When is $A^t+B^{1-t}$ for $t \in[0,1]$ maximized?

Question

When is $A^t+B^{1-t}$ for $t \in[0,1]$ maximized. Suppose that $A,B \in \mathbb{R}^{+}$. This is very similar to convex combination but only in exponents.

Answer 1

Since $$ f(t) = A^t+B^{1-t} = e^{t\log A}+e^{(1-t)\log B} $$ and: $$ f'(t) = \log A\cdot e^{t\log A}-\log B\cdot e^{(1-t)\log B} $$ the stationary point of $f(t)$ occurs when: $$ \log A\cdot e^{t(\log A+\log B)} = B\log B $$ or: $$ t\log(AB) = \log(B)+\log(\log B)-\log(\log A) $$ $$ t = \frac{\log(B)+\log(\log B)-\log(\log A)}{\log A+\log B}.$$ Anyway, this point is a point of minimum (and it does not always fall in $[0,1]$), since $f''(t)\geq 0$ gives that $f(t)$ is a convex function, hence it is maximized on $[0,1]$ when $t=0$ or $t=1$.

Answer 2

The function given is convex (sum of two convex functions), so the maximum is at $t=0$ or $t=1$. Just check the two endpoints.