## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - 9.3 Maxima and Minima - 9.3 Exercises - Page 488: 10

#### Answer

The critical point is: $(10,-3)$. There is a saddle point at $(10,-3)$

#### Work Step by Step

$f(x,y)=4y^{2}+2xy+6x+4y-8$ $f_x(x,y)=2y+6=0 \rightarrow y=-3$ $f_y(x,y)=8y+2x+4=0$ Use the substitution method to solve the system of equations $f_y(x,y)=8y+2x+4=0 \rightarrow 8(-3)+2x+4=0$ $x=10$ The critical point is: $(10,-3)$ To identify any extrema: $f_{xx}(x,y)=0$ $f_{yy}(x,y)=8$ $f_{xy}(x,y)=2$ For $(10,-3)$: $f_{xx}(10,-3)=0$ $f_{yy}(10,-3)=8$ $f_{xy}(10,-3)=2$ $D=0\times8 - (2)^{2}=-4 \lt 0 \rightarrow$ There is a saddle point at $(10,-3)$

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