precalc
posted by ssassie on .
Suppose the function p= 1/8 x + 100 (0 grtr&= x ~12) relates the selling price p of an item to the quantity x that is sold. Assume that p is in dollars. What is the maximum revenue possible in this situation?

Suppose the function p= 1/8 x + 100
(0 grtr&= x ~12) relates the selling
price p of an item to the quantity x
that is sold. Assume that p is in
dollars. What is the maximum
revenue possible in this situation? 
You should use the formula for revenue R(x)=p*q.
In this case q is x. So you do R(x)= 1/8 x + 100(x). This results in R(x)= 1/8x^2 + 100x.
To find the max, set this to 0 and complete the square:
0=1/8(x^2800x)
0=1/8(x400)^2+20000
20,000 is your answer