business applications integrals

Question

A semiconductor company has decided that the marginal revenue of its newest computer chip (in dollars per chip) is given by R'(x) = −5x² + 60x + 180 and the marginal cost (in dollars per chip) is given by C'(x) = 20x + 80 where x is the number of chip manufactured. What would be the maximized profit of this chip for the company?

Answer 1

The profit per chip will be P(x) = R(x) - C(x).

To maximise profit solve P'(x) = 0. P'(x0 = R'(x) - C'(x)

So you need R'(x) = C'(x)

-5x² + 60x + 180 = 20x + 80

5x² - 40x - 100 = 0

x² - 8x - 20 = 0

(x-10) (x+2) = 0

x = 10 or -2

Clearly x>0 so only possible solution is x=10

To find the maximised profit you need to find R(x) and C(x) then find R (10) - C(10)

To find R(x) integrate R'(x) giving (-5/3) x³ + 30x² + 180x

And similarly C(x) = 10x² + 80x

Now you can find R(10) - C(10).