Minimum of a sum of two fractions

Text

Find the value of $x\in(0,6)$ that minimizes $y=\dfrac{9}{x}+\dfrac{9}{6-x}$, and the minimum value.

Solution

$y'=-\dfrac{9}{x^{2}}+\dfrac{9}{(6-x)^{2}}=0\Rightarrow (6-x)^{2}=x^{2}\Rightarrow x=3$ (in the interval). The value is $y(3)=3+3=6$; it is a minimum (at the endpoints $y\to+\infty$). $\boxed{x=3,\qquad y_{min}=6}$