<PRE><B>Question 81322</B>
The price p and the quantity x sold of a certain product obey the demand equation: 
x = -20p + 500 
p(x)=(500-x)/20 = 25-(1/20)x
0 less than or equal to p which is less than or equal to 25 
a) express the revenue R as a function of x
Revenue(x) = (price)*(quantity)
Rev(x) = (25-(1/20)x)(x
Rev(x) = 25x-(1/20)x^2
b) what is the revenue if 20 units are sold?
Rev(20) = 25*20-(1/20)20^2
Rev(20) = 500-20
Rev(20) = $480
-----------------
c) what quantity x maximizes revenue? what is the maximum revenue?
Rev(x) is a quadratic with a=-1/20, b=25
max occurs at x=-b/2a = [-25/(-2/20)] = [250]
=-------------------------

d) what price should the company charge to maximize revenue? 
p(x) = 25-(1/20)x
p(250) = 25-(1/20)(250)
p(250) = 25-25/2
p(250) = $12.50
========================
Cheers,
Stan H.
</PRE>