Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.3 Maxima and Minima - 9.3 Exercises - Page 489: 30


The correct answer is (e)

Work Step by Step

$f(x,y)=y^{2}-2x^{2}y+4x^{3}+20x^{2}$ $f_x(x,y)=-4xy+12x^{2}+40x$ $f_y(x,y)=2y-2x^{2}$ $f_{xx}(x,y)=-4y+24x+40$ $f_{yy}(x,y)=2$ $f_{xy}(x,y)=-4x$ For $(-2,4)$: $f_{xx}(-2,4)=-24$ $f_{yy}(-2,4)=2$ $f_{xy}(-2,4)=8$ $D=-24\times2 - (8)^{2}=-112 \lt 0$ There is a saddle point at $(15,-8)$ For $(0,0)$: $f_{xx}(0,0)=40$ $f_{yy}(0,0)=2$ $f_{xy}(0,0)=0$ $D=40\times2 - (0)^{2}=80 \gt 0$ and $f_{xx}(0,0)=40 \gt0 \rightarrow$ There is a minimum at $(15,-8)$ For $(5,25)$: $f_{xx}(5,25)=60$ $f_{yy}(5,25)=2$ $f_{xy}(5,25)=-20$ $D=60\times2 - (-20)^{2}=-280 \gt 0 \rightarrow$ There is a saddle point at $(5,25)$ The right choice is e.
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