Calculus with Applications (10th Edition)

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

Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 51


There is a saddle point at $(\frac{3}{2},\frac{-11}{8})$

Work Step by Step

We are given $z=f(x,y)=x^{3}+2xy-4x-3y-2$ $f_{x}(x,y)=3x^{2}+2y-4$ $f_{y}(x,y)=2x-3$ Set the partial derivative equal to 0 $2x-3=0$ $x=\frac{3}{2}$ Now substitute for x in the other equation and solve for y $3(\frac{3}{2})^{2}+2y-4=0$ $y=\frac{-11}{8}$ The critical point is $(\frac{3}{2},\frac{-11}{8})$ $f_{xx}(x,y)=6x$ $f_{yy}(x,y)=0$ $f_{xy}(x,y)=2$ $f_{xx}(\frac{3}{2},\frac{-11}{8}) =9$ $f_{yy}(\frac{3}{2},\frac{-11}{8})=0$ $f_{xy}(\frac{3}{2},\frac{-11}{8})=2$ $D=9\times0-(2)^{2}=-4$ Since D<0, there is a saddle point at $(\frac{3}{2},\frac{-11}{8})$
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