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

We use the inequality:  
\( AM \geq GM \geq HM \) 
Which means: 
\( \frac{x + y}{2} \geq \sqrt{xy} \geq \frac{2}{\frac{1}{x} + \frac{1}{y}} \)

Given: \( x + y = 102 \) 
Using AM ≥ GM: 
\( \sqrt{xy} \leq \frac{102}{2} = 51 \) 
So: \( xy \leq 51^2 = 2601 \) 
Hence: \( \frac{1}{xy} \geq \frac{1}{2601} \)

Using HM inequality: 
\( \frac{1}{x} + \frac{1}{y} \geq \frac{2}{51} \)

Now consider the expression: 
\( 2601 \left(1 + \frac{1}{x} \right)\left(1 + \frac{1}{y} \right) \) 
Expand this: 
\( = 2601 \left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \right) \) 
\( = 2601 \left( 1 + \left( \frac{1}{x} + \frac{1}{y} \right) + \frac{1}{xy} \right) \)

Using the minimum values from earlier: 
\( = 2601 \left( 1 + \frac{2}{51} + \frac{1}{2601} \right) \)

Simplifying: 
\( = 2601 \left( 1 + \frac{2}{51} + \frac{1}{2601} \right) = 2601 \times \left( \frac{2601 + 102 + 1}{2601} \right) = 2601 \times \frac{2704}{2601} = 2704 \)

∴ The minimum value is \( \boxed{2704} \).
Correct option is (C): 2704

Answer 2

Given: 
$x + y = 102$

We know the inequality:
Arithmetic Mean (AM) $\ge$ Geometric Mean (GM)
$\Rightarrow \frac{x + y}{2} \ge \sqrt{xy}$

Substitute the given value:
$\frac{102}{2} \ge \sqrt{xy}$
$51 \ge \sqrt{xy}$
$\Rightarrow 2601 \ge xy$

We have to find the maximum value of:
$2601\left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right)$

Use identity: $(1 + \frac{1}{x})(1 + \frac{1}{y}) = \frac{xy + x + y + 1}{xy}$
So, the expression becomes:
$= 2601 \cdot \frac{xy + x + y + 1}{xy}$

Substitute:
$x + y = 102$
$xy \le 2601$ (maximum when $xy = 2601$)

So, max value is:
$= 2601 \cdot \frac{2601 + 102 + 1}{2601}$
$= 2601 \cdot \frac{2704}{2601}$
$= \mathbf{2704}$

Final Answer: Option (C): 2704