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