calculus problem!!!!

Question

 A company manufacturers and sells x electric drills per month. The monthly cost and price-demand equations are

C(x)=73000+80x,

p=190−x/30, 0≤x≤5000.

(A) Find the production level that results in the maximum profit.

Production Level = 

(B) Find the price that the company should charge for each drill in order to maximize profit.

Price =

Answer 1

Max Profit = $197,750

with Price = p = $95

and output = x = 2,850

C(x) = 73000 -80x

p = 190-x/30

R=px = 190x -x^2/30

Profit = P=R-C

= 190x-x^2/30 - (73000-80x)

= 190x -x^2/30 -73000+ 80x

= 270x -x^2/30 -73000

take the derivative of P with respect to x, and set = 0

P'(x) = 270 -x/15 =0

x = 15(270) = 4050 = profit maximizing out put number of electric drills per month, which is less than 5000

p = 190-x/30 = 190-4050/30 = 190- 405/3 = 190-135 = $55 = profit maximizing price

max Profit

= 190x -x^2/30 -73000

= 190(4050) -(4050)^2/30 -73000

= 769500-546750-7300

=$149,750 = most Profit possible, per month