University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.7 - Extreme Values and Saddle Points - Exercises - Page 737: 28

Answer

Local maximum of $f(-2,0)=\dfrac{4}{e^2}$ and Saddle point at $(0,0)$

Work Step by Step

Since, we have $f_x(x,y)=e^x(x^2-2x+y^2)=0, f_y(x,y)=-2ye^{x}=0$ After solving the above two equations, we get The critical point are: $(0, 0)$ and $(-2,0)$ As per the second derivative test, for $(0,0)$ we have $D=f_{xx}f_{yy}-f^2_{xy}=-4$ and $D=-4 \lt 0$ Thus, we have: Saddle point at $(0,0)$ As per the second derivative test for $(-2,0)$, we have $D=f_{xx}f_{yy}-f^2_{xy}=4$ and $D=\dfrac{4}{e^4} \gt 0$ and $f_{xx}=\dfrac{-2}{e^2} \lt 0$ Thus, we have: Local maximum of $f(-2,0)=\dfrac{4}{e^2}$ Hence: Local maximum of $f(-2,0)=\dfrac{4}{e^2}$ and Saddle point at $(0,0)$
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