The sum of two positive numbers is 20.If the product of the square of one number and the cube of other is maximum then the numbers are

$\begin{array}{l}\mathrm{x}+\mathrm{y}=20\Rightarrow \mathrm{y}=20-\mathrm{x}\\ \mathrm{p}={\mathrm{x}}^2{\mathrm{y}}^3\Rightarrow \mathrm{p}={\mathrm{x}}^2(20-\mathrm{x}{)}^3\\ \frac{\mathrm{dp}}{\mathrm{dx}}={\mathrm{x}}^2\cdot 3(20-\mathrm{x}{)}^2(-1)+(20-\mathrm{x}{)}^3\left(2\mathrm{x}\right)\Rightarrow (20-\mathrm{x}{)}^2\left[-3{\mathrm{x}}^2+40\mathrm{x}-2{\mathrm{x}}^2\right]\\ \frac{\mathrm{dp}}{\mathrm{dx}}=(20-\mathrm{x}{)}^2\left(40\mathrm{x}-5{\mathrm{x}}^2\right)\\ For min or max \frac{\mathrm{dp}}{\mathrm{dx}}=0\\ (20-\mathrm{x}{)}^2\left(40\mathrm{x}-5{\mathrm{x}}^2\right)=0\\ \mathrm{x}=0,20,8\\ \frac{{\mathrm{dp}}^2}{{\mathrm{dx}}^2}=(20-\mathrm{x}{)}^2(40-10\mathrm{x})+\left(40x-5{\mathrm{x}}^2\right)2(20-\mathrm{x}\left)\right(-1)\\ {\left[\frac{{\mathrm{dp}}^2}{{\mathrm{dx}}^2}\right]}_{\mathrm{x}=8}=-\mathrm{ve}<0 function as maxvalue at x=8\\ \Rightarrow \mathrm{x}=8,\mathrm{y}=12\end{array}$