Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 886: 4

$4$

We are given that $S=x+2y~~~(a)$ and $xy=2~~~(b)$ or, $y=\dfrac{2}{x}~~~(c)$ So, $S=x+(2) \dfrac{2}{x}=x+\dfrac{4}{x}$ For minimum, we must have $\dfrac{dS}{dx}=0$ or, $1-\dfrac{4}{x^2}=0 \\ x^2=4 \\ x=\pm 2$ Because $x \gt 0$, so $x=2$ Now, equation $c$ becomes: $y=\dfrac{2}{2}=1$ Thus, our equation (a) becomes: $S_{min}=2+(2)(1)=4$3$