1. Finding the Price to Maximize Revenue Ted Mitchell

2. Introduction To Revenue TJM

3. Sales Revenue R = P x Q where R = Sales Revenue P = Price per Unit Q = Quantity Sold (Demanded)

4. Demand Function The Quantity, Q, that will be sold is also determined by the price per unit, P Q = ƒ(P) If R = P x Q R = P x ƒ(P) If ƒ(P) = (a-bP) Then R = P x (a-bP)

5. We have seen that Demand Q = a – bP is a slope-intercept version of a two-factor marketing machine where b = the meta-conversion rate from two observations, ∆Q/∆P a = y-intercept

6. Higher Price Sells Fewer Units Price per Unit $50 100 $75 Quantity Sold ??? Demand Equation Q = a - bP TJM

7. Higher Price Sells Fewer Units Price per Unit $50 100 $75 Quantity Sold ??? Demand Equation Q = a - bP TJM

8. Higher Price Sells Fewer Units Price per Unit $50 100 $75 Quantity Sold ??? Demand Equation Q = a - bP TJM Revenue is an area! R=PQ

9. How To Calculate Demand? Demand equation is represented by Q = a – bP Market Research provides estimates of the a & b

10. How To Calculate Demand? Demand equation is represented by Q = a - bP Q = Quantity That Will Be Demanded At Any Given Price, P

11. How To Calculate Demand? Demand equation is represented by Q = a - bP a = quantity that could be given away at a zero price

12. How To Calculate Demand? Demand equation is represented by Q = a - bP b = number of units or lost sales if the price is increased by one unit

13. How To Calculate Demand? Demand equation is represented by Q = a - bP P = Price Per Unit

14. Data on Prices and Quantities Price per Unit $200 500 Quantity Sold Demand Equation Q = a - bP TJM X X X X X X X X XX

15. Get an estimate of a & b Price per Unit $200 500 Quantity Demand Equation Q = a - bP TJM X X X X X X X X XX a

16. Example Market research has estimated Demand Quantity sold = a – b(Price) a = 1,500 and b = 5, your demand function is estimated to be Q = 1500 -5P The current price is $200 per unit. What quantity is being demanded?

17. Example Market research has estimated your demand to be Q = 1500 -5P The current price is $200 per unit. What quantity is being Demanded? Q = 1500 -5(200) Q = 1500 -1000 = 500 units

18. Get an estimate of a & b Price per Unit $200 500 Quantity Demand Equation Q = 1,500 - 5P TJM X X X X X X X X XX a =1,500

19. The Revenue Equation – Revenue, R = P xQ Q = a - bP Substitute Q = (a-bP) R = P(a - bP) R = aP - bP 2

20. Revenue looks like R = aP - bP 2 Revenue Price 0 TJM

21. $6 & $4 Examples The Demand For Your Product Changes with The Price You Charge As » Q = 5000 - 500P » R = PQ » R = P(5000 - 500P)

22. $6 Example R = PQ R = P(5000 - 500P) What is your total sales revenue if your price is six dollars?

23. $6 Example R = PQ R = P(5000 - 500P) – substitute

24. $6 Example R = PQ R = P(5000 - 500P) – Substitute P = $6 R = $6(5000 - 500($6)) R =$12,000

25. $4 Example R = PQ R = P(5000 - 500P) What is your total sales revenue if your price is four dollars?

26. $4 Example R = PQ R = P(5000 - 500P) – Substitute $4 = P

27. $4 Example R = PQ R = P(5000 - 500P) – Substitute $4 = P R = 4(5000 - 500(4)) R = $12,000

28. How Can The $12,000 Revenue At P = $6 Be The Same As The Revenue At P = $4?

29. Because Revenue looks like R = 5,000P - 500P 2 Revenue Price 0 $12,000 $4$6

30. What Happens At $5.00 R = 5,000P - 500P 2 Revenue Price 0 $12,000 $4$6 $5 $?

31. $5 Example *** R = PQ R = P(5000 - 500P) What is your total sales revenue if your price is five dollars?

