Two spheres of equal radius are taken out by cutting from a solid cube with a side of (12 + 4√3) cm. What is the maximum volume (in cm3) of each sphere?
My approach: suppose side of cube =1 and let the sphere be of unequal size with diameter D and d,
$$\sqrt3=D+d+x+y....(1 )$$ enter image description here where x and y are corner distances.

$$\implies x=D/\sqrt2-D/2 $$ similarly $$y=d/\sqrt2-d/2$$
Putting in 1 and putting $$d=D$$
$$\sqrt3=2D+D\sqrt2-D=D(1+\sqrt2)$$ $$Radius =\sqrt3/2(1+\sqrt2)$$
scaling it by 12+4$\sqrt3$,
$$radius =(12\sqrt3+12)/2(1+\sqrt2)$$
Where am i getting it wrong ?

  • 3
    $\begingroup$ That line segment $x$ does not lie in the flat plane of the picture you drew. It is a segment of the main diagonal of the cube. If you include the missing third dimension, it has length $r\sqrt{3}-r$. $\endgroup$ Sep 26, 2019 at 8:30
  • $\begingroup$ It is however easiest to divide that main diagonal in three parts: from cube corner, to sphere centre 1, to sphere centre 2, to opposite cube corner. So don't start with adding diameters, but with radii. $\endgroup$ Sep 26, 2019 at 8:33
  • $\begingroup$ @JaapScherphuis the parts are not equal, how does it help me ? are you suggesting that x=corner distance =$\sqrt3$r-r? $\endgroup$ Sep 26, 2019 at 11:28
  • $\begingroup$ @JaapScherphuis how do you get $\sqrt3$ instead of $\sqrt2$ $\endgroup$ Sep 26, 2019 at 11:34
  • $\begingroup$ Calculate the distance from the corner of the cube to the centre of the sphere near that corner. This is a distance of $r$ along all three axes, so $r\sqrt{3}$. The distance $x$, which is from a corner of a cube to the surface of the sphere is $r$ less than that. (But you don't really need to calculate $x$ since it is easier to use the original distance of $r\sqrt{3}$ to the sphere's centre when solving the rest of the problem.) $\endgroup$ Sep 26, 2019 at 11:43

2 Answers 2


enter image description here

The maximum volume is achieved when the two spheres are placed along diagonal line of the cube.

Let $r$ be the radius, $d$ the diagonal of the cube and $\theta$ the angle formed by the diagonal line and the face of the cube. It follows that

$$ \cos\theta = \frac{\sqrt 2}{\sqrt 3}, \>\>\> \sin\theta = \frac{1}{\sqrt 3}$$

Then, the diagonal line calculated from the spheres is,

$$ d = 2r + 2\frac{r}{\sin\theta}$$

Given that $d= \sqrt 3 (12+4\sqrt 3)$, we get

$$2(1+\sqrt 3)r= \sqrt 3 (12+4\sqrt 3)$$

Solve for the radius

$$r= 6$$

  • $\begingroup$ how did you get value of cos(@) = root(2)/root(3)? $\endgroup$ Jan 6, 2020 at 9:58
  • $\begingroup$ @AngelusMortis - Note that the diagonal of the cube is $c=\sqrt3$ and the diagonal of the face is $a=\sqrt2$. Thus, $\cos\theta = \frac ac = \frac{\sqrt2}{\sqrt3}$ $\endgroup$
    – Quanto
    Jan 6, 2020 at 13:57
  • $\begingroup$ but isn't the diagonal of face of cube is (12 + 4√3)*root(2) $\endgroup$ Jan 6, 2020 at 14:43
  • $\begingroup$ @AngelusMortis - To be more precise. Let the side s = (12 + 4√3). Then, a = √2s and c = √3s. So, $\cos\theta=\frac{\sqrt2 s}{\sqrt3 s}=\frac{\sqrt2}{\sqrt3}$ $\endgroup$
    – Quanto
    Jan 6, 2020 at 15:06
  • $\begingroup$ Got it, Thanks a lot:) $\endgroup$ Jan 8, 2020 at 12:26

If it was 1 sphere, the radius would be $x$ inside the cube. And so the side of the cube would be $2x$. And so the diagonal of this cube would be $\sqrt{3} \cdot 2x$

And so the, going from one diagonal to the other would be [$(\sqrt{3}-1)x$] [$x$] [$x$] [$(\sqrt{3}-1)x$] ...the [$x$][$x$] part would be the 2 radii making the diameter of sphere.

If there were two spheres of same size, it would be this: [$(\sqrt{3}-1)x] [$x$] [$x$] [$x$] [$x$] [$(\sqrt{3}-1)x]. But now the length of the diagonal is $(2 \sqrt{3}+2)x$

Since the original side length was supposed to be $12+4 \sqrt{3}$, our diagonal is $(12+4 \sqrt{3})(\sqrt{3})=12 \sqrt{3}+12$.

And since the diagonal is also $(2 \sqrt{3}+2)x$, then the $x=6$. So the radius is $6$ and therefore volume is $288 \pi cm^3$

  • $\begingroup$ i can not understand your notation.plz edit your answer $\endgroup$ Sep 26, 2019 at 11:25

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