NDA 2022 Maths · Mathematics · Application of Derivatives · medium

The least value of 2x+8/x for x>0 is

Single correct option

The least value of

2x+\dfrac8x

for

x>0

Show answer and solution

Answer: (B) $8$

f'(x)=2-\frac8{x^2}=0

gives

x=2

and

f''>0

there, so the minimum is

f(2)=4+4=8

. (AM-GM gives the same:

2x+\frac8x\ge2\sqrt{16}=8

.) Putting

x=1

gives

10

, which is not the minimum.

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