SOLUTION: Maximizing profit. The total profit (in dollars) for sales of x rowing machines is given by P(x) = 0.2x^2 + 300x - 200. What is the profit if 500 are sold? For what value of x will

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Maximizing profit. The total profit (in dollars) for sales of x rowing machines is given by P(x) = 0.2x^2 + 300x - 200. What is the profit if 500 are sold? For what value of x will      Log On


   

Question 37677: Maximizing profit. The total profit (in dollars) for sales of x rowing machines is given by P(x) = 0.2x^2 + 300x - 200. What is the profit if 500 are sold? For what value of x will the profit be at a maximum?
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Given the profit equation
P(x) = 0.2x^2 + 300x - 200
we merely plug in x = 500 and get
P(500) = .2(500)^2 + 300(500) - 200 = 50000 + 150000 - 200 = $199800
The x-coordinate of the vertex, and hence of the maximum, is found by -b/2a, but in this case there is no maximum...
I have a feeling that this P(x) equation is written incorrectly since it has no maximum...