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

The Correct Option is C

\(AM≥GM≥HM\)
\(\frac {x+y}{2}≥\sqrt {xy}≥\frac {2}{\frac 1x+\frac 1y}\)
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

Answer 2

Given,
\(x+y=102\)
As we know,
Arithmetic Mean = \(\frac{x+y}{2}\), Geometric Mean = \(\sqrt{xy}\) and AM≥GM

\(\frac{x+y}{2} \ge \sqrt{xy}\)

\(\frac{102}{2} \ge \sqrt{xy}\)
\(51 \ge \sqrt{xy}\)
\(51^2 \ge xy\)
\(2601 \ge xy\)
Now, we have to find maximum possible value of  \(2601\bigg(1+\frac{1}{x}\bigg)\bigg(1+\frac{1}{y}\bigg)\)

\(=2601(\frac{xy+y+x+1}{xy})\)

put the value of x+y and xy
\(=2601(\frac{2601+102+1}{2601})\)
\(=2704\)

So, the correct option is (C): 2704