Answer
Maximum:$f(\pm 1,0)=1$, Minimum: $f(0, \pm 1)=-1$
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y)=\lambda \nabla g(x,y)$
This yields $\nabla f=\lt x,-y \gt$ and $\lambda \nabla g=\lambda \lt x,y \gt$
Using the constraint condition $x^2+y^2=1$ we get, $x=\lambda x, -y=\lambda y$
After solving, we get $x=0$ and $\lambda =1$
Since, $g(x,y)=x^2+y^2=1$ yields $y=\pm 1$
When $\lambda =1$, then $x=\pm 1, y=0$
Hence, Maximum:$f(\pm 1,0)=1$, Minimum: $f(0, \pm 1)=-1$
