Find the minimum value of 2   x + 3   y when x y = 6 .

Given: $xy=6$

$\Rightarrow y=\frac{6}{x}$

$\Rightarrow f\left(x,y\right)=2x+3y$

$\Rightarrow f\left(x\right)=2x+3\times \frac{6}{x}$

$\Rightarrow f\left(x\right)=2x+\frac{18}{x}$

$\Rightarrow f^{\text{'}}\left(x\right)=2-\frac{18}{x^2}$

Put $f^{\text{'}}\left(x\right)=0$ for maximum or minimum

$\Rightarrow 0=2-\frac{18}{x^2}$

$\Rightarrow x^2=9$

$\Rightarrow x=\pm 3$

$\Rightarrow f^{\text{'}\text{'}}\left(x\right)=0-\left(\frac{-18\times 2}{x^3}\right)$

$\Rightarrow f^{\text{'}\text{'}}\left(3\right)=\frac{36}{3^3}>0$

$\Rightarrow f^{\text{'}\text{'}}\left(3\right)>0$

At $x=3 ,f\left(x\right)$ is minimum

The minimum value of $f\left(3\right)=2\left(3\right)+3\left(2\right) =12$.