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Question
Let x, y be two variables and x>0, xy=1, then minimum value of x+y is _______________ .
Options
1
2
\[2\frac{1}{2}\]
\[3\frac{1}{3}\]
Solution
2
\[\text { Given }: xy = 1\]
\[ \Rightarrow y = \frac{1}{x}\]
\[f\left( x \right) = x + \frac{1}{x}\]
\[ \Rightarrow f'\left( x \right) = 1 - \frac{1}{x^2}\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 1 - \frac{1}{x^2} = 0\]
\[ \Rightarrow x^2 - 1 = 0\]
\[ \Rightarrow x^2 = 1\]
\[ \Rightarrow x = \pm 1\]
\[ \Rightarrow x = 1 ..............\left( \text { Given }: x > 1 \right)\]
`rArry=1`
\[\text { Now,} \]
\[f''\left( x \right) = \frac{2}{x^3}\]
\[ \Rightarrow f''\left( 1 \right) = 2 > 0\]
\[\text { So, x = 1 is a local minima } . \]
\[ \therefore \text { Minimum value of } f\left( x \right) = f\left( 1 \right) = 1 + 1 = 2\]
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