10 Using the demand function given in 8 q 167p 66667 a Sta

10. Using the demand function given in #8, q = -1.67p + 666.67,

a. State the revenue as a function of price.

b. What price should the company charge to maximize revenue?

c. What is the maximum revenue?

Solution

given demand function  q = -1.67p + 666.67

Where p is the price (in dollars) per unit when q units are demanded

q -666.67 = -1.67p

1.67p = -q + 666.67

p = -q/1.67 + 666.67 /1.67

p = -0.6q + 399.20

now  revenue functionR(q) = pq

= (-0.6q + 399.20)q

= -0.6 q^2 + 399.20q

now to maximize this we have to differentiate the revenue function and make it as zero

R\' (q) = 2(-0.6p) + 399.20

0 = -1.2p +666.67

1.2q = 399.20

q =333.67

p = -0.6q +339.2

P = -0.6 X199.6 +339.2

P = -199.6 +339.2

p = 139.6

wiil maximize the revenue

maximum revenue R(199.6) = -0.6 q^2 + 399.20q

= 0.6(333.67)^2 +399.20 x333.67

= -66801.4 +113180.9

= 46,379.5