Question 1204672
The price p(in dollars) and the quantity x sold of a certain product obey the demand equation      p=-1/5x+20 , 0<=p<=20
A.express the revenue R as a function of x

R= p*x

R= ((-1/5)x+20)*x

R = (-1/5)x^2+20x

B. What is the revenue if 15 units are sold?

R =  (-1/5)x^2+20x

R=  (-1/5)*15^2+20*15
R=$255

C.what quantity x maximizes revenue? 

R =  (-1/5)x^2+20x
dR/dx= -(2/5)*x+20
To maximise set the derivative to 0
 -(2/5)*x+20=0
2/5*x=20
x=100/2
x= 50

D. What price should the company charge to maximize revenue?
Maximum units 50

P= -1/5*50+20
-10+20
=10
The company should charge $10 to maximize revenue when selling 50 units.