Find-the-smallest-value-5x-16-x-21-over-positive-value-of-x-

Question Number 141136 by bobhans last updated on 16/May/21

F i n d t h e s m a l l e s t v a l u e

5 x + \frac{16}{x} + 21 o v e r p o s i t i v e

v a l u e o f x

Answered by iloveisrael last updated on 16/May/21

\Rightarrow 5 x & \frac{16}{x} h a v e a c o n s t a n t p r o d u c t

80 h e n c e t h e m i n i m u m i s a c h i e v e d

b y e q u a t i n g t h e s e 5 x = \frac{16}{x} o r

x = \sqrt{\frac{16}{5}} = \frac{4}{\sqrt{5}} . O u r c o n c l u s i o n

i s t h a t s m a l l e s t v a l u e o f

5 x + \frac{16}{x} + 21 = 10 x + 21

= \frac{40}{\sqrt{5}} + 21 = 8 \sqrt{5} + 21

Answered by EDWIN88 last updated on 16/May/21

say f (x) = 5 x + {16 x}^{- 1} + 21

f^{'} (x) = 5 - \frac{16}{x^{2}} = 0 \Rightarrow x^{2} = \frac{16}{5}

since x > 0 then x = \sqrt{\frac{16}{5}} = \frac{4 \sqrt{5}}{5}

minimum f (\frac{4 \sqrt{5}}{5}) = 4 \sqrt{5} + \frac{16^{4} . 5}{4 \sqrt{5}} + 21 = 8 \sqrt{5} + 21

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