The minimum value of ax + by, where xy = r2, is (r, ab > 0)

The correct answer is Option 1: $2 r \sqrt{a b}$

Let $f(x)=a x+\frac{b r^2}{x}$

$f'(x)=a-\frac{b r^2}{x}=0$

$x=\frac{\sqrt{b}}{\sqrt{a}} r$

$f\left(\frac{\sqrt{b}}{\sqrt{a}} r\right)=\frac{a \sqrt{b} r}{\sqrt{a}}+\frac{b r^2}{\sqrt{b} r} \sqrt{a}=2 r \sqrt{a b}$