Results 1 to 10 of 10
Like Tree4Thanks
  • 1 Post By romsek
  • 2 Post By JeffM
  • 1 Post By JeffM

Thread: Lagrange multiplier

  1. #1
    Newbie
    Joined
    Nov 2014
    From
    Lithuania
    Posts
    20

    Exclamation Lagrange multiplier

    A firm that manufactures speciality bicycles has a profit function
    π = 5x^2 − 10xy + 3y^2 + 240x
    where x denotes the number of frames and y denotes the number of wheels. Find the maximum profit
    assuming that the firm does not want any spare frames or wheels left over at the end of the produc-
    tion run.

    PLEASE HELP ME TO SOLVE THIS EXCERSISE
    Last edited by Dagne3; Nov 3rd 2014 at 07:25 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Feb 2014
    From
    United States
    Posts
    1,734
    Thanks
    808

    Re: Lagrange multiplier

    You have a function, in this case a profit function. To optimize a function subject to one or more constraints, you must construct a different function, let's call it the L function, and optimize that. What are the elements of the L function?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2014
    From
    Lithuania
    Posts
    20

    Re: Lagrange multiplier

    in fact I already got the L function. The constraint is 2x-y=0 so L(x,y,λ)=5x^2-10xy+3y^2+240x+ λ(2x-y)
    But the problem is that after I apply 2nd derivative test to see if the critical point I got is really local maximum point,I get the discriminant less than zero, which mean that its a saddle point,not a local maximum point..Maybe I missed something?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2014
    From
    Lithuania
    Posts
    20

    Re: Lagrange multiplier

    Does it mean that its impossible to solve the excersise?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    5,781
    Thanks
    2421

    Re: Lagrange multiplier

    If the maximum point is not in the interior (i.e. no critical points) then it lies on the boundary of your constraint function.

    In this case you are so constrained, you have to use all your parts, that there is really only one possible solution. The one where you use all your parts.

    In this case as you noted $2x=y$

    thus your profit function is simply

    $5x^2 − 10xy + 3y^2 + 240x|_{y=2x} = 5x^2-10x(2x) + 3(2x)^2 + 240x$

    It's trivial to find the critical point here and a 2nd derivative check shows that it is indeed a maximum.
    Thanks from Dagne3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Feb 2014
    From
    United States
    Posts
    1,734
    Thanks
    808

    Re: Lagrange multiplier

    Quote Originally Posted by Dagne3 View Post
    in fact I already got the L function. The constraint is 2x-y=0 so L(x,y,λ)=5x^2-10xy+3y^2+240x+ λ(2x-y)
    But the problem is that after I apply 2nd derivative test to see if the critical point I got is really local maximum point,I get the discriminant less than zero, which mean that its a saddle point,not a local maximum point..Maybe I missed something?
    $profit = P = 5x^2 - 10xy + 3y^2 + 240x.$

    $L = P + \lambda (2x - y) = 5x^2 - 10xy + 3y^2 + 240x + \lambda (2x - y) \implies$

    $\dfrac{\delta L}{\delta x} = 10x - 10y + 240 + 2\lambda;$

    $\dfrac{\delta L}{\delta y} = -\ 10x + 6y - \lambda;\ and$

    $\dfrac{\delta L}{\delta \lambda} = 2x - y.$

    So first order conditions are:

    $10x - 10y + 240 + 2\lambda = 0 \implies 2\lambda = 10y - 10x - 240 \implies \lambda = 5y - 5x - 120;$

    $-\ 10x + 6y - \lambda = 0 \implies \lambda = 6y - 10x \implies 6y - 10x = 5y - 5x - 120 \implies y = 5x - 120;\ and$

    $2x - y = 0 \implies 2x = y \implies 2x = 5x - 120 \implies 3x = 120 \implies x = 40 \implies y = 80.$

    I presume that is what you got. The second order conditions for L are not relevant because we are REALLY trying to maximize P, not L. We must go back to P.

    $x = 40\ and\ y = 80 \implies P = 5(40^2) - 10(40)(80) + 3(80^2) + 240(40) = 8000 - 32000 + 19200 + 9600 = 4800.$

    $x = 40 + a\ and\ y = 80 + 2a \implies$

    $P = 5(40 + a)^2 - 10(40 + a)(80 + 2a) + 3(80 + 2a)^2 + 240(40 + a) =$

    $5(1600 + 80a + a^2) - 10(3200 + 160a + 2a^2) + 3(6400 +320a + 4a^2) + 9600 + 240a =$

    $8000 + 400a + 5a^2 - 32000 - 1600a - 20a^2 + 19200 + 960a + 12a^2 + 9600 + 240a = 4800 - 3a^2 < 4800.$

    This kind of analysis is important because you may have constraints that are inequalities rather than equalities, in which case you may find that the inequalities are not real constraints.

