The maximum value of ${\mathrm{x}}^4{\mathrm{e}}^{-{\mathrm{x}}^2}$ is 

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4{\mathrm{e}}^{-{\mathrm{x}}^2}$
or $$\begin{gathered} {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=4{\mathrm{x}}^3{\mathrm{e}}^{-{\mathrm{x}}^2}+{\mathrm{x}}^4{\mathrm{e}}^{-{\mathrm{x}}^2}(-2\mathrm{x}) \\ =2{\mathrm{x}}^3{\mathrm{e}}^{-{\mathrm{x}}^2}\left(2-{\mathrm{x}}^2\right) \end{gathered}$$
Sign scheme of $f^{\text{'}}\left(x\right)$ is as follows:
 

Hence, $f\left(x\right)$ is maximum at $\mathrm{x}=\pm \sqrt{2}$.Thus, maximum value $=4{\mathrm{e}}^{-2}$.