Explanation

(a)
Let $f(x) = 2x + 3y$ and $xy = 6$
$\Rightarrow \ \ \ f(x) = 2x + \frac{18}{x}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = 2 - \frac{18}{x^{2}}$
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 of $f(x)$ is
$f(3) = 2(3) + 3(2) = 12$