Express Revenue R as function of X

Question

Demand Equation. The price (in dollars) and the quantity x sold of certain product obey the demand equation:
p= - 1/10x + 150
Revenue is x*p.
(a)Express the revenue R as a function of x.
(b)What quantity x maximizes revenue? What is the maximum revenue?

Answer 1

a) In the revenue (R) equation, we can replace p by -1/10 x + 150

 

So R = x(-1/10 x + 150) = -x² + 150x

 

b) This is a quadratic equation (parabola), so the value of x = "-b/2a" = -150/2(-1) = 75 SOLD FOR MAX REVENUE

 

    The max value of R = -(75)²  + 150(75) = $5625  MAX REVENUE