Question

Find the profit function for the given marginal profit and initial condition. Marginal Profit Initial Condition dP dx = −20x + 290 P(5) = $670 P(x) =

Answer 1

Answer:

$P(x) = -10x^2+290x-530$    

Step-by-step explanation:

We are given the following in the question:

$\dfrac{dP}{dx} = -20x + 290$

Initial condition:

$P(5) = 670$

Solving the given differential equation, we get,

$\dfrac{dP}{dx} = -20x + 290\\\\dP=(-20x + 290)dx\\\text{Integrating both sides}\\\\\displaystyle\int dP = \int (-20x + 290)dx\\\\P(x) = -20\dfrac{x^2}{2} + 290x + C\\\\\text{where C is constant of integration}\\P(x) = -10x^2+290x + C\\P(5) = 670\\670 = -10(5)^2+290(5) + C\\670 = 1200 + C\\\Rightarrow C = -530\\P(x) = -10x^2+290x-530$

is the required profit function.