32. $5 Example R = P(Q) R = P(5000 - 500P) – Substitute P=$5 R = $5(5000 - 500($5)) R = $12,500 TJM

33. $5 Price Maximizes Revenue??? R = 5,000P - 500P 2 Revenue Price 0 $12,000 $4$6 $12,500 $5 TJM

34. Find the optimal Price that Maximizes Revenue Establish the Demand Function, Q = 5000 -500P Establish the Revenue function R = P x Q = P x (5000 -500P) = 5000P -500P 2 Find the first derivative wrt price dR/dP = 5000 – 2(500)P = 5000 -1000P Set the first derivative equal to zero and solve for P 5000 – 1000P = 0 P = 5000/1000 = $5

35. The maximum revenue is Revenue = P x Q = P x (5,000 – 500P) Substitute optimal P = $5 Max revenue = 5,000P – 500P 2 Max revenue = 5(5,000 – 500(5 2 )) Max Revenue = 25,000 – 12,500 = $12,500

36. Fortunately there is a general solution!

37. Revenue looks like R = aP - bP 2 Revenue Price 0 R P TJM

38. Slope of Revenue Curve is Revenue Price 0 R P TJM

39. Slope Of Revenue Curve Is Zero Revenue Price 0 R P TJM

40. Find The First Derivative of The Revenue Curve and Set It Equal to Zero R = aP - bP 2 Revenue Equation TJM

41. Find The First Derivative of The Revenue Curve and Set It Equal to Zero R = aP - bP 2 TJM

42. Find The First Derivative of The Revenue Curve and Set It Equal to Zero R = aP - bP 2 TJM

43. Find The First Derivative of The Revenue Curve and Set It Equal to Zero R = aP - bP 2 TJM dR/dP = a – 2bP dR/dP set equal to 0

44. Solve for P dR/dP = a – 2bP = 0 a – 2bP = 0 -2bP = -a P = a/2b

45. The Price that maximizes revenue is

46. The optimal price for max revenue The Demand For Your Product Changes with The Price You Charge As » Q = 5000 - 500P

47. $6 & $4 Examples The Demand For Your Product Changes with The Price You Charge As Q = 5000 - 500P P = 5,000/2(500) = $5

48. Rule of Thumb The Price That Maximizes Revenue Is Give-Away-Quantity Divided by Twice The Number Of Lost Sales For Any Dollar Increase In Price TJM

49. Simple Exam Question What Is The Maximum Revenue That Can Be Generated If The Demand For The Product Is » Q = a - bP

50. Simple Exam Question What Is The Maximum Revenue That Can Be Generated If The Demand For The Product Is » Q = a - bP P = a/2b R = aP - bP 2

51. Optimal Price Max Rev Price per Unit a/2b Quantity Sold a/2 Demand Equation Q = a - bP TJM

52. Optimal price Max Rev Price per Unit a/2b = 5000/2(500) = $5 Quantity Sold a/2 = 5000/2= 2,500 Demand Equation Q = 5000 – 500P TJM

53. Price per Unit $4$5 Quantity Sold 2,500 Demand Equation Q = a - bP TJM 3,000 $4 x 3,000 =12,000

54. Lower Price Sells More Units Price per Unit $4$5 Quantity Sold 2,500 Demand Equation Q = a - bP TJM 3,000 $4 x 3,000 =12,000

55. Exam Question The Revenue equation has been estimated as R = P(5,000 -500P) R = 5,000P -500P 2 Find the first derivative = dR/dP dR/dP = 5,000 -1,000P Set dR/dP equal to zero and solve for P* 5,000 -1,000P* = 0 P* = 5,000/1,000 = $5

56. Slope Of Revenue Curve Is Zero, dR/dP = 0 Revenue Price 0 R P* TJM

57. What we are learning? There is a optimal price, P* that maximizes sales revenue Too High a selling price reduces revenue Too Low a selling price reduces revenue