Explanation

(c)
Let two numbers be x and y, then
$\ \ x + y = 6$ …(i)
Let $z = \frac{1}{x} + \frac{1}{y}$ …(ii)
$\Rightarrow \ \ \ \ z = \frac{1}{x} + \frac{1}{6 - x}$
On differentiating w. r. t. x, we get
$\frac{dz}{dx} = - \frac{1}{x^{2}} + \frac{1}{(6 - x)^{2}}$
Put $\frac{dz}{dx} = 0 \Rightarrow - \frac{1}{x^{2}} + \frac{1}{(6 - x)^{2}} = 0$
$\therefore\ - (6 - x)^{2} + x^{2} = 0$
$\Rightarrow \ \ \ \ 12x - 36 = 0$
$\Rightarrow \ \ \ \ \ \ x = 3$
Now, $\frac{d^{2}z}{dx^{2}} = \frac{2}{x^{3}} + \frac{2}{(6 - x)^{3}}$
At $x = 3,\ \ \frac{d^{2}z}{dx^{2}} > 0,\ minimum$
Hence, minimum value at $x = 3$ is
$z = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$