Find the minimum value of (ax + by), where xy = c^2.

The correct answer is Option (2) → $2c\sqrt{ab}$

Given $xy = c^2⇒y=\frac{c^2}{x}$,

then $ax + by = ax + b. \frac{c^2}{x}$.

Let $f(x) = ax +\frac{bc^2}{x}$, so we need to find the minimum value of $f(x)$.

$D_f = R - \{0\}$ and $f$ is differentiable for all $x ∈ D_f$.

Differentiating w.r.t. x, we get

$f'(x) = a. 1+ bc^2. (-1) . x^{-2} = a -\frac{bc^2}{x}$ and

$f''(x) = -bc^2. (-2). x^{-3} = \frac{2bc^2}{x^3}$.

For critical points, $f'(x)=0⇒ -\frac{bc^2}{x}=0$

$⇒x^2=\frac{bc^2}{a}⇒x=±c\sqrt{\frac{b}{a}}$.   (Assuming a, b, c are all positive)

Therefore, the points where the extremum may occur are

$x=c\sqrt{\frac{b}{a}}$ and $x=-c\sqrt{\frac{b}{a}}$.

$f''\left(c\sqrt{\frac{b}{a}}\right)=\frac{2bc^2}{\left(c\sqrt{\frac{b}{a}}\right)^3}=\frac{2a}{c}\sqrt{\frac{b}{a}}>0$ $⇒f$ has local minima at $x=c\sqrt{\frac{b}{a}}$

Local minimum value = $f\left(c\sqrt{\frac{b}{a}}\right)=a.c\sqrt{\frac{b}{a}}+\frac{bc^2}{c\sqrt{\frac{b}{a}}}$

$=c\sqrt{ab}+\sqrt{ab}=2\sqrt{ab}$.

$f''\left(-c\sqrt{\frac{b}{a}}\right)=\frac{2bc^2}{\left(-c\sqrt{\frac{b}{a}}\right)^3}=-\frac{2a}{c}\sqrt{\frac{b}{a}}<0$

$⇒f$ has local maxima at $x=-c\sqrt{\frac{b}{a}}$.

Hence, the minimum value of $f(x)$ i.e. $ax+by$ is $2c\sqrt{ab}$