Finding maximum of a function with unknown constants I have a function in the form:
$$y = \frac{ax}{b + \frac{x^2}{c} + x}$$
Supposedly, the maximum of this function is equal to $\sqrt{bc}$.
I've tried substituting in $\sqrt{bc}$ for $x$, but I don't really know what any of my answers mean in relation to the maximum value for $y$.
All of the constants are positive.
How can I prove that $\sqrt{bc}$ is the maximum? can I just take the derivative of the function and assume that terms without $x$ coefficients go to 0?
 A: First of all, if $c\geq4b$ then the polynomial in denominator is zero at $x=\frac{1}{2}(-c\pm\sqrt{c^2-4bc})$. Thus, we assume that $c<4b$ (if this is not the OK, check the limits as $x$ approaches these zeros)
Since $y(x)\to0$ at $x\to\pm\infty$ and $y$ attains positive values and is differentiable it suffices to check points where $y'(x)=0$.
We have
$$
y'(x)=\frac{ac(bc-x^2)}{(bc+x(c+x))^2}
$$
so $y'(x)=0$ only for $x=\pm \sqrt{bc}$. Now
$$
y(\sqrt{bc})-y(-\sqrt{bc})=\frac{4a\sqrt{bc}}{4b-c}>0
$$
since $4b>c$. Thus the maximal value is attained at $x=\sqrt{bc}$.
A: You'll either need to second derivative test, to confirm that you, in fact, have a maximum, or you'll need to show that the functions behavior changes from increasing to decreasing at the critical point.
A: Hint: The terms $a,b$ and $c$ are constants. Using the quotient rule: $$\frac{{\rm d}y}{{\rm d}x}(x) = \frac{a\left(b+\frac{x^2}{c}+x\right) - ax\left(\frac{2x}{c}+1\right)}{\left(b+\frac{x^2}{c}+x\right)^2}.$$
Check (or solve for it, whatever) that $\frac{{\rm d}y}{{\rm d}x}(\sqrt{bc}) = 0$ and that $\frac{{\rm d}^2y}{{\rm d}x^2}(\sqrt{bc}) < 0$. You will compute $\frac{{\rm d}^2y}{{\rm d}x^2}$ in the same fashion as $\frac{{\rm d}y}{{\rm d}x}$.
A: The limit of $y$ as $x$ approaches $\pm\infty$ is zero, so if there is only one critical point and $y$ is positive there it must be a maximum. If, as in this case, there are two critical points, and $y$ is positive at only one of them, that one must be a maximum.
