Given a is a constant (positive integer), $0 \leq x \leq \sqrt{3}$. Find the minimum and the maximum of:
$$P = \sqrt{4a^2+\frac{a^2 x^2}{4a^2-2a\sqrt{3}x+x^2}}$$
The problem seems hard for me since I can't even simplify $P$ (due to $a$ is an unknown constant). I took many attempts but I can't find a clean solution to this. How do I approach this?
On
When $P$ is a maximum, $P^2$ is also a maximum. So let $Q=P^2$.
$Q=4a^2+\dfrac{a^2 x^2}{4a^2-2a\sqrt{3}x+x^2}$
$\dfrac {\mathrm dQ}{\mathrm dx}=-\dfrac{(4 \sqrt 3 - 2 a) a^3 x}{(4 a^2 - 2 \sqrt 3 a + x^2)^2}$
So our turning point lies on $x=0$. As there are no other turning points, $x=\sqrt 3$ is the point to take in consideration as for $x \ge 0$, our function is either strictly increasing or strictly decreasing.
$P(0)=2a$ and $P(\sqrt3)= \dfrac {a\sqrt {217}}7$ so the minimum value of $P$ is $P=2a$ and the maximum value of $P$ is $P= \dfrac {a\sqrt {217}}7$.
Maximizing $P$ or $P^2$ is same for $x$. And to maximize for minimze $P$ , since $4a^2$ is constant and thus we only need to maximize/minimize for the fraction term : $k=\frac{a^2 x^2}{4a^2-2a\sqrt{3}x+x^2}$
Differentiating wrt $x$ we obtain $\frac{dk}{dx}=\frac{8a^4x-2a^3\sqrt3x^2}{(4a^2-2a\sqrt{3}x+x^2)^2}$.Now , $\frac{dk}{dx} \ge0$ for $0\le x \le \sqrt3$. Hence maximum value is obtained at $x=\sqrt3$ and minimum value is obtained at $x=0$. Now, you can proceed from here to calculate the maximum/minimum values of $P$.