Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.7 Maximum and Minimum Values - 14.7 Exercises - Page 1009: 43


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

Work Step by Step

Given: Equation of cone: $z^2=x^2+y^2$ ; point $(4,2,0)$ Need to apply Lagrange Multipliers Method to determine the shortest distance from the point $(4,2,0)$ to the plane $z^2=x^2+y^2$ we have $\nabla f=\lambda \nabla g$ Rewrite the equation of cone as: $z=\sqrt {x^2+y^2}$ The closet distance can be found as: $d=\sqrt{x-l)^2+(y-m)^2+(z-n)^2}$ or, $d=\sqrt{(x-4)^2+(y-2)^2+(x^2+y^2)}$ $f(x,y)=d^2=(x-4)^2+(y-2)^2+(x^2+y^2)$ Also, $f_x=4x-8, f_y=4y-4$ Simplify to get the values for $x$ and $y$. we have $x=2$ and $y=1$ Therefore, $z=(2)^2+(1)^2=\pm \sqrt {5}$ Hence, our result is: $(2,1,\sqrt {5}), (2,1,-\sqrt {5})$
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