Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.8 Exercises - Page 942: 15

Answer

$${\text{The function }}f\left( {x,y} \right){\text{ has a relative minimum at }}\left( {0,0,0} \right)$$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} \cr & {\text{Let }}z = f\left( {x,y} \right) \cr & f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} \cr & {\text{Calculate the first partial derivatives of }}f\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\sqrt {{x^2} + {y^2}} } \right] \cr & {f_x}\left( {x,y} \right) = \frac{{2x}}{{2\sqrt {{x^2} + {y^2}} }} = \frac{x}{{\sqrt {{x^2} + {y^2}} }} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\sqrt {{x^2} + {y^2}} } \right] \cr & {f_y}\left( {x,y} \right) = \frac{{2y}}{{2\sqrt {{x^2} + {y^2}} }} = \frac{y}{{\sqrt {{x^2} + {y^2}} }} \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {f_x}\left( {x,y} \right) = 0,{\text{ }}{f_y}\left( {x,y} \right) = 0 \cr & \frac{x}{{\sqrt {{x^2} + {y^2}} }} = 0,{\text{ }}\frac{y}{{\sqrt {{x^2} + {y^2}} }} = 0 \cr & {\text{Solving the system of equations, we obtain}} \cr & x = 0,{\text{ }}y = 0 \cr & {\text{The critical point is }}\left( {0,0} \right) \cr & {\text{The function }}\sqrt {{x^2} + {y^2}} {\text{ is always positive for all real numbers.}} \cr & {\text{Because the critical point is }}\left( {0,0} \right){\text{, we can conclude that this is a}} \cr & {\text{relative minimum }} \cr & f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} \cr & f\left( {0,0} \right) = \sqrt 0 = 0 \cr & {\text{The function }}f\left( {x,y} \right){\text{ has a relative minimum at }}\left( {0,0,0} \right) \cr} $$
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