1. ## Minimize

What's the minimum of $K=x^6+y^6$ if $x-y=1$?

2. Originally Posted by james_bond
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

3. Thanks for your answer but is there any other smarter (sorry) solution? This problem is between trigonometric and logarithmic/exponential problems...

4. I completely agree with you : my solution is not very beautiful

5. Originally Posted by running-gag
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

6. Originally Posted by james_bond
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 $K = \cosh^{12}(\alpha) + \sinh^{12}(\alpha)$

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

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

$K'(\alpha) = 12 \cos(\alpha) \:\sin(\alpha)\\sin^{10}(\alpha) - \cos^{10}(\alpha))" alt="K'(\alpha) = 12 \cos(\alpha) \:\sin(\alpha)\\sin^{10}(\alpha) - \cos^{10}(\alpha))" />

7. Originally Posted by IluvMath
To me it looks like Running-gag's solution seems the simplest but I would be interested in seeing another solution
Me too :P

8. 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.