The maximum value of ${\sin }^{2}x {\cos }^{3}x$ is

$$\begin{gathered} y={\sin }^{2}x {\cos }^{3}x \\ \therefore y^{\mathrm{\prime }}=2\sin x\cos x{\cos }^{3}x+{\sin }^{2}x3{\cos }^{2}x(-\sin x) \\ \text{ fer maximum or minimum value }{y}^{\mathrm{\prime }}=0 \\ \therefore \sin x{\cos }^{2}x\left(2{\cos }^{2}x-3{\sin }^{2}x\right)=0 \\ \Rightarrow x=0\text{ or }x=\pi /2\text{ or, }\tan x=\sqrt{\frac{2}{3}} \\ \begin{array}{r}{y}^{\prime \prime }=2{\cos }^{5}x-8{\sin }^{2}x{\cos }^{3}x-9{\sin }^{2}x{\cos }^{3}x\\ +8\sin { }^{4}x\cos x\end{array} \\ \begin{array}{l}for x=0\\ \Rightarrow y^{\prime \prime }\left(0\right)=2\end{array} \\ forx=\pi /2\Rightarrow y^{\prime \prime }(\pi /2)=0 \end{gathered}$$
$$\begin{gathered} \begin{array}{c}for \tan x=\sqrt{\frac{2}{3}} \\ \Rightarrow y^{\prime \prime }\left(\tan x=\sqrt{\frac{2}{3}}\right)=2\left\{{\left(\frac{\sqrt{3}}{\sqrt{5}}\right)}^3-8{\left(\frac{\sqrt{2}}{\sqrt{5}}\right)}^2\cdot {\left(\frac{\sqrt{3}}{\sqrt{5}}\right)}^3\right.\\ \therefore -9{\left(\frac{\sqrt{2}}{\sqrt{5}}\right)}^2\cdot {\left(\frac{\sqrt{3}}{\sqrt{5}}\right)}^3+6{\left(\frac{\sqrt{2}}{\sqrt{5}}\right)}^4\sqrt{\frac{3}{5}}\\ \therefore y^{\prime \prime }\left(\tan =\sqrt{\frac{2}{3}}\right)<0\end{array} \end{gathered}$$

$$\begin{gathered} \therefore y_{\text{max }}={\left(\frac{\sqrt{2}}{\sqrt{5}}\right)}^2{\left(\frac{\sqrt{3}}{\sqrt{5}}\right)}^3 \\ =\frac{6\sqrt{3}}{25\sqrt{5}} \end{gathered}$$