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Math Help - Minimium value

  1. #1
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    Minimium value

    if x+y=1 then what is the min. value of x^3+y^3+xy ?
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    Re: Minimium value

    Min =1/2 when x =1/2
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  3. #3
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    Re: Minimium value

    Step-by-step solution please.
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  4. #4
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    Re: Minimium value

    Quote Originally Posted by RuyHayabusa View Post
    if x+y=1 then what is the min. value of x^3+y^3+xy ?
    y = 1 - x

    Substitute this into the second equation, then evaluate where the derivative of the second function is 0.
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  5. #5
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    Re: Minimium value

    Another way to do it: \nabla x+ y= <1, 1>, \nabla x^3+ y^3+ xy= <3x^2+ y, 3y^2+ x>.

    We must have 3x^2+ y= \lambda, 3y^2+ x= \lambda for some number \lambda, as well as x+y= 1.

    So 3x^2+ y= 3y^2+ x which says that 3x^2- 3y^2+ y- x= 3(x- y)(x+ y)- (x- y)= 0.
    If x= y, from x+y= 1 we have x= y= 1/2.

    If x\ne y then x- y\ne 0 so we can divide by it to get 3(x+ y)- 1= 0 which says x+y= 1/3 contradicting x+y= 1.

    Thus x= y= 1/2 is the only solution.
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  6. #6
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    Re: Minimium value

    Yet another way, and probably simplest:
    From x+ y= 1, y= 1- x. Putting that into [tex]x^3+ y^3+ xy[/itex] as Prove It suggested, we have x^3+ (1- x)^3+ x(1- x)= x^3+ 1- 3x+ 3x^2- x^3+ x- x^2= 2x^2- 2x+ 1. We can write that as 2(x^2- x)+ 1 and, completing the square, 2(x^2- x+ 1/4-1/4)+ 1= 2(x^2- x+ 1/4)- 1/2+ 1= 2(x- 1/2)^2+ 1/2. Since a square is never negative, that will have its minimum value 2(1/2- 1/2)+ 1/2= 2(0)+ 1/2= 1/2 when x= 1/2.
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