Results 1 to 4 of 4

Math Help - Maximum, function with three variables

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    60

    Maximum, function with three variables

    Hi everybody!
    I have a function, which encompassed formerly two variables and three constants. I found the maximum of this function. For some analysis I'm doing, I have to make one of the constants a variable though.
    If i do the partial derivative regarding the new variable, i get (B-2m)/A (A,B = constants; m = variable; A,B,m >0; m<B). I wonder now, where the maximum of this new function with three variables is.
    What i did is i set (B-2m)/A = 0, what's true for m=B/2. This itself is a linear function with a negative slope, but the function is always positive before m=B/2. It starts from B and does down linearly untill m=B/2. So theoretically the overall function increases untill m=B/2, after this point, the partial derivative gets negative ( ((B-2m)/A)< 0 for m>B/2) and henceforth starts from then on decreasing the overall function.

    Is it correct that i thus say, the maximum of the function with three varibales is the same as the one with two, except for m=B/2?

    Thank you

    Schdero
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Schdero View Post
    Hi everybody!
    I have a function, which encompassed formerly two variables and three constants. I found the maximum of this function. For some analysis I'm doing, I have to make one of the constants a variable though.
    If i do the partial derivative regarding the new variable, i get (B-2m)/A (A,B = constants; m = variable; A,B,m >0; m<B). I wonder now, where the maximum of this new function with three variables is.
    Partial derivative of what?

    What i did is i set (B-2m)/A = 0, what's true for m=B/2. This itself is a linear function with a negative slope, but the function is always positive before m=B/2. It starts from B and does down linearly untill m=B/2. So theoretically the overall function increases untill m=B/2, after this point, the partial derivative gets negative ( ((B-2m)/A)< 0 for m>B/2) and henceforth starts from then on decreasing the overall function.

    Is it correct that i thus say, the maximum of the function with three varibales is the same as the one with two, except for m=B/2?

    Thank you

    Schdero
    You need to post sufficient information so that we can see what exactly you are talking about.

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    60
    Overall Function=  M((b-M)/a)>((b-d)(ad+bc))/(a+c)^2

    Maximum of the function f(a,c) is at a and c = infinity, as then fa=0 fc=0, (faa)(fcc)-fac^2 > 0.

    Is there any further information you need to judge the corectness of my implication, that the maximum of f(a,c,m) lies at a and c = infinity, m=b/2?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Schdero View Post
    Overall Function=  M((b-M)/a)>((b-d)(ad+bc))/(a+c)^2

    Maximum of the function f(a,c) is at a and c = infinity, as then fa=0 fc=0, (faa)(fcc)-fac^2 > 0.

    Is there any further information you need to judge the corectness of my implication, that the maximum of f(a,c,m) lies at a and c = infinity, m=b/2?
    What is the function you are trying to maximise, and what are the vaiables and constraints?

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 15th 2011, 07:21 AM
  2. Replies: 10
    Last Post: September 3rd 2011, 01:03 PM
  3. Replies: 1
    Last Post: April 2nd 2011, 12:49 PM
  4. Maximum of Function with three variables
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 5th 2011, 11:23 AM
  5. Replies: 2
    Last Post: June 4th 2009, 07:57 PM

Search Tags


/mathhelpforum @mathhelpforum