Question 76

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

Correct Answer: 2704


Now we have $$2601\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=2601\left(\frac{xy+y+x+1}{xy}\right)$$

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


Maximum value of xy  can be found out by AM>= GM relationship

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

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

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

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