Answer
Maximum: $f(1,\sqrt 2,-\sqrt 2)=1+2\sqrt 2$ and Minimum: $f(1,-\sqrt 2,\sqrt 2)=1-2\sqrt 2$
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y,z)=\lambda_1 \nabla g(x,y,z)+\lambda_2 \nabla h(x,y,z)$
This yields $\nabla f=\lt 1,2,0 \gt$ and $\lambda_1 \nabla g(x,y,z)+\lambda_2 \nabla h(x,y,z)= \lt \lambda_1,-\lambda_1+2\lambda_2y,-2\lambda_2 z \gt$
Using the constraint condition $y^2+z^2=4$we get, $y=-z$
If we use the above value of $y$ in second constraint , we get $y=z=\pm \sqrt 2$
Hence, Maximum: $f(1,\sqrt 2,-\sqrt 2)=1+2\sqrt 2$ and Minimum: $f(1,-\sqrt 2,\sqrt 2)=1-2\sqrt 2$
