Chapter 3 - Applications of Differentiation - 3.7 Optimization Problems - 3.7 Exercises - Page 264: 3

$x=10$, $y=10$

Let $x,\ y $ two positive numbers such that $$xy=100$$ Then $y=\dfrac{100}{x} $ Since the sum given by $$S=x+y=x+\frac{100}{x}$$ The goal is to minimize $S$ , since $$ S'(x)=1-\frac{100}{x^2}$$ Then $S'(x)=0$ for $x=-10,\ x=10$ , since $x>0$ then we reject $x=-10$ and $S'(x)<0$ for $x<10$ , $S'(x)>0$ for $x>10$, hence $S(x)$ has minimum at $x=10$ and $y=100/x=10$