## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.7 - Extreme Values and saddle Points - Exercises 14.7 - Page 843: 8

#### Answer

Local minimum of $f(1,0)=-6$

#### Work Step by Step

Given: $f_x(x,y)=2x-2y-2=0, f_y(x,y)=-2x+4y+2=0$ Simplify the given two equations. This implies that $x=1,y=0$ Critical point : $(1,0)$ In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$. 1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum. 2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum. 3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point. $D=f_{xx}f_{yy}-f^2_{xy}=4$ and $D=4 \gt 0$ and $f_{xx}=2 \gt 0$ Hence, Local minimum of $f(1,0)=-6$

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