University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Practice Exercises - Page 279: 127


$x=\dfrac{1}{\sqrt 2}$ and $y=\dfrac{1}{\sqrt e}$ and $A=\frac{1}{\sqrt{2e}}$

Work Step by Step

Let the area of a rectangle be $a=xy$ Here, $a= xe^{-x^2}$ Then $a'=e^{-x^2}-2x^2e^{-x^2}=0$ This implies that $x=\dfrac{1}{\sqrt 2}$ Now, $y=e^{-x^2}=e^{(-1/\sqrt 2)^2}=\dfrac{1}{\sqrt e}$ Hence, $x=\dfrac{1}{\sqrt 2}$ and $y=\dfrac{1}{\sqrt e}$ The maximum area is: $A=x\times y=\frac{1}{\sqrt{2e}}$
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