A hemisphere is inscribed in a cube Finding the largest cube inscribed in a hemisphere has been considered here previously.  So let's consider the reverse relationship:

A hemisphere is inscribed in a cube with an edge of $1m$. What is the maximum radius of the hemisphere ?

Obviously a whole sphere of radius $\frac{1}{2}$ can be inscribed in the cube, but could a hemisphere of larger radius fit in there somehow?
 A: Extending my comment on Win Vineeth's answer:
Imagine a flat circular disk $D$ of radius $1$ centered on the line $t\mapsto(t,t,t)$, and lying in a plane  orthogonal to this line, such that it just touches the coordinate planes in three points forming an equilateral triangle. 

Doing a little coordinate geometry one computes $$M=\sqrt{2\over3}(1,1,1)\ ,$$ and the $(x,y)$-plane is touched at $\bigl(\sqrt{3\over2},\sqrt{3\over2},0\bigr)$. Erect a hemiball of radius $1$ on the "far side" of $D$. This hemiball will then fit into the cube
$$\left[0,\sqrt{2\over3}+1\right]^3\ .$$
If the cube is required to have side length $1$ the admissible radius $\rho$ of the hemiball is therefore given by  $$\rho={1\over \sqrt{2/3}+1}=3-\sqrt{6}\doteq0.5505\ .$$
A: I forward and edit this proof from a middle school math teacher
Note that a sphere inscribe to a cubiod is tangent to all the six faces of the cuboid, then a hemisphere inscribe to a cuboid should be tangent to at least three cuboid-faces adjacent to each other.
This means, the (sphere) center of the hemisphere locates at a diagonal line of the cuboid. 
Since a hemisphere is tangent to at least three cuboid faces adjacent-to-each other, and the three cuboid faces are perpendicular to each other and they have a common intersection point, let it be $O$, we can set up a 3D Descartes coordiate with $O$ as origin, and the three intersection lines of the three tangent faces as $x,y,z$-axes. 
Suppose such a hemisphere is half of a sphere with radius $r$ centered at $A$. Then the coordinate of $A$ is $(r,r,r)$; the implicit equation of the sphere/hemisphere is $(x-r)^2+(y-r)^2+(z-r)^2=r^2$.
Suppose a plane $ax+by+cz=1$ (then $ar+br+cr=1$, $a>0,b>0,c>0$)passes through $A$ and separates sphere $A$ into two halfs, one of which is what we expect to find.
**Here the $W$ is the key, but I don't really understand **
why $W$ and $a\le b, c$ can be assumed together. from the picture at the bottom, $W$ has three possibilities, for each one, we have to assume different inequalities to prove that $3-\sqrt{6}$ is the maximum $r$.
Without loss of generality, we can assume $a\le b, c$, then $a^2+b^2+c^2\le \dfrac{3}{2}(b^2+c^2)$
and it is easy to verify that point $W(x_w,y_w,z_w)=$
 $\left(\dfrac{r \sqrt{b^2+c^2}}{\sqrt{a^2+b^2+c^2}}+r,\;r-\dfrac{a b r}{\sqrt{b^2+c^2}
   \sqrt{a^2+b^2+c^2}},\;r-\dfrac{a c r}{\sqrt{b^2+c^2} \sqrt{a^2+b^2+c^2}}\right)$
is on both the plane $ax+by+cz=1$ and the hemisphere.
Easy to know that: 
$$\left(1+\sqrt{\dfrac{2}{3}}\right)r\le x_w=\dfrac{r \sqrt{b^2+c^2}}{\sqrt{a^2+b^2+c^2}}+r\le1$$
which means:
$$r\le \dfrac{1}{1+\sqrt{\dfrac{2}{3}}}=3-\sqrt{6}$$
$r$ maximizes when $a=b=c=\dfrac{1}{3r}=\dfrac{1}{3}+\dfrac{\sqrt{6}}{9}$
The results can be visualized as:


A: Maximum radius without any doubt is $\frac 12m$  
Imagine the hemisphere with the base on a side of the cube.
