Explanation

(b)
Given curve is
$xy = c^{2}\ \ \Rightarrow y = \frac{c^{2}}{x^{2}}$
Let $f(x) = ax + by = ax + \frac{bc^{2}}{x^{2}}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = a - \frac{bc^{2}}{x^{2}}$
For a maximum or minima, put $f^{'}(x) = 0$
$\Rightarrow \ \ ax^{2} - bc^{2} = 0$
$\Rightarrow \ \ \ x^{2} = \frac{bc^{2}}{a}\ \Rightarrow \ \ x = \pm c\sqrt{\frac{b}{a}}$
Again on differentiating, we get $f^{''}(x) = \frac{2bc^{2}}{x^{3}}$
At6 $x = c\sqrt{\frac{b}{a}},\ \ f^{''}(x) > 0$
$\therefore\ \ \ \ f(x)$ is minimum at $x = c\sqrt{\frac{b}{a}}$
The minimum value at $x = c\sqrt{\frac{b}{a}}$ is
$\therefore\ f\left( c\sqrt{\frac{b}{a}} \right) = a.c\sqrt{\frac{b}{a}} + \frac{bc^{2}}{c}.\ \sqrt{\frac{a}{b}}$
$= \frac{abc + abc}{\sqrt{ab}} = \frac{2abc}{\sqrt{ab}} = 2c\sqrt{ab}$