cat 2020 Question 7

Question

If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1 + $\frac{1}{x}$ )(1 + $\frac{1}{y}$ ) is

Accepted Answer

Now we have

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

Now we know that $x + y = 102$. Substituting it in the above equation:

$2601\left(\frac{xy + y + x + 1}{xy} ight) = 2601\left(\frac{103}{xy} + 1 ight)$

Maximum value of $xy$ can be found out by AM ≥ GM relationship:

$\frac{x + y}{2} \ge \sqrt{xy} \quad ext{or} \quad \sqrt{xy} \le 51 \quad ext{or} \quad xy \le 2601$

Hence the maximum value of "xy" is 2601. Substituting in the above equation we get:

$2601\left(\frac{103 + 2601}{2601} ight) = 2704$

Question 7.