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# Find the Least Value of F(X) = a X + B X , Where A>0, B>0 and X>0 . - CBSE (Science) Class 12 - Mathematics

#### Question

Find the least value of f(x) = $ax + \frac{b}{x}$, where a > 0, b > 0 and x > 0 .

#### Solution

$\text { We have },$

$f\left( x \right) = ax + \frac{b}{x}$

$\Rightarrow f'\left( x \right) = a - \frac{b}{x^2}$

$\text { For a local maxima or a local minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow a - \frac{b}{x^2} = 0$

$\Rightarrow x^2 = \frac{b}{a}$

$\Rightarrow x = \sqrt{\frac{b}{a}}, - \sqrt{\frac{b}{a}}$

$\text { But, }x > 0$

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

$\text { Now },$

$f''\left( x \right) = \frac{2b}{x^3}$

$\text { At }x = \sqrt{\frac{b}{a}}$

$f''\left( \sqrt{\frac{b}{a}} \right) = \frac{2b}{\left( \sqrt{\frac{b}{a}} \right)^3} = \frac{2 a^\frac{3}{2}}{b^\frac{1}{2}} > 0 .....................\left[ \because a > 0 \text{ and }b > 0 \right]$

$\text { So }, x = \sqrt{\frac{b}{a}} \text { is a point of local minimum }.$

$\text { Hence, the least value is }$

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

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Solution Find the Least Value of F(X) = a X + B X , Where A>0, B>0 and X>0 . Concept: Graph of Maxima and Minima.
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