Answer
Maximum value is $\sqrt 2$
Minimum value is $-\sqrt 2$
Work Step by Step
$f(x,y)=\frac{1}{x}+\frac{1}{y}$; $\frac{1}{x^2}+\frac{1}{y^2}$
According to Lagrange multipliers, we have the following equations:
$\frac{1}{x^2}= \lambda (\frac{-2}{x^3})$ ...(1)
$\frac{1}{y^2}= \lambda (\frac{-2}{y^3})$... (2)
From equations (1) by (2), we get
$x=y$
Thus, $x^2=y^2=2$
Therefore,
Maximum value is $\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}=\sqrt 2$
Minimum value is $-\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}=-\sqrt 2$
