Find the radius of the largest possible sphere inside a cube when two sphere of equal radii are made to sit inside a cube of side $x$. 
My approach: since it is easier to start with single sphere, do that and find the distance between the sphere and corner of the cube.
Symmetry mandates that the corner distance will not change even if there are two spheres and they will touch each other at center and we are done.
However, if someone can provide a better solution less dependent on symmetry or calculus, that will be more elegant and convincing .
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George carlin
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Your argument is incomplete, there is no prior reason the optimal solution is symmetric.
Instead, consider two spheres of radius $r$. If they can fit inside a cube of side $x$, their centers fit inside a cube of side $x - 2r$. This means the distance between the centers is at most $(x-2r)\sqrt{3}$. Since this need to be at least $2r$, we have
$$(x-2r)\sqrt{3} \ge 2r\quad\implies\quad r \le \frac{3-\sqrt{3}}{4}x$$
It is possible to fit two spheres of radius $\frac{3-\sqrt{3}}{4}x$ inside the cube by placing the centers along a diagonal. As a result, $\frac{3-\sqrt{3}}{4}x \simeq 0.316987 x$ is the largest radius and the configuration you have is indeed the optimal one.
achille hui
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suppose the spheres are along the diagonals of a cube of side x , then why should the distance between their center be ($x$-2$r$)$\sqrt {3}$? – George carlin Sep 07 '19 at 08:23
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@GeorgeCarlin if you pick two points from a cube of side $y$, the largest distance they can have is $\sqrt{3} y$. There are only 4 possible configuration that achieve this distance. Namely, when those two points are the endpoints a diagonal. So $(x-2r)\sqrt{3}$ is an upper bound. Since you want to maximize $r$, one obvious thing to check is whether placing the center at the vertices of the smaller cube works or not. It does work. – achille hui Sep 07 '19 at 08:39