Find two numbers whose sum is $24$ and whose product is as large as possible.

Let the two numbers be $x$ and $24 - x$
Let $p = x(24 - x) \Rightarrow p = 24x - {x^2}$

$\Rightarrow \cfrac{{dp}}{{dx}} = 24 - 2x$

For $p$ to be largest $\cfrac{{dp}}{{dx}} = 0 \Rightarrow 24 - 2x = 0 \Rightarrow x = 12$ and
$\cfrac{{{d^2}p}}{{d{x^2}}} = - 2,{\left( {\cfrac{{{d^2}p}}{{d{x^2}}}} \right)_{x = 12}} = - 2 < 0$

$\Rightarrow p$ has maximum value at $x = 12$.
So, the required parts are $12{\rm{ and }}24 - 12$ i.e., $12{\rm{ and }}12$.