −−−−−− ++++++ d2f/dx2 Sign x Critical (Break)Points a b c Inflection NOInflection Inflection

45. P3.4-58 Plot by MuPAD