Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Practice Exercises - Page 440: 116

Answer

$10e$ is maximum $0$ is minimum

Work Step by Step

$\frac{dy}{dx}$ = $10x(\frac{-1}{x})+(2-\ln{x})(10)$ $\frac{dy}{dx}$ = $10-10\ln{x}$ $\frac{dy}{dx}$ = $0$ $10-10\ln{x}$ = $0$ $x$ = $e$ if $x$ = $e$ then $y$ = $10e(2-\ln{e})$ = $10e$ is maximum if $x$ = $e^{2}$ then $y$ = $10e^{2}(2-\ln{e^{2}})$ = $0$ is minimum
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