Results 1 to 3 of 3

Math Help - Maximum revenue problem.

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    57

    Maximum revenue problem.

    Hi,

    I have revenue:
    r(x)=100x-.0001x^2

    and cost:
    c(x)=360+80x+.002x^2+.00001x^3

    I must maximise profit.
    profit is: r(x)-c(x)=-.00001x^3-.0021x^2+20x-360

    p'(x)=-.00003x^2-.0042x+20.

    The answer is 749 units.

    I cannot seem to factor p' to get this result.
    I have tried completing the square.

    x^2+140x-\frac{20}{.00003}=0

    (x+70)^2=70^2+\frac{20}{.00003}

    x=-70\pm \sqrt{70^2+\frac{20}{.00003}}

    Calculating that out doesn't get me near 749.

    I make it x\approx{3431.3998}

    Thanks
    Craig.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2006
    Posts
    77
    p'(x)=-.00003x^2-.0042x+20=0
    0=3x^2+420x-2000000
    x=\frac{-420\pm\sqrt{420^2+24000000}}{6}
    x=\frac{-420\pm4916.95...}{6}

    Gives 749.491...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    77
    Might be going crazy but can't find the edit button suddenly. Anyway your answer is fine, perhaps you're typing it in wrong?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. revenue problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 11th 2010, 06:46 PM
  2. Need help with a maximum revenue problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 4th 2010, 02:42 PM
  3. Replies: 1
    Last Post: November 24th 2009, 01:43 PM
  4. Replies: 1
    Last Post: November 18th 2009, 08:16 PM
  5. Maximum Weekly Revenue
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 9th 2008, 03:31 PM

Search Tags


/mathhelpforum @mathhelpforum