Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 886: 3

$6$

We are given that $S=x+y~~~(a)$ and $xy=9~~~(b)$ or, $y=\dfrac{9}{x}~~~(c)$ So, $S=x+\dfrac{9}{x}$ For minimum, we must have $\dfrac{dS}{dx}=0$ or, $1-\dfrac{9}{x^2}=0 \\ x^2=9 \\ x=\pm 3$ Because $x \gt 0$, so $x=3$ Now, equation $c$ becomes: $y=\dfrac{9}{3}=3$ Thus, our equation (a) becomes: $S_{min}=3+3=6$