    EDIT: Romsek has given you a simpler analysis, which comes to the same thing, that will always apply if your constraints are all equalities.
    Last edited by JeffM; Nov 3rd 2014 at 09:43 AM.
    Thanks from romsek and Dagne3
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2014
    From
    Lithuania
    Posts
    20

    Re: Lagrange multiplier

    Quote Originally Posted by JeffM View Post
    $profit = P = 5x^2 - 10xy + 3y^2 + 240x.$

    $L = P + \lambda (2x - y) = 5x^2 - 10xy + 3y^2 + 240x + \lambda (2x - y) \implies$

    $\dfrac{\delta L}{\delta x} = 10x - 10y + 240 + 2\lambda;$

    $\dfrac{\delta L}{\delta y} = -\ 10x + 6y - \lambda;\ and$

    $\dfrac{\delta L}{\delta \lambda} = 2x - y.$

    So first order conditions are:

    $10x - 10y + 240 + 2\lambda = 0 \implies 2\lambda = 10y - 10x - 240 \implies \lambda = 5y - 5x - 120;$

    $-\ 10x + 6y - \lambda = 0 \implies \lambda = 6y - 10x \implies 6y - 10x = 5y - 5x - 120 \implies y = 5x - 120;\ and$

    $2x - y = 0 \implies 2x = y \implies 2x = 5x - 120 \implies 3x = 120 \implies x = 40 \implies y = 80.$

    I presume that is what you got. The second order conditions for L are not relevant because we are REALLY trying to maximize P, not L. We must go back to P.

    $x = 40\ and\ y = 80 \implies P = 5(40^2) - 10(40)(80) + 3(80^2) + 240(40) = 8000 - 32000 + 19200 + 9600 = 4800.$

    $x = 40 + a\ and\ y = 80 + 2a \implies$

    $P = 5(40 + a)^2 - 10(40 + a)(80 + 2a) + 3(80 + 2a)^2 + 240(40 + a) =$

    $5(1600 + 80a + a^2) - 10(3200 + 160a + 2a^2) + 3(6400 +320a + 4a^2) + 9600 + 240a =$

    $8000 + 400a + 5a^2 - 32000 - 1600a - 20a^2 + 19200 + 960a + 12a^2 + 9600 + 240a = 4800 - 3a^2 < 4800.$

    This kind of analysis is important because you may have constraints that are inequalities rather than equalities, in which case you may find that the inequalities are not real constraints.

    EDIT: Romsek has given you a simpler analysis, which comes to the same thing, that will always apply if your constraints are all equalities.
    Okay,I got what you did here,I got the same, only I used 2nd derivative test which is not neccessary as I see. But can you explain how you get x = 40 + a and y = 80 + 2a ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Feb 2014
    From
    United States
    Posts
    1,734
    Thanks
    808

    Re: Lagrange multiplier

    Quote Originally Posted by Dagne3 View Post
    Okay,I got what you did here,I got the same, only I used 2nd derivative test which is not neccessary as I see. But can you explain how you get x = 40 + a and y = 80 + 2a ?
    The idea is this. Consider any arbitrary pair of numbers that differs from 40 and 80 while retaining the constraint that one of the pair is double the other. So if I change 40 by an arbitrary a, I must also change 80 by 2a to continue to meet that constraint.
    Thanks from Dagne3
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Aug 2017
    From
    Antartica
    Posts
    2

    Re: Lagrange multiplier

    Quote Originally Posted by romsek View Post
    If the maximum point is not in the interior (i.e. no critical points) then it lies on the boundary of your constraint function.

    In this case you are so constrained, you have to use all your parts, that there is really only one possible solution. The one where you use all your parts.

    In this case as you noted $2x=y$

    thus your profit function is simply

    $5x^2 − 10xy + 3y^2 + 240x|_{y=2x} = 5x^2-10x(2x) + 3(2x)^2 + 240x$

    It's trivial to find the critical point here and a 2nd derivative check shows that it is indeed a maximum.
    -------

    Hi, I would really appreciate if you could clarify why the constraint is 2x=y or 2x-y=0?? Doesn't x refer to the frames and y to the wheels? Hence, I thought that a bycicle would have 2 wheels and 1 frame, meaning that the constraint is 2y-x=0?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Aug 2017
    From
    Antartica
    Posts
    2

    Re: Lagrange multiplier

    What if the profit function was: π = 5x^2 − 10xy + 3y^2 = 240x (just: =240x, changes), Its a question from a book.... Thanks in advance
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lagrange multiplier
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Sep 15th 2012, 12:57 AM
  2. Lagrange Multiplier help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 2nd 2010, 12:49 PM
  3. Lagrange Multiplier
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 29th 2009, 06:37 AM
  4. lagrange multiplier
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 9th 2008, 08:54 PM
  5. Lagrange multiplier
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 16th 2008, 05:34 PM

Search Tags


/mathhelpforum @mathhelpforum