The length, width and height of the rectangular parallelepiped form a geometric progression

The length, width and height of the rectangular parallelepiped form a geometric progression, the volume of the parallelepiped is 216 m3, and the sum of the lengths of all its edges is 104 m. Find the dimensions of the box.

Let A be the smallest or largest of the dimensions of a given parallelepiped, and B the denominator of the geometric progression that the dimensions of the parallelepiped form.

Then the other two dimensions of the parallelepiped are A * B and A * B ^ 2.

The volume of a parallelepiped is equal to the product of its three dimensions:

A * A * B * A * B ^ 2 = 216;

(A * B) ^ 3 = 216;

A * B = 6;

B = 6 / A.

The sum of the lengths of the edges of the parallelepiped 104:

4 * (A + A * B + A * B ^ 2) = 104;

4 * (A + 6 + 6B) = 104;

A + 6 + 6B = 26;

A – 20 + 6B = 0;

A – 20 + 36 / A = 0;

A ^ 2 – 20A + 36 = 0;

By the Vieta converse theorem A1 = 2; A2 = 18.

Then B1 = 6/2 = 3; B2 = 6/18 = 1/3.

Let’s find other dimensions of the parallelepiped for both cases:

1) A1 * B1 = 2 * 3 = 6;

A1 * B1 ^ 2 = 2 * 3 ^ 2 = 18;

2) A1 * B1 = 18 * 1/3 = 6;

A1 * B1 ^ 2 = 18 * (1/3) ^ 2 = 2.

So, the measurements of the parallelepiped are 2 m; 6 meters and 18 meters.

Answer: 2 m; 6 meters and 18 meters.