$\displaystyle f(x)=\frac {x-5}{x^2-5x+1992}$For which $\displaystyle x$ does it have its maximum?

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- Apr 1st 2008, 04:55 AMjames_bondFind the maximum$\displaystyle f(x)=\frac {x-5}{x^2-5x+1992}$For which $\displaystyle x$ does it have its maximum?
- Apr 1st 2008, 05:15 AMmr fantastic
- Apr 1st 2008, 05:47 AMjames_bond
No but unfortunately I need a solution without using calculus. Maybe with using some well-known inequalities (AM-GM, Cauchy, etc.) and some tricks...

- Apr 3rd 2008, 07:51 AMJaneBennet
Although you need to find a solution without using calculus, that doesn’t mean you can’t use calculus to find what the maximum value might be. The problem should be easier if you know what the maximum value is.

Using calculus, I found that the maximum value occurs at $\displaystyle x=5+\sqrt{1992}$. So the problem is to use non-calculus methods to prove that $\displaystyle \frac{x-5}{x^2-5x+1992}\leq\mathrm{f}\left(5+\sqrt{1992}\right)$ for all $\displaystyle x\in\mathbb{R}$.

$\displaystyle \mathrm{f}\left(5+\sqrt{1992}\right)$ should simplify to $\displaystyle \frac{1}{5+2\sqrt{1992}}$. So the inequality becomes

$\displaystyle \color{white}.\quad.$ $\displaystyle \frac{x-5}{x^2-5x+1992}\ \leq\ \frac{1}{5+2\sqrt{1992}}$

$\displaystyle \Leftrightarrow\ (x-5)(5+2\sqrt{1992})\ \leq\ x^2-5x+1992$ (note that $\displaystyle x^2-5x+1992>0$ for all*x*)

$\displaystyle \Leftrightarrow\ 5x-25+2x\sqrt{1992}-10\sqrt{1992}\ \leq\ x^2-5x+1992$

$\displaystyle \Leftrightarrow\ 0\ \leq\ x^2-2(5+\sqrt{1992})x+25+10\sqrt{1992}+1992\ =\ \left(x-[5+\sqrt{1992}]\right)^2$

QED. - Apr 5th 2008, 11:20 AMjames_bond
The simple answer was:

$\displaystyle f(x)=\frac {x-5}{x^2-5x+1992}=\frac{1}{x+\frac{1992}{x-5}}=\frac{1}{x-5+\frac{1992}{x-5}+5}\le\frac{1}{4\sqrt{498}+5}$ by AM-GM

Equality occurs when $\displaystyle x-5=\frac{1992}{x-5}\implies x=2\sqrt{498}+5$