Chapter 14 - Partial Derivatives - Review - Exercises - Page 994: 61

Maximum value is $1$ Minimum value is $-1$

$f(x,y,z)=xyz$; $x^2+y^2+z^2=3$ According to Lagrange multipliers, we have the following equations: $yz= \lambda (2x)$ ...(1) $xz= \lambda (2y)$... (2) Divide equation (1) by (2), we get $y^2=x^2$ and $x^2=z^2$ Thus, $x^2=y^2=z^2$ Therefore,$x^2+x^2+x^2=3$ or $x^2=1$ Maximum value is $1$ Minimum value is $-1$