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Math Help - Lagrange Multiplier

  1. #1
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    Lagrange Multiplier

    I'm stuck on a pretty straightforward LM type of problem.

    Here's the problem statement.
    a,b,c are positive constants. x,y,z are positive and that ayz+bzx+cxy=3abc. Show that xyz =< abc.

    So I've approached it by setting f(x,y,z)=xyz subject to ayz+bzx+cxy=3abc

    so F(x,y,z,lamda)=xyz-lamda(ayz+bzx+cxy-3abc)
    then I kinda got lost...

    Any pointers? Thanks.
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  2. #2
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by geegee View Post
    I'm stuck on a pretty straightforward LM type of problem.

    Here's the problem statement.
    a,b,c are positive constants. x,y,z are positive and that ayz+bzx+cxy=3abc. Show that xyz =< abc.

    So I've approached it by setting f(x,y,z)=xyz subject to ayz+bzx+cxy=3abc

    so F(x,y,z,lamda)=xyz-lamda(ayz+bzx+cxy-3abc)
    then I kinda got lost...

    Any pointers? Thanks.
    I think what you did is good. Why don't you find the maximum of F(x,y,z,\lambda)? Take the partial derivatives of F with respect to x, y and z and equal them to 0. Then solve for each variable, except maybe for \lambda which doesn't really matter.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    You should obtain (L is lambda)

    1) yz/(bz+cy)= L

    2) xz/(az+cx) = L

    3) xy/(ay+bx) = L


    Using 1 and 2 x = (a/b)y

    Using 2 and 3 z = (c/b)y

    Use these results in ayz+bzx+cxy=3abc

    You should obtain y = +b

    Then x = +a

    and z = +c

    The max of f = xyz then is abc
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