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I am trying to fit a pyramid inside of a cuboid, but maximize the dimensions of the pyramid while still remaining inside of the cuboid.

Given the dimensions of the cuboid (length, width, height), how could I calculate the dimensions of the pyramid in order to maximize it's volume inside of the cuboid?

I am looking to have the pyramid originate in the position shown in the photos below. I think the angle of rotation of the pyramid is important too. But I am not really sure how to perform the math for these calculations.

enter image description here

enter image description here

Gary
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    Wouldn't the base of the pyramid have to be a side of the cuboid? In this case the maximum volume would be $\frac 13$ of the volume of the cuboid. – Steven Alexis Gregory Oct 11 '17 at 19:25
  • I think that makes sense. If the pyramid pointing down is 0 degrees, would 45 degrees be the optimal angle for the pyramid to be pointing toward? – Gary Oct 11 '17 at 20:43
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    I suppose that is a right pyramid with square base, right? – Intelligenti pauca Oct 11 '17 at 21:23
  • The base can be on any face and the vertex can be at any point on the face of the opposite side. The volume will be $\frac13$ x base x altitude = $\frac 13 abc$. – Steven Alexis Gregory Oct 11 '17 at 21:41
  • Thanks for the volume calculation, but could you help me translate this into height and base dimensions of the pyramid? And @Aretino, yes it is a right pyramid with a square or rectangular base. – Gary Oct 12 '17 at 12:42

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There are two possible candidates for the largest right pyramid, with rectangular base, inscribed in a cuboid. The most obvious is the "upright" one, having as base a face of the cuboid and as vertex the center of the opposite face. The volume of this pyramid is ${1\over3}abc$, where $a$, $b$ and $c$ are the cuboid dimensions.

The other possible candidate is the "slanted" one, which reaches its greatest volume when its vertex $V$ is the midpoint of an edge (see diagram below: of course you need $FB\ge BC$). But it turns out that the volume of such a pyramid is, once again, ${1\over3}abc$.

enter image description here

The reason for that can also be seen in the plane: blue and red isosceles triangles in figure below have the same area, for any rectangle. Indeed, if blue triangle has base $a$ and altitude $b$, then red triangle has base $b$ (right side of the rectangle) and altitude $a$.

enter image description here

And inscribed isosceles triangles not having a side in common with the rectangle have lower area, as can be seen in the two examples above: if we divide each triangle into two smaller triangles with the dashed line, taken as common base, then the sum of the altitudes is $\le a$ and the base is $\le b$.

Intelligenti pauca
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Lets take a cube.

Now on visualizing, it becomes clear that the enclosed pyramid would have its one side along diagonal of the cube's face, which means one of the other two vertices would be on the opposite side facing downwards and remaining would be hanging or not touching any side/plane.

Which means if the cube has side of unit length then the pyramid/tetrahedron will be having that of SQRT2

And so the last or hanging vertex will be along the opposite diagonal of the cube's face mentioned above but at a distance of (SQRT3 /2) COS(PI -ATAN(SQRT2) -ACOS(1÷3) or 1/2 ​from the center and at a depth of (SQRT3 /2) SIN(ATAN(SQRT2) + ACOS(1÷3)) or 1/SQRT2 from the face of the cube.

So the converse calculations show that for a pyramid/tetrahedron of side SQRT2, the smallest cuboid or cube will be of dimensions 1 *1 *1/SQRT2

Anshuman Dwivedi
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