MCQ
If $xy = {c^2},$ then minimum value of $ax + by$ is
- A$c\sqrt {ab} $
- ✓$2c\sqrt {ab} $
- C$ - c\sqrt {ab} $
- D$ - 2c\sqrt {ab} $
Differentiate with respect to $x$ $f'(x) = a - \frac{{b{c^2}}}{{{x^2}}}$
Put $f'(x) = 0$ ==> $a{x^2} - b{c^2} = 0$
==> ${x^2} = \frac{{b{c^2}}}{a}$ ==> $x = \pm \,\,c\sqrt {b/a} $
At $x = + \,c\sqrt {b/a,} \,\,ax + by$ will be minimum.
The minimum value $f\,\,\left( {c\sqrt {\frac{a}{b}} } \right) = a.c\sqrt {\frac{a}{b}} + \frac{{b{c^2}}}{c}.\sqrt {\frac{b}{a}} $
$= 2c\sqrt {ab} $.
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