Curiosity on function maxima

Question

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.

Answer 1

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 $$x=\frac{a}{b}$$

Answer 2

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.

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

Answer 3

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 $$\frac a{2x\sqrt x}=\frac b{2\sqrt x}.$$