I need to find the maximum of the function $f(s,h) = 100h^{2/3}s^{1/3}$ with the condition that $20h+2000s = 20000$.
How I approached it is I derrvied $h$ from the condition and substituted it in the function $f(s,h)$ in order to have a function of one variable, so
$$20h+2000s = 20000 \quad \rightarrow \quad h = 1000-100s$$
So I received $f(s) = 100(1000-100s)^{2/3}s^{1/3}$. I am confused as to what to do next, because when I take steps as calulating the derivative $d/ds f(s) = 0$ I receive $s=10/3$ and i am not sure what to do furthermore. Can you pleasse help, thanks!
The maximum of $f$ is attained at the point where the maximum of $\left(\frac f{100}\right)^3$ is attained. But$$\left(\frac{f(s,h)}{100}\right)^3=h^2s.$$So, since $h=1000-100s$, you are after the maximum of $100^2(10-s)^2s$. Again, you can forget the factor $100^2$. And if $g(s)=(10-s)^2s=s^3-20 s^2+100 s$, then $g'(s)=100-40s+s^2$, which is equal to $0$ when (and only when) $s=10$ or $s=\frac{10}3$. But $g''(10)=20$ and $g''\left(\frac{10}3\right)=-20$. So, $g$ has a local maximum at $\frac{10}3$ and a local minimum at $10$. However, since $\lim_{s\to\pm\infty}g(s)=\pm\infty$, $g$ has no global maximum (and therefore neither does $f$).