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 August 1st, 2013, 02:50 PM #1 Newbie   Joined: Jun 2013 Posts: 7 Thanks: 0 Help with a minimum I can´t find the minimum of followin function. Let $x \in [0;8];\;a(x)= sqrt{x^2+16}\; and \; b(x)= sqrt{(8-x)^2+49}.\; a(x)\; and\; b(x)\; are\; hypotenuses,\; so\; are\; greater\; than\; 0$. $f(x)=[a(x)]^2+ [b(x)]^2\; has\; a\; minimum\; at\; x=4$ The question is: $h(x)=a(x)+ b(x)\; has\; the\; same\; minimum\; (at\; x=4)$ Thanks
 August 2nd, 2013, 12:05 PM #2 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 Re: Help with a minimum Have you tried elementary calculus - find the zeros of the derivative?
 August 2nd, 2013, 03:27 PM #3 Newbie   Joined: Jun 2013 Posts: 7 Thanks: 0 Re: Help with a minimum thanks mathman. Since h'(4) $\neq 0$ h(x) don´t have a minimum at x=4. Then, How to find the minimum of h(x)?
 August 2nd, 2013, 04:30 PM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Help with a minimum Okay, h'(4) is NOT 0. Where is it 0? That should be fairly easy to solve.
 August 2nd, 2013, 10:00 PM #5 Newbie   Joined: Jun 2013 Posts: 7 Thanks: 0 Re: Help with a minimum Let a=a(x) and b=b(x). $a= \sqrt{x^2+16};\;\;\;$ $b= \sqrt{(8-x)^2+49};\;\;\;$ $\partial a= \dfrac {x}{\sqrt{x^2+16}};\;\;\;$ $\partial b= -\dfrac {8-x}{\sqrt{(8-x)^2+49}}$. $\partial\partial a= \dfrac {16\sqrt{x^2+16}}{(x^2+16)^2};\;\;\;$ $\partial\partial b= \dfrac {49}{[(8-x)^2+49]\sqrt{(8-x)^2+49}}$. I found: $\;\;\;\partial a +\partial b= 0\;$ at $\;x=\dfrac{32}{11}.\;$ From the expression of $\;\partial\partial a\;$ and $\;\partial\partial b\;$ follows that $\;\partial\partial a>0\;$ and $\;\partial\partial b>0 \;\;\forall{x \in \mathbb{R}}$. Then $\;\;\;\partial\partial a +\partial\partial b>0.$ So a+b have a minimum at $\;\;\;x= \dfrac {32}{11}.$ Thank you HallsofIvy and mathman.

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