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Math Help - maximum value

  1. #1
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    maximum value

    Given that 2u+3v+4x is equal to 12. Then find the maximum value of u^2v^3x^5.
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  2. #2
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    Re: maximum value

    Hey geniusgarvil.

    If you perform a trinomial expansion on your 2u+4x+4x, you will be able to get a term in regards to u^2*v^3*x^5 if you raise it to the power of (2+3+5) = 10.

    If you assume that the highest term contains all the information, then you can verify for yourself by setting one value to the highest and setting the rest equal to zero what the highest value of this term will be, by looking at the relevant coffecient that corresponds to that trinomial term.

    So for a term of u^2*v^3*x^5, this will have a trinomial term in the same that a binomial has an nCr term before x^r*y(n-r) and this term is called the multinomial coeffecient:

    Multinomial theorem - Wikipedia, the free encyclopedia
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: maximum value

    While I am aware that you posted in the Algebra forum, a method from differential calculus may be used, called Lagrange multipliers for optimization with a constraint.

    If we define:

    f(u,v,x)=u^2v^3x^5

    subject to the constraint:

    g(u,v,x)=2u+3v+4x-12=0

    Then we find:

    2uv^3x^5=2\lambda

    3u^2v^2x^5=3\lambda

    5u^2v^3x^4=4\lambda

    Using this system, along with the constraint, you can now determine the optimal values of the 3 variables.
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  4. #4
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    Re: maximum value

    can you solve it in a simpler way. i m no pro in algebra
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: maximum value

    What methods have you been using for similar problems? I would use the method of Lagrange, but I don't know what you are being taught to use.
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