Explanation

(a)
Let the numbers be $x,\ y.,$
$\therefore\ \ \ x + y = 20\ \ \ \Rightarrow \ \ \ y = 20 - x$
Let $S = x^{3}\ (20 - x)^{2}$
$\Rightarrow \ \ \ \frac{dS}{dx} = 2x^{3}\ (20 - x)( - 1) + 3x^{2}(20 - x)^{2}$
For maximum or minimum, put $\frac{dS}{dx} = 0$
$\Rightarrow \ \ x^{2}(20 - x)( - 2x + 60 - 3x) = 0$
$\Rightarrow \ \ \ \ \ x = 0,\ \ x = 12\$ or $x = 20$
$x$ cannot be 0 or 20
Now $\frac{dS}{dx} = x^{2}(20 - x)^{2}\lbrack 60 - 5x\rbrack$
For $0 < x < 12,\frac{dS}{dx} > 0$ and $12 < x < 20,\ \frac{dS}{dx} < 0$
$\therefore\ \ \ S$ is maximum at $x = 12$
For $x = 12$, then $y = 20 - 12 = 8$
The required numbers are 12, 8