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Question 7 : 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

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Video Explanation

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Explanatory Answer

To find the minimum possible value keep x and y as close as possible
So, x = y = 51
= 2601(1 + \\frac{1}{51}))(1 + \\frac{1}{51}))
= 2601(\\frac{52}{51}))(\\frac{52}{51}))
= 52 × 52
= 2704

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The question is "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"

Hence, the answer is, " 2704"