jmc

Let $u=x^2+2y^2.$ By AM-GM, $$u=x^2+2y^2\ge 2\sqrt{x^2\cdot 2y^2}=2xy\sqrt{2},$$so $xy\le \frac{u}{2\sqrt{2}}.$

Let $xy=ku,$ so $k\le \frac{1}{2\sqrt{2}}.$ Then from the equation $x^2-xy+2y^2,$ $$u(1-k)=8,$$and $$x^2+xy+2y^2=u(1+k)=8\cdot \frac{1+k}{1-k}.$$This is an increasing function of $k$ for $k<1,$ so it is maximized at $k=\frac{1}{2\sqrt{2}}.$ Hence, the maximum value of $x^2+xy+2y^2$ is $$8\cdot \frac{1+\frac{1}{2\sqrt{2}}}{1-\frac{1}{2\sqrt{2}}}=\frac{72+32\sqrt{2}}{7}.$$The final answer is $72+32+2+7=\boxed{113}.$