Step 1:

Given

Sum of two no$\rightarrow$ 24

Product $\rightarrow$ as large as possible

Let the required number be $x$ and $(24-x)$

Their product $P=x(24-x)$

$\qquad\qquad\quad\;\;\;=24x-x^2$

Step 2:

For maximum value

Now $\large\frac{dP}{dx}$$=0$

$P=24x-x^2$

Differentiating with respect to x we get

$\large\frac{dP}{dx}$$=24-2x$

$24-2x=0$

$-2x=-24$

$x=12$

$\large\frac{d^2P}{dx^2}$$=-2 < 0$

$\Rightarrow P$ is minimum at $x=12$

Hence the required numbers are 12 and $(24-12)\Rightarrow (i.e)12$