Explanation

(b)
Let $f(x) = 2x + 3y$
$\therefore\ \ \ \ \ f(x) = 2x + \frac{18}{x}$ [$\because xy = 6$, given]
$\Rightarrow \ \ \ \ \ \ \ f^{'}(x) = 2 - \frac{18}{x^{2}}$
Put $f^{'}(x) = 0$ for maxima or minima
$\Rightarrow \ \ 0 = 2 - \frac{18}{x^{2}}\ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ x = \pm 3$
And $f^{''}(x) = \frac{36}{x^{3}}\ \ \ \ \ \ \Rightarrow \ f^{''}(3) = \frac{36}{3^{3}} > 0$
$\therefore$ At $x = 3,\ f(x)$ is minimum.
The minimum value is $f(3) = 12$