Let the numbers are $\mathrm{x}$ and $\mathrm{y}$. So, $x+y=20$; Let $P=x^{2} y^{3} \quad$ (As given) $=x^{2}(20-x)^{3}$
Differentiating w. r. t. $\mathrm{x}$ $\begin{aligned} \frac{d P}{d x} &=x^{2} \cdot 3(20-x)^{2}(-1)+(20-x)^{3} \cdot 2 x \\ &=(20-x)^{2}\left[-3 x^{2}+40 x-2 x^{2}\right] \\ &=(20-x)^{2}\left[40 x-5 x^{2}\right] \\ \frac{d^{2} P}{d x^{2}} &=(20-x)^{2}[40-10 x]+\left(40 x-5 x^{2}\right) 2(20-x)(-1) \end{aligned}$
$\frac{d p}{d x}=0$ for maxima or minima. So, $(20-x)^{2}\left[40 x-5 x^{2}\right]=0$
$\Rightarrow(20-\mathrm{x})^{2} \times(\mathrm{x})(40-5 \mathrm{x})=0 \Rightarrow \mathrm{x}=20,0,8$
We get, $\left(\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=8} < 0 ;\left(\frac{\mathrm{d}^{2} \mathrm{p}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=0}>0$ and
$\left(\frac{\mathrm{d}^{2} \mathrm{p}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=20}=0$
Hence. $\mathrm{P}$ is maximumat $\mathrm{x}=8$
and, Numbers are 12 and 8 .