jmc

Completing the square in $x$ and $y,$ the expression becomes $$\sqrt{x^2+400}+\sqrt{y^2+900}+\sqrt{(x-40{)}^2+(y-50{)}^2}=\sqrt{x^2+400}+\sqrt{y^2+900}+\sqrt{(40-x{)}^2+(50-y{)}^2}.$$By QM-AM, $$\begin{array}{rl}\sqrt{\frac{x^2+400}{2}}& \ge \frac{x+20}{2},\\ \sqrt{\frac{y^2+900}{2}}& \ge \frac{y+30}{2},\\ \sqrt{\frac{(40-x{)}^2+(50-y{)}^2}{2}}& \ge \frac{(40-x)+(50-y)}{2},\end{array}$$so $$\begin{array}{rl}& \sqrt{x^2+400}+\sqrt{y^2+900}+\sqrt{(40-x{)}^2+(50-y{)}^2}\\ & \ge \sqrt{2}\cdot \frac{x+20}{2}+\sqrt{2}\cdot \frac{y+30}{2}+\sqrt{2}\cdot \frac{(40-x)+(50-y)}{2}\\ & =70\sqrt{2}.\end{array}$$Equality occurs when $x=20$ and $y=30,$ so the minimum value is $\boxed{70\sqrt{2}}.$