Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.7 - Optimization Problems - 4.7 Exercises: 4

Answer

128

Work Step by Step

Let the numbers be $x$ and $y$, so $x+y=16$, and $y=16-x$. Call the sum of the squares $S$, so $S=x^{2}+(16-x)^{2}$. $S'=4x-32$, so $8$ is a critical value and minimizes the function. Therefore the minimum value of the squares is $8^{2}+(16-8)^{2}=64+64=128.$
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