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Math Help - constraint maximization

  1. #1
    Senior Member Sambit's Avatar
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    Question constraint maximization

    Given x+y=2. Then what will be the maximum value of x^3y^3(x^3+y^3)?

    Intuitively, I think it will be maximum when x=y=1. But can not put it mathematically. Any help?
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  2. #2
    Moo
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    Hello,

    Do you know Lagrange's multiplier ?
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  3. #3
    MHF Contributor

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    One method is to use the first equation to replace y with 2- x. Then the function to be minimized becomes x^3(2- x)^3(x^2+ (2- x)^3). You could either multiply that out or use the product rule to differentiate it.

    If you are used to working with two or more variables and know "Lagrange's multiplier method", you can say that grad(x+ y)= <1, 1> while [itex]grad(x^3y^3(x^3+ y^3))= <3x^2y^3(x^3+ y^3+ 3x^5y^3, 3x^3y^2(x^3+ y^3)+ 3x^3y^5>[/tex]. "Lagrange's multiplier method" says that the max or min of one function subject to the constraint that the other be a constant occurs where one of those gradients is a multiple of the other. Since grad(x+ y)= <1, 1>, any multiple is of the form <\lambda, \lambda> so, here, that just says that the two components of grad(x^3y^3(x^3+ y^3) are equal:
    3x^2y^3(x^2+ y^3)+ 3x^5y^3= 3x^3y^2(x^3+ y^3)+ 3x^3y^5. That, together with the constraint, x+ y= 2, gives you two equations to solve for x and y.

    I would start by saying that, if neither x nor y is 0, I can divide through by x^2y^2.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Sambit View Post
    Given x+y=2. Then what will be the maximum value of x^3y^3(x^3+y^3)?

    Intuitively, I think it will be maximum when x=y=1. But can not put it mathematically. Any help?
    If you do not know about Lagrange multipliers as Moo suggests you can substitute x=2-y from the constraint into the objective differentiate the result and set to zero then solve for x. This will give you the constrained local stationary points ..

    CB
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  5. #5
    Senior Member Sambit's Avatar
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    As far as I know, Lagrange's fucntion is x^3y^3(x^3+y^3)+\lambda(x+y-2) where \lambda is Lagrange's multiplier. Differentiating this function partially wrt x and y equating them with 0, we get 3x^2y^3(3x^2)+\lambda=0 and 3x^3y^2(3y^2)+\lambda=0. Solving, we have x=y=1. Is this correct?
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