Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.8 Exercises - Page 988: 31

Answer

$(2,1,\sqrt {5}), (2,1,-\sqrt {5})$

Work Step by Step

Use Lagrange Multipliers Method: $\nabla f=\lambda \nabla g$ From the question, we have $z=\sqrt {x^2+y^2}$ ....(1) Distance $d=\sqrt{x-l)^2+(y-m)^2+(z-n)^2}$ Thus, $d=\sqrt{(x-4)^2+(y-2)^2+(x^2+y^2)}$ [From equation (1)] $F=d^2=(x-4)^2+(y-2)^2+(x^2+y^2)$ This implies, $F_x=4x-8, F_y=4y-4$ After solving for $x$ and $y$, we get $x=2$ and $y=1$ Thus, $z=x^2+y^2=2^2+1^2=\pm \sqrt {5}$ The required points are: $(2,1,\sqrt {5}), (2,1,-\sqrt {5})$
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