A rectangular box $P$ is inscribed in a sphere of radius $r$. The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$?
$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$
Box P has dimensions $l$, $w$, and $h$. 
Its surface area is \[2lw+2lh+2wh=384,\] 
and the sum of all its edges is \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\]
The diameter of the sphere is the space diagonal of the prism, which is \[\sqrt{l^2 + w^2 +h^2}.\]
Notice that \[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,\] so the diameter is
\[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.\] The radius is half of the diameter, so
\[r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.\]