Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.7 Optimization Problems - 3.7 Exercises - Page 264: 4

Answer

$$x=8,\ y=8$$

Work Step by Step

Let $x,\ y $ two positive numbers such that $$x+y=16$$ Then $y=16-x $ Since the sum of squares given by $$S=x^2+y^2=x^2+(16-x)^2=2x^2-32x+256$$ The goal is to minimize $S$ , since $$ S'(x)=4x-32$$ Then $S'(x)=0$ for $x=8$ , since $S'(x)<0$ for $x<8$ , $S'(x)>0$ for $x>8$, hence $S(x)$ has minimum at $x=8$ and $y=16-x=8$
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