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6. Application of derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths6. Application of derivatives1 Mark
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} $
✓

Answer

Correct option: B.
$2c\sqrt {ab} $
b
(b) $xy = {c^2}$ ==> $y = \frac{{{c^2}}}{x}$ ==> $f(x) = ax + by = ax + \frac{{b{c^2}}}{x}$

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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