x > 0, y > 0 and xy = 1, find the least value of x – Class 12 Mathematics

Question 38 of 58

Q. x > 0, y > 0 and xy = 1, find the least value of x + y.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2022Subjective· 2mImportance★★★★★

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Reduce $x+y$ to a single-variable function using $xy=1$ , then minimise it using the first- and second-derivative tests.

Given $x>0$ , $y>0$ , and $xy=1$ , so $y=\dfrac{1}{x}$ .

Set up the function to minimise:

$S(x) = x+y = x+\dfrac1x, \quad x>0$

Differentiate and find critical points:

$\dfrac{dS}{dx} = 1 - \dfrac{1}{x^2}$

Set $\dfrac{dS}{dx}=0$ : $1-\dfrac{1}{x^2}=0 \Rightarrow x^2=1 \Rightarrow x=1$ (taking $x=1$ since $x>0$ ).

Second-derivative test:

$\dfrac{d^2S}{dx^2} = \dfrac{2}{x^3}$

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Q.x > 0, y > 0 and xy = 1, find the least value of x + y.