Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - Review - Exercises - Page 1024: 65


$P(2-\sqrt 3),\dfrac{P(3-\sqrt 3)}{6},\dfrac{P(2\sqrt 3-3)}{3}$

Work Step by Step

From the given figure, we have $c= b \cos \theta$ Area of a pentagon is: $A=b \cos \theta (b \sin \theta +2a)$ Perimeter of a pentagon is: $P=2 (a+b+b \cos \theta)$ Need to apply Lagrange Multipliers Method to determine the dimensions of a rectangular box of maximum volume. we have $\nabla f=\lambda \nabla g$ Using constraint condition, we get $a=b(\dfrac{\sqrt 3+1}{2})$ and $\theta = \pi/6$ Thus, $P=2 (a+b+b \cos \theta) \implies b=\dfrac{P}{2\sqrt 3+3}$ The third side can be found as: $2c =2b \cos \theta$ $=2(\dfrac{P}{2\sqrt 3+3}) \cos (\pi/6)$ $=P(2-\sqrt3)$ Hence, Sides of a pentagon are: $P(2-\sqrt 3),\dfrac{P(3-\sqrt 3)}{6},\dfrac{P(2\sqrt 3-3)}{3}$
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