Maximum of $A^tB^{1-t}$

Question

What is the maximum of $A^tB^{1-t}$ for positive $A$ and $B$ and for $t \in [0,1]$. I think the maximum occurs at end points either $t=0$ or $t=1$, right?

Answer 1

This is maximum when its log is, that is when $$ t\log A + (1-t)\log B $$is. The max is obtained at $t=0$ or $t=1$ (depending on which of $A$ and $B$ is bigger), as you thought it was.

Answer 2

$$ \ln A^tB^{1-t} = t \ln A + (1-t) \ln B $$ justifies your assessment

Answer 3

You can write the given expression as $$A^t \times B^{1-t}=\left(\frac{A}{B}\right)^tB$$ Taking the derivative (with respect to $t$) you have that $$\frac{d}{dt}\left(\frac{A}{B}\right)^tB=\left(\frac{A}{B}\right)^t B\left(\ln A - \ln B\right)=\begin{cases}>0, \quad \text{ if } A>B\\=0, \quad \text{ if } A=B \\<0, \quad \text{ if } A<B\end{cases}$$ Thus, if $A>B$ the maximum is attained for $t=1$, if $A<B$ for $t=0$ and if $A=B$ the function is constant in $t$.