Revenue function, profit function, and domain

Question

Suppose the demand price for selling out a production run of x DVDs is given by p(x)=-0.0005x2+60 dollars per DVD. Further suppose the weekly cost of producing these DVDs is given by C(x)=-0.001x^2+18x+4000 dollars to produce that many DVDs.

(i) Write the Revenue function R(x). And be sure to state its DOMAIN.

(ii) Use (i) to write the profit function P(x). Again, be sure to state its DOMAIN.

(iii) Draw a signs diagram for P’(x), the marginal profit function. Here you will first need to solve P’(x) =0 by using the quadratic formula.

I am very confused as to how to get the domain in these problems because initially I thought it was (-infinity, infinity) for both revenue function and profit function, but my professor said that mathematically this is correct, but since we're talking about the real world it would be something else, and I would like an explanation as to how to get the domain.

Answer 1

p(x) =- 0.0005x^2 + 4,000

Revenue = x time Price = R(x) = xp(x) = -0.0005x^3 +4000x

domain of Revenue is x = or > 0

realistically, there is some upper bound as well, in the Long Run revenues > cost or the business goes bankrupt

find the upper bound for Profit, then find the corresponding revenue

Profit = P(x) = R(x)-C(x) = -0.0005x^3 +4000x - (0.001x^2 + 18x +4000) = -0.0005x^3 -0.001x^2 + 3982x

domain is any non negative x value

realistically there is also some upper bound as well

in the Long Run, the business shuts down or goes bankrupt if profits are negative, so any x value that makes profits negative is a Long Run upper bound for the domain. x= -.002/.001+or- (1/.001)sqr(.001^2 +.002(3982))

but in the Short Run, profits can be negative, if believed to be temporary

set P=0 solve for x

the Long Run is defined as when you can vary your fixed cost, as you're stuck with them, in the Short Run

in the Short Run if revenue covers your variable costs, it pays to stay in business.

P'(x) = R'(x)-C'(x) = -0.0015x^2 - 0.002x + 3982= Marginal Profit