Respuesta :

-> R (y) = (10y -3) 2y = 20y² - 6y -> R' (y) = 40y - 6 Optimal level of output when R' (y) = C'(y) -> 40y - 6 = 2 => 40y = 8 => y = 8/40 = .2 by using differential calculus w got its optimal level and price