Revenue function, profit function, and drawing a signs diagram.

Question

This is the question:

Suppose the demand price for selling 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 oof 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.

iv. use (iii) to find the production level x for which the profit is a maximum.

v. so what is the maximum profit the DVD manufacturer can hope to make? Show your work.

I am having trouble with part 3, because when I use the quadratic formula, I get x= -166.7 and x=168, in which I then use the intervals to be able to draw the signs diagram: (-Infiniti, -166.67) (-166.67, 168) (168, Infiniti), but I'm not sure this is correct because when I use a test point in the interval (-166.67, 168), for example -150 and 1, I plug it in into p'(x), but one of them gives me a negative number while the other gives me a positive number, and I can't tell whether this interval is increasing or decreasing. I may have set this up wrong, or I may have done something wrong before this step.

Answer 1

If R(x) = xp(x)

then P(x) = xp(x) - c(x) = -.0005x³ + .001x² + 42x - 4000

P'(x) = -.0015x² +.002x + 42 and I get 166.7 as the positive root.

The negative root is inadmissible because x must be > 0. The sign changes from one side to the other for the maximum (>0 to the left and <0 to the right) . You can also check the second derivative: -.003x +.002 which is clearly <0 for x= 166.7 (indicating max)

Answer 2

Revenue = Px = (.0005x^2 + 60)x = .0005x^3 + 60x

R(x) =- 0.0005x^3 + 60x

=.0005x(x^2 - 120,000)

domain is 0<x < sqr120000

0< x < about 346.2

C(x) = 0.001x^2 + 18x + 4000

profit function, P(x) = R(x)-C(x) = -.0005x^3 -.001x^2 + 42x -4000

max profit is when the derivative = 0

P'(x) = -.0015x^2 -.002x + 42 = 0

multiply by -10000, move the decimal to the right by 4 digits and change the signs

15x^2 + 20x - 420,000 = 0

divide by 5

3x^2 + 4x - 85,000 = 0

use the quadratic formula

x = -8/6 + or - (1/6)sqr(16+12(85000)) (ignore the negative square root)

x = -4/3 + (1010)6 = about 167

max profit is P(167) = -0.0005(167)^3 - 0.001(167)^2 + 42(167) - 4000

= -2,328.7 -27.9 + 7,014 -4000 = about $657.4