Find the production level that will maximaze the profit

Question

For the given cost function

C(x)=3000+640x+1.1x² and the demand function p(x)=1920 (i.e. the price p at the given volume of demand x) .
Find the production level that will maximaze the profit P(x)=R(x)-C(x), where R(x) is the revenue function.

Answer 1

The problem is not well stated.  The equation for R(x) is not given so I'm guessing that R(x) = 1920x (revenue = price*demand)?  If that's not correct, please provide the equation for R(x).

 

P(x) = 1920x - (3000+ 640x + 1.1x²)

 

P(x) = -1.1x² + 1280x - 3000

 

Take the derivative of P(x), set it to zero, and solve for x:

 

dP(x)/dx = -2.2x + 1280

 

0 = -2.2x + 1280

 

Solve for x.