## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - 9.3 Maxima and Minima - 9.3 Exercises - Page 488: 3

#### Answer

f(x) has a relative minimum at $(-3,-3)$. This relative minimum is $f(-3,-3)=-19$

#### Work Step by Step

We are given $f(x,y)=3x^{2}-4xy+2y^{2}+6x-10$ $f_{x}(x,y)=6x -4y + 6 = 0$ $f_{y}(x,y)=-4x +4y=0 \rightarrow x=y$ Use the substitution method to solve the system of equations $6(y)-4y+6=0 \rightarrow y=-3$ $x=-3$ The critical point is $(-3,-3)$ $f_{xx}(x,y)=6$ $f_{yy}(x,y)=4$ $f_{xy}(x,y)=-4$ $D=6*4-(-4)^{2}=8$ Since $D\gt0$ and $f_{xx}(-3,-3)=6\gt0$ , f(x) has a relative minimum at $(-3,-3)$. This relative minimum is $f(-3,-3)=-19$

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