Practicing Success
Practicing Success
CUET Preparation Today
The minimum value of ax + by, where xy = r2, is (r, ab > 0) |
$2 r \sqrt{a b}$ $2 ab \sqrt{r}$ $-2 r \sqrt{a b}$ None of these |
$2 r \sqrt{a b}$ |
Let $f(x)=a x+\frac{b r^2}{x}$ $f'(x)=a-\frac{b r^2}{x}=0$ $x=\frac{\sqrt{b}}{\sqrt{a}} r$ $f\left(\frac{\sqrt{b}}{\sqrt{a}} r\right)=\frac{a \sqrt{b} r}{\sqrt{a}}+\frac{b r^2}{\sqrt{b} r} \sqrt{a}=2 r \sqrt{a b}$ |
