Results 1 to 2 of 2

Math Help - Profit function!

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    4

    Profit function!

    P(X) = -60X2 + 1500X - 4000 - 1 x 25

    This is the profit function and its domain,

    can you please help me find
    1) the value of X where there is a breakeven point
    2) the value of X where profit be maximum

    thankyou so very much!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by karimwahab View Post
    P(X) = -60X2 + 1500X - 4000 - 1 x 25

    This is the profit function and its domain,

    can you please help me find
    1) the value of X where there is a breakeven point
    2) the value of X where profit be maximum

    thankyou so very much!!!
    The breakeven point is the point where there is no profit and no deficit. In other words, the poitns when  P(x) = 0 . You find x values of this by finding the roots of the equation. Using the quadratic formula!

     P(x) = -60x^2+1500x-4000 = 0

     x = \frac{-1500 \pm \sqrt{1500^2-4(-60)(-4000)}}{2(-60)}

     x = \frac{-1500 \pm \sqrt{1290000}}{-120}

     x = 3.035 or  x = 21.964

    For the maximum, the first derivative of the function is 0. Solve for x, and then plug that x value into the original equation to find P(x) for that value of x. The 2nd derivative test will tell you whether you have a maximum or a minimum. (Although you will have a maximum, since your coefficients of x^2 is negative!)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 20
    Last Post: November 27th 2012, 06:28 AM
  2. Replies: 14
    Last Post: May 7th 2011, 06:58 PM
  3. determining a profit function
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 25th 2011, 06:01 PM
  4. Replies: 1
    Last Post: February 23rd 2011, 06:15 PM
  5. maximum profit? quadratic function??
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 6th 2010, 12:04 PM

Search Tags


/mathhelpforum @mathhelpforum