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Thread: Minimize

  1. #1
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    Minimize

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

    $\displaystyle K(x)=x^6+(x-1)^6$

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

    By successive derivation of $\displaystyle L(x)=x^4-2x^3+4x^2-3x+1$ 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

    $\displaystyle K(x)=x^6+(x-1)^6$

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

    By successive derivation of $\displaystyle L(x)=x^4-2x^3+4x^2-3x+1$ 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 $\displaystyle y \geq 0$ then you can set $\displaystyle y = \sinh^2(\alpha)$ where $\displaystyle \alpha \in [0,+\infty]$

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

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

    $\displaystyle K'(\alpha) = 12 \cosh(\alpha) \:\sinh(\alpha)\\cosh^{10}(\alpha) + \sinh^{10}(\alpha))$

    Therefore $\displaystyle K'(\alpha) \geq 0$

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



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

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

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

    $\displaystyle K'(\alpha) = 12 \cosh(\alpha) \:\sinh(\alpha)\\cosh^{10}(\alpha) + \sinh^{10}(\alpha))$

    Therefore $\displaystyle K'(\alpha) \leq 0$

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


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

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

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

    $\displaystyle K'(\alpha) = 12 \cos(\alpha) \:\sin(\alpha)\\sin^{10}(\alpha) - \cos^{10}(\alpha))$
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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:

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

    $\displaystyle 6x^5=-6(x-1)^5$

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

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

    $\displaystyle x=-x+1$

    $\displaystyle 2x=1$

    $\displaystyle x=\frac{1}{2}$

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