jmc

By Cauchy-Schwarz $$\left(\frac{49}{4}+4+4\right)(4x^2+y^2+16z^2)\ge (7x+2y+8z{)}^2.$$Since $4x^2+y^2+16z^2=1,$ $$(7x+2y+8z{)}^2\le \frac{81}{4}.$$Hence, $7x+2y+8z\le \frac{9}{2}.$

For equality to occur, we must have $\frac{2x}{7/2}=\frac{y}{2}=\frac{4z}{2}$ and $4x^2+y^2+16z^2=1.$ We can solve, to find $x=\frac{7}{18},$ $y=\frac{4}{9},$ and $z=\frac{1}{9},$ so the maximum value of $7x+2y+8z$ is $\boxed{\frac{9}{2}}.$