Supremum of $\alpha^2 z^\gamma/(\alpha + z)^2$ for $\gamma \in (0, 2)$

Question

I want to calculate $\sup\limits_{z} \left( \dfrac{\alpha}{\alpha + z} \right)^2 z^{\gamma}$ for $\gamma \in (0, 2)$. I know that the solution is $z = \dfrac{\gamma \alpha}{2 - \gamma}$ but I don't know how to get it. My idea is to build the first derivative. So if we define $f(z) = \left( \dfrac{\alpha}{\alpha + z} \right)^2 z^{\gamma}$, we have $f ' (z) = \left( \dfrac{\alpha}{\alpha + z} \right)^2 \gamma z^{\gamma - 1} - \dfrac{2 \alpha^2 z^{\gamma}}{(\alpha + z)^3}$. Now we can set $f'(z) = 0$ and solve it for $z$. But this seems difficult with all the $\gamma$...

Answer 1

This is just (changing the notation a bit) $f(x) =\dfrac{z^g}{(a+z)^2} $.

All we need here is the quotient rule.

$\begin{array}\\ f'(z) &=\dfrac{(a+z)^2gz^{g-1}-2(a+z)z^g}{(a+z)^4}\\ &=z^{g-1}\dfrac{(a+z)g-2z}{(a+z)^3}\\ \end{array} $

This has an extreme when $0 =(a+z)g-2z =ag+z(g-2) $ so $z =\dfrac{ag}{2-g} $.