Answer
Maximum:$f(3,1)=10$, Minimum: $f(-3,-1)=-10$
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y)=\lambda \nabla g(x,y)$
This yields $\nabla f=\lt 3,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y \gt$
Using the constraint condition $x^2+y^2=10$ we get, $3=\lambda 2x, 1=\lambda 2y$
After solving, we get $x=\pm 3$
Since, $g(x,y)=x^2+y^2=10$ gives $y=\pm 1$
Hence, Maximum:$f(3,1)=10$, Minimum: $f(-3,-1)=-10$