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Math Help - Minimize

  1. #1
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    Minimize

    What's the minimum of K=x^6+y^6 if x-y=1?
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  2. #2
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    Quote Originally Posted by james_bond View Post
    What's the minimum of K=x^6+y^6 if x-y=1?
    Hi

    K(x)=x^6+(x-1)^6

    K'(x)=6x^5+6(x-1)^5=6(2x-1)(x^4-2x^3+4x^2-3x+1)

    By successive derivation of L(x)=x^4-2x^3+4x^2-3x+1 you can show that L(x) is decreasing up to x=1/2 and then increasing and is always positive

    The minimum of K(x) is therefore obtained for x=1/2
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  3. #3
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    Thanks for your answer but is there any other smarter (sorry) solution? This problem is between trigonometric and logarithmic/exponential problems...
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  4. #4
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    I completely agree with you : my solution is not very beautiful
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  5. #5
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    Quote Originally Posted by running-gag View Post
    Hi

    K(x)=x^6+(x-1)^6

    K'(x)=6x^5+6(x-1)^5=6(2x-1)(x^4-2x^3+4x^2-3x+1)

    By successive derivation of L(x)=x^4-2x^3+4x^2-3x+1 you can show that L(x) is decreasing up to x=1/2 and then increasing and is always positive

    The minimum of K(x) is therefore obtained for x=1/2
    To me it looks like Running-gag's solution seems the simplest but I would be interested in seeing another solution
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  6. #6
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    Quote Originally Posted by james_bond View Post
    Thanks for your answer but is there any other smarter (sorry) solution? This problem is between trigonometric and logarithmic/exponential problems...
    Let's try this

    If y \geq 0 then you can set y = \sinh^2(\alpha) where \alpha \in [0,+\infty]

    Then x = 1 + y = \cosh^2(\alpha)

    And K = \cosh^{12}(\alpha) + \sinh^{12}(\alpha)

    \cosh^{10}(\alpha) + \sinh^{10}(\alpha))" alt="K'(\alpha) = 12 \cosh(\alpha) \:\sinh(\alpha)\\cosh^{10}(\alpha) + \sinh^{10}(\alpha))" />

    Therefore K'(\alpha) \geq 0

    And K(x) is increasing over [1,+\infty]



    If y \leq -1 then you can set y = -\cosh^2(\alpha) where \alpha \in ]-\infty,0]

    Then x = 1 + y = -\sinh^2(\alpha)

    And K = \sinh^{12}(\alpha) + \cosh^{12}(\alpha)

    \cosh^{10}(\alpha) + \sinh^{10}(\alpha))" alt="K'(\alpha) = 12 \cosh(\alpha) \:\sinh(\alpha)\\cosh^{10}(\alpha) + \sinh^{10}(\alpha))" />

    Therefore K'(\alpha) \leq 0

    And K(x) is decreasing over ]-\infty,0]


    If -1 \leq y \leq 0 then you can set y = -\sin^2(\alpha) where \alpha \in [0,\frac{\pi}{2}]

    Then x = 1 + y = \cos^2(\alpha)

    And K = \cos^{12}(\alpha) + \sin^{12}(\alpha)

    \sin^{10}(\alpha) - \cos^{10}(\alpha))" alt="K'(\alpha) = 12 \cos(\alpha) \:\sin(\alpha)\\sin^{10}(\alpha) - \cos^{10}(\alpha))" />
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  7. #7
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    Quote Originally Posted by IluvMath View Post
    To me it looks like Running-gag's solution seems the simplest but I would be interested in seeing another solution
    Me too :P
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  8. #8
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    I think I've got a simpler solution:

    K'(x)=6x^5+6(x-1)^5=0

    6x^5=-6(x-1)^5

    (\frac{x}{x-1})^5=-1

    \frac{x}{x-1}=-1

    x=-x+1

    2x=1

    x=\frac{1}{2}

    Then by using the second derivative test it's easy to prove that it is indeed a min.
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