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I was recently working with an equation of the form: $$ \frac{\sqrt{x}}{a+bx+c\sqrt{x}} $$ And I realized that the maxima (only considering positive real numbers) would always be at the point where: $$ x=\frac ab $$ This is straightforward to prove by finding where the first derivative equals 0. Given this 'easy' result, I tried to find the logic behind it, which should probably be something easy, but I do not find it (I'm evidently no expert in mathematics, just curious).

My question is, should it be evident that the function has a maxima at that point without having to calculate the derivative? In the case it should, could someone explain me the reasoning behind it?

Thank you in advance. Kind regards, J.

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  • $\begingroup$ If your question is simply the logic behind setting the derivative equal to 0, that is because, if the derivative, at a given point, is positive, the function is increasing- we could get a larger value by increasing x a little so the given point is NOT a maximum. Similarly, if the derivative at a give point is negative, the function is decreasing- we could get a larger value by decreasing x a little so the given point is NOT a maximum. This says that in order to be a (local) maximum, the derivative must be 0. The other way- that is the derivative is 0, the function has a maximum, is FALSE. $\endgroup$ – user247327 Mar 29 '17 at 15:40
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Write it as:$$ \frac{1}{b\sqrt{x}+\cfrac{a}{\sqrt{x}}+c} $$

Then by AM-GM:

$$ b\sqrt{x}+\cfrac{a}{\sqrt{x}} \ge 2 \sqrt{ab} $$

Also by AM-GM, equality holds when $b\sqrt{x}=\cfrac{a}{\sqrt{x}} \iff x = \cfrac{a}{b}\,$, which thus gives the minimum of the denominator, which in turn gives the maximum of the fraction.


[ EDIT ]   The above interpretes the "only considering positive real numbers" stated condition to mean that $a,b,c$ and $x$ are strictly positive numbers.

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  • $\begingroup$ What happens if $0$ is in the domain of $f$? Any way, nice approach. $\endgroup$ – Bumblebee Mar 29 '17 at 15:44
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    $\begingroup$ @Nil I took only considering positive real numbers to mean strictly positive. $\endgroup$ – dxiv Mar 29 '17 at 15:46
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    $\begingroup$ Ahhh... Yes you are right. +! $\endgroup$ – Bumblebee Mar 29 '17 at 15:50
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    $\begingroup$ What I particularly like in your approach is that the first line of equation emphasize the non-dependency of the solution on $c$. $\endgroup$ – Surb Mar 29 '17 at 16:03
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    $\begingroup$ Thank you! It is elegant and I learnt something new today. $\endgroup$ – JIbab Mar 29 '17 at 17:13
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The reciprocal function is

$$\frac a{\sqrt x}+b\sqrt x+c$$ and the position of its extrema is independent of $c$.

We can factor out $b$ and get

$$b\left(\frac ab\frac 1{\sqrt x}+\sqrt x\right)+c,$$ which shows that the position can only depend on $\dfrac ab$.

The term $\dfrac a{\sqrt x}$ is decreasing and $b\sqrt x$ is increasing, the extremum is achieved when their slopes are opposite, which occurs when $$\frac a{2x\sqrt x}=\frac b{2\sqrt x}.$$

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the first derivative is given by $$f'(x)=1/2\,{\frac {-bx+a}{ \left( a+bx+c\sqrt {x} \right) ^{2}\sqrt {x}}}$$ the searched extrema ( if they exist) are located at $$x=\frac{a}{b}$$

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  • $\begingroup$ Also $x=0$ should be a suspicious extreme point. $\endgroup$ – Bumblebee Mar 29 '17 at 15:40
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    $\begingroup$ OP explicitly asks for a solution without computing derivatives.. $\endgroup$ – Surb Mar 29 '17 at 16:01

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