# Two spheres of equal radius are taken out by cutting from a solid cube

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,
$$\sqrt3=D+d+x+y....(1 )$$ where x and y are corner distances.

$$x=r\sqrt2-r$$
$$\implies x=D/\sqrt2-D/2$$ 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 ?

• 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 $r\sqrt{3}-r$. Sep 26, 2019 at 8:30
• 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. Sep 26, 2019 at 8:33
• @JaapScherphuis the parts are not equal, how does it help me ? are you suggesting that x=corner distance =$\sqrt3$r-r? Sep 26, 2019 at 11:28
• @JaapScherphuis how do you get $\sqrt3$ instead of $\sqrt2$ Sep 26, 2019 at 11:34
• 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.) Sep 26, 2019 at 11:43

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

$$r= 6$$

• how did you get value of cos(@) = root(2)/root(3)? Jan 6, 2020 at 9:58
• @AngelusMortis - Note that the diagonal of the cube is $c=\sqrt3$ and the diagonal of the face is $a=\sqrt2$. Thus, $\cos\theta = \frac ac = \frac{\sqrt2}{\sqrt3}$ Jan 6, 2020 at 13:57
• but isn't the diagonal of face of cube is (12 + 4√3)*root(2) Jan 6, 2020 at 14:43
• @AngelusMortis - To be more precise. Let the side s = (12 + 4√3). Then, a = √2s and c = √3s. So, $\cos\theta=\frac{\sqrt2 s}{\sqrt3 s}=\frac{\sqrt2}{\sqrt3}$ Jan 6, 2020 at 15:06
• Got it, Thanks a lot:) 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 $$288 \pi cm^3$$

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