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?

Answer 1

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?

Answer 2

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^2-800x)
0=-1/8(x-400)^2+20000
20,000 is your answer