The price p and the quantity x sold of a certain product obe

The price p and the quantity x sold of a certain product obey the demand equation below, x = -6p + 120, 0 lessthanorequalto p lessthanorequalto 20 Express the revenue R as a function of x What is the revenue if 30 units are sold? What quantity x maximizes revenue? What is the maximum revenue? What price should the company charge to maximize revenue? What price should the company charge to earn at least $450 in revenue? R(x)= The revenue if 30 units are sold is (Simplify your answer) The quantity units maximizes revenue. Simplify your answer.) The maximum revenue is (Simplify your answer) The company should charge s to maximize revenue (Simplify your answer.) The company should charge at least $ to at most $ to earn at least $-450 in revenue.

Solution

demand equation

x = -6p + 120

price : p = (120 -x)/6

a) revenue function is given by x* p

therefore, R(x) = x*(120 -x)/6 = -x^2/6 +120x

b) revenue if 30 units are sold , plug x = 30

R(x) = -(30)^2/6 +120*30 = $ 3450

revenue = -6(15)^2 + 120 (15) = -1350 + 1800 = 450

c) maximum revenue occurs at p :

R(x) = -x^2/6 +120x

maximum revenue : x = -b/2a = -(120)/(2*-1/6) = 120*3 = 360 units

d) R(360) = -(360)^2/6 +120*360 = $ 21600