Question

If x and y are positive real numbers satisfying $x+y=102$, then the minimum possible value of $2601(1+\frac 1x)(1+\frac 1y)$ is
[This Question was asked as TITA]

• A. 2432
• B. 2807
• C. 2704
• D. 2605

Answer 1

Answer: (C)

2704

Answer 2

$AM≥GM≥HM$
$\frac {x+y}{2}≥\sqrt {xy}≥\frac {2}{\frac 1x+\frac 1y}$
Given $x+y=102$
$⇒ xy≤512 or \frac {1}{xy}≥\frac {1}{2601}$

$⇒ \frac 1x+\frac 1y≥\frac {2}{51}$
The minimum value of $2601(1+\frac 1x)(1+\frac 1y)=(2601)(1+\frac 1x+\frac 1y+\frac {1}{xy})$
$= 2601(1+\frac {2}{51}+\frac {1}{2601})$
$= 2704$

So, the correct option is (C): 2704