Answer
Maximum: $f(\dfrac{3}{\sqrt 2},- \dfrac{3}{\sqrt 2})=9+12 \sqrt2 $ and Minimum: $f(-2,2)=-8$
Work Step by Step
Our aim is to calculate the extreme values with the help of Lagrange Multipliers Method subject to the given constraints. For this, we have:$\nabla f(x,y)=\lambda \nabla g(x,y)$
This yields $\nabla f(x,y)=\lt 2x+4,2y-4 \gt$ and $\lambda g(x,y)= \lt 2x,2y \gt$
Using the constraint condition $x^2+y^2 \leq 9$ we get, $x^2=\pm \dfrac{9}{2}$ and $x= \pm \dfrac{3}{\sqrt 2}$
$\nabla f(x,y)=\lt 2x+4,2y-4 \gt$ and $\lambda g(x,y)= \lt 2x,2y \gt$ we get $y=-x$
and $y=-\dfrac{3}{\sqrt 2}, \dfrac{3}{\sqrt 2}$
Hence, Maximum: $f(\dfrac{3}{\sqrt 2},- \dfrac{3}{\sqrt 2})=9+12 \sqrt2 $ and Minimum: $f(-2,2)=-8$
