profitable function

Question

 Given revenue equation R(x) = 27x and cost equation C(x) = .1x^2 + 9x +300, production x > 0, answer the following.

 Profitable production level
 Maximum profit and related production level
Maximum average profit and related production level
Average change in profit between x = 80 and x = 90

Answer 1

Profit is Revenue - Cost

Profit = 27x - (0.1x² + 9x + 300)

Profit > 0 when -0.1x² + 18x - 300 > 0

Use the quadratic formula to solve for the two zeros of this parabolic function, which are

18.6 and 161.4

so production levels between 18.6 and 161.4 are profitable (too little production and too much production are unprofitable.)

 

You can find the maximum profit by differentiating the profit function and setting it equal to zero

that is

-0.2x + 18 = 0 

x = 90 at this point the profit(x=90) = -0.1*(90*(90) + 19*90 -300 = 510

 

average change in profit = [profit (90) - profit(80)]/[90-80) = (510-500)/(90-80) = 10/10 = 1