1708\hspace{-16}$Here $\bf{y=2(a-x).\left(x+\sqrt{x^2+b^2}\right)}$\\\\\\ Let $\bf{\left(\sqrt{x^2+b^2}+x\right)=t.........(1)}$\\\\\\ Multiply both side by $\bf{\left(\sqrt{x^2+b^2}-x\right)}$\\\\\\ $\bf{\left(\sqrt{x^2+b^2}+x\right).\left(\sqrt{x^2+b^2}-x\right)=t.\left(\sqrt{x^2+b^2}-x\right)}$\\\\\\ $\bf{\left(\sqrt{x^2+b^2}-x\right)=\frac{b^2}{t}.......(2)}$\\\\\\\ Now Sub.. $\bf{(1)-(2)}$\\\\\\ $\bf{2x=\left(t-\frac{b^2}{t}\right)}$\\\\\\ So $\bf{y=(2a-2x).t=\left(2a-t+\frac{b^2}{t}\right).t}$\\\\\\ $\bf{y=-\left(t^2-2at-b^2\right)=(a^2+b^2)-(t-a)^2\leq a^2+b^2}$\\\\\\ So $\boxed{\boxed{\bf{y_{Max.}=a^2+b^2}}}$
