Additional Topic for Ch.16: Optimal Price for Maximizing Revenue Warin Chotekorakul

How to get an optimal price? Two main ways: Two main ways: Calculus: taking the 1 st derivative Calculus: taking the 1 st derivative Algebra: substituting constants Algebra: substituting constants

1 st way: Calculus Taking only the 1 st derivative of the revenue function, then we can get the optimal price. Taking only the 1 st derivative of the revenue function, then we can get the optimal price. y = f(x) = 3x 2 + 9x + 4 y = f(x) = 3x 2 + 9x + 4 y = f(x) = -4x 2 + 9x + 5 y = f(x) = -4x 2 + 9x + 5 y = f(x) = 7x 2 - 6x + 6 y = f(x) = 7x 2 - 6x + 6

2 nd way: Algebra Substituting constants or parameters into the formulae: Substituting constants or parameters into the formulae: ( x = -b, y = 4ac – b 2 ) ( x = -b, y = 4ac – b 2 ) 2a 4a

The demand for the product of a firm varies with the price that the firm charges for the product. The firm estimates that annual total revenue R (stated in $1,000s) is a function of the price p (stated in dollars). Specifically, The demand for the product of a firm varies with the price that the firm charges for the product. The firm estimates that annual total revenue R (stated in $1,000s) is a function of the price p (stated in dollars). Specifically, R = f(p) = - 50p p a) Determine the price which should be charged in order to maximize total revenue. b) What is the maximum value of annual total revenue? E XAMPLE

The 1 st derivative of R = f(p) = - 50p p is The 1 st derivative of R = f(p) = - 50p p is R’ = f’(p) = -100p If we set f’(p) = 0, then -100p = p = p = p = -500 p = 5 a) Therefore, a max revenue occurs at price = 5 a) Therefore, a max revenue occurs at price = 5 S OLUTION: Calculus

The max value of R is found by substituting p = 5 into revenue function b) The max value of R is found by substituting p = 5 into revenue function R = f(p) = -50p p R = f(p) = -50p p R = f(5) = -50(5) (5) = -1, ,500 = 1,250 = -1, ,500 = 1,250 Thus, annual total revenue is expected to be maximized at $1,250 (1,000s) or 1.25 million when the firm charges $5 per unit. S OLUTION (2) : Calculus

S OLUTION (3) : Algebra R = f(p) = - 50p p Substitute constants or parameters of the function into the formulae: R = f(p) = - 50p p Substitute constants or parameters of the function into the formulae: ( x = -b, y = 4ac – b 2 ) ( x = -b, y = 4ac – b 2 ) 2a 4a x = -(500), y = 4(-50)(0) – (500) 2 x = -(500), y = 4(-50)(0) – (500) 2 2(-50) 4(-50) 2(-50) 4(-50) Then, p = 5; R = 1,250

What if, the question asks for max profit? We knew Revenue = Price * Quantity sold We knew Revenue = Price * Quantity sold Quantity sold = Revenue/Price Then, we can calculate max profit by Then, we can calculate max profit by Profit = (P-Cost)* Quantity sold