jmc

First, complete the square as follows: $$z=x^2+2y^2+6x-4y+22=\left(x^2+6x\right)+2\left(y^2-2y\right)+22.$$To complete the square, we need to add ${\left(\frac{6}{2}\right)}^2=9$ after the $6x$ and ${\left(\frac{2}{2}\right)}^2=1$ after the $-2y.$ So we have $$z+9+2(1)=\left(x^2+6x+9\right)+2\left(y^2-2y+1\right)+22.$$This gives $$z={\left(x+3\right)}^2+2{\left(y-1\right)}^2+11.$$Now, since ${\left(x+3\right)}^2\ge 0$ and ${\left(y-1\right)}^2\ge 0,$ the minimum value is when both squared terms are equal to $0.$ So the minimum value is $$z={\left(x+3\right)}^2+2{\left(y-1\right)}^2+11=0+2\cdot 0+11=\boxed{11}.$$