Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 12 - Functions of Several Veriables - 12.8 Maximum/Minimum Problems - 12.8 Exercises - Page 948: 18

Answer

\[\left( {1,y} \right){\text{ and }}\left( {x,0} \right)\,\,\,{\text{Where }}x{\text{ and }}y \in \,\,\mathbb{R}\]

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = {e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr & {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{{x^2}{y^2} - 2x{y^2} + {y^2}}}} \right] \cr & {f_x}\left( {x,y} \right) = \left( {2x{y^2} - 2{y^2}} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{e^{{x^2}{y^2} - 2x{y^2} + {y^2}}}} \right] \cr & {f_y}\left( {x,y} \right) = \left( {2{x^2}y - 4xy + 2y} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr & {\text{Set }}{f_x}\left( {x,y} \right) = 0{\text{ and }}{f_y}\left( {x,y} \right) = 0 \cr & {f_x}\left( {x,y} \right) = \left( {2x{y^2} - 2{y^2}} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr & \left( {2x{y^2} - 2{y^2}} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} = 0 \cr & {e^{{x^2}{y^2} - 2x{y^2} + {y^2}}}{\text{ is allways positive}}{\text{, then}} \cr & 2x{y^2} - 2{y^2} = 0 \cr & 2{y^2}\left( {x - 1} \right) = 0 \cr & x = 1,\,\,\,\,\,y = 0 \cr & \cr & and \cr & \cr & {f_y}\left( {x,y} \right) = \left( {2{x^2}y - 4xy + 2y} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr & \left( {2{x^2}y - 4xy + 2y} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} = 0 \cr & 2{x^2}y - 4xy + 2y = 0 \cr & \cr & {\text{We know that }}x = 1{\text{ and }}y = 0 \cr & {\text{For }}x = 1 \cr & 2{\left( 1 \right)^2}y - 4\left( 1 \right)y + 2y = 0 \cr & 0 = 0 \cr & {\text{Infinite solutions}}{\text{, then}} \cr & \left( {1,y} \right){\text{ Where }}y{\text{ is any real number}} \cr & and \cr & {\text{For }}y = 0 \cr & 2{x^2}\left( 0 \right) - 4x\left( 0 \right) + 2\left( 0 \right) = 0 \cr & 0 = 0 \cr & {\text{Infinite solutions}}{\text{, then}} \cr & \left( {x,0} \right){\text{ Where }}x{\text{ is any real number}} \cr & \cr & {\text{The critical point occurs at}} \cr} $$ \[\left( {1,y} \right){\text{ and }}\left( {x,0} \right)\,\,\,{\text{Where }}x{\text{ and }}y \in \,\,\mathbb{R}\]
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