The idea would be to suppose 2 variables $x$ and $y$ such that $x+y=20$. And let $f(x,y)=x^2y^3$(assuming $y>x$). Now convert the 2 variable function, i.e. $f(x,y)$ to a single variable function, i.e. $g(x)$ by substituting $y=20-x$ in $x^2y^3$.
Then, you proceed to find $g '(x)$ and set it to zero to obtain critical values for maximum and minimum.
Here, $g(x)=x^2(20-x)^3$ and $g '(x)= 2x(20−x)^3-3x^2(20−x)^2$.
Now, $g '(x)= 2x(20−x)^3-3x^2(20−x)^2$= $0$, which implies $x=0$ and $x=20$
Since $20$ is split into $2$ parts and $x^2y^3$ is maximum, so $x≠20$ or $ x≠0$ as the product wouldn't be maximum in that case and will equal $0$.
So, I am not able to obtain a critical value for $x$. Do I need to simplify $g '(x)$ further and then equate it to zero to obtain critical values? Would further simplification be of any significance?
Ultimately, How do I proceed further to find $x$ and $y$?
Note: I am also open to other kinds of solutions that don't involve calculus. Looking forward to other methods for answering this question?
you were correct with $$ g′(x)= 2x(20−x)^3−3x^2(20−x)^2 = 0 $$ but $ 0 $ and $ 20 $ aren't the only solutions. $$ 2x(20−x)^3−3x^2(20−x)^2 = x(20-x)^2(2(20-x) - 3x) = 0 $$ which gives you three solutions. $x = 0$, $(20-x)^2 = 0,$ with $x = 20 $ and $2(20-x) - 3x = 0$, the simplification of which gives us $40 - 2x - 3x = 0$ leading us to x = 8. $$ $$ So the answer is x = 8, y = 12.