## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - Review - Exercises - Page 1024: 62

#### Answer

Minimum value of $f(x,y,z)=x^2+2y^2+3z^2$ is $\dfrac{33}{23}$.

#### Work Step by Step

$f(x,y,z)=x^2+2y^2+3z^2$; $x+y+z=1$;$x-y+2z=2$ $x=\frac{18}{23}$ $y=-\frac{6}{23}$ $z=\frac{11}{23}$ Minimum value of $f(x,y,z)=x^2+2y^2+3z^2$ is $(\frac{18}{23})^2+2(-\frac{6}{23})^2+3(\frac{11}{23})^2=\frac{33}{23}$

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