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Finding the Price to Maximize Revenue

Finding the Price to Maximize Revenue. Ted Mitchell. Introduction To Revenue. TJM. Sales Revenue. R = P x Q where R = Sales Revenue P = Price per Unit Q = Quantity Sold (Demanded). Demand Function.

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Finding the Price to Maximize Revenue

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  1. Finding the Price to Maximize Revenue Ted Mitchell

  2. Introduction ToRevenue 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, PQ = 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 Quantity Sold Demand Equation Q = a - bP 100 ??? Price per Unit $50 $75 TJM

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

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

  9. How To Calculate Demand? Demand equation is represented byQ = 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 Quantity Sold Demand Equation Q = a - bP X 500 X X X X X X X X X Price per Unit $200 TJM

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

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

  17. Example • Market research has estimated your demand to be Q = 1500 -5PThe 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 Quantity Demand Equation Q = 1,500 -5P a =1,500 X 500 X X X X X X X X X Price per Unit $200 TJM

  19. The Revenue Equation • Revenue, R = P xQ • Q = a - bP Substitute Q = (a-bP) • R = P(a - bP) • R = aP - bP2

  20. Revenue looks likeR = aP - bP2 Revenue 0 Price 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 likeR = 5,000P - 500P2 Revenue $12,000 0 $4 $6 Price

  30. What Happens At $5.00R = 5,000P - 500P2 Revenue $? $12,000 0 $4 $6 $5 Price

  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 - 500P2 Revenue $12,500 $12,000 0 $4 $5 $6 Price 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 -500P2 • 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 – 500P2 • Max revenue = 5(5,000 – 500(52)) • Max Revenue = 25,000 – 12,500 = $12,500

  36. Fortunately there is a general solution!

  37. Revenue looks likeR = aP - bP2 Revenue R 0 P Price TJM

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

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

  40. Find The First Derivative of The Revenue Curve and Set It Equal to Zero R = aP - bP2 Revenue Equation TJM

  41. Find The First Derivative of The Revenue Curve and Set It Equal to Zero R = aP - bP2 TJM

  42. Find The First Derivative of The Revenue Curve and Set It Equal to Zero R = aP - bP2 TJM

  43. Find The First Derivative of The Revenue Curve and Set It Equal to Zero R = aP - bP2 dR/dP = a – 2bP dR/dP set equal to 0 TJM

  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 AsQ = 5000 - 500P P = 5,000/2(500) = $5

  48. Rule of ThumbThe Price That Maximizes Revenue Is Give-Away-QuantityDivided 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 - bP2

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