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: 49


There is a saddle point at $(\frac{13}{3},\frac{-5}{3})$

Work Step by Step

We are given $z=f(x,y)=\frac{1}{2}x^{2}+\frac{1}{2}y^{2}+2xy-x-7y+10$ $f_{x}(x,y)=x+2y-1$ $f_{y}(x,y)=y+2x-7$ Set the partial derivative equal to 0 $x+2y-1=0$ $x=1-2y$ Now substitute for x in the other equation and solve for y $y+2x-7=0$ $y+2(1-2y)-7=0$ $y=\frac{-5}{3} \rightarrow x=\frac{13}{3}$ The critical point is $(\frac{13}{3},\frac{-5}{3})$ $f_{xx}(x,y)=1$ $f_{yy}(x,y)=1$ $f_{xy}(x,y)=2$ $f_{xx}(\frac{13}{3},\frac{-5}{3}) =1$ $f_{yy}(\frac{13}{3},\frac{-5}{3})=1$ $f_{xy}(\frac{13}{3},\frac{-5}{3})=2$ $D=1\times1-(2)^{2}=-3$ Sinc D<0, there is a saddle point at $(\frac{13}{3},\frac{-5}{3})$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.