I've the following optimization problem:$$\max f(R,z)=R^2(a+z)$$$$\operatorname{sub}\begin{cases}R^2+z^2=a^2\\0\le z \le a\end{cases}$$ Once solved it gives $z=a/3$, ...
Consider now the optimization problem: $$\max f(R,z)=R(a+z)$$$$\operatorname{sub}\begin{cases}R^2+z^2=a^2\\0\le z \le a \\R\ge 0\end{cases}$$
whose solutions are $z=a/2$, ...
My question is: why does the latter problem differ from the first problem, that is why don't they give the same solutions?
The second doesn't have $R$ squared in the objective, so doesn't "feel the same pull" to make $R$ large.
If we set $a$ to $1$ (which just sets the scale) and use your constraint, the first objective is $(1-z^2)(1+z)$ while the second is $\sqrt{1-z^2}(1+z)$ This Alpha plot shows the comparison. The red curve has the square root, so pulls the maximum for $z$ higher.