Find the minimum volume box?

Steiner starts out the way I would do. Solve the continuous version of the problem, then search for, or find a way to "lock onto," the integer solution.

Three variables, two constraints -- suggests the method of Lagrange multipliers.

Let x, y, and z be the 3 edge lengths; choose any vertex, P, and:

A = 4E, where E = x + y + z = sum of the lengths of the 3 edges that meet at P

B = 2F, where F = xy + yz + zx = sum of the areas of the 3 faces that meet at P

Goal: Given E and F, minimize

V = xyz

∂/∂x(V + λF + µE) = ∂/∂y(V + λF + µE) = ∂/∂z(V + λF + µE) = 0

-- these 3, plus the 2 constraints, make 5 equations in the 5 unknowns: x, y, z, λ, µ

yz + λ(y + z) + µ = xz + λ(x + z) + µ = xy + λ(x + y) + µ = 0

xy + λ(E - z) = xz + λ(E - y) = yz + λ(E - x) = -µ

Add all three of those:

F + 2λE = -3µ

-µ = ⅓(F + 2λE)

xy + λ(E - z) = xz + λ(E - y) = yz + λ(E - x) = ⅓(F + 2λE)

xy - λz = xz - λy = yz - λx = ⅓(F - λE)

... TBC

I think this is a fine question with actual solutions. Here's why.

With dimensions x, y, z, clearly A/4 = x+y+z, B/2 = xy+yz+zx, and V = xyz. These are elementary symmetric polynomials, so we have a relation between them, namely

(x+y+z)^3 = 3(x+y+z)(xy+yz+zx) - 3xyz + x^3 + y^3 + z^3

=> (A^3)/64 = 3AB/8 - 3V + (x^3+y^3+z^3)

=> 3V = 3AB/8 + (x^3+y^3+z^3) - (A^3)/64.

For fixed A, B, we then minimize V by minimizing (x^3+y^3+z^3). Assuming there is an integer solution at all, we're reduced to a finite computation to see what the smallest V is--check all positive integer triples (x, y, z) with sum of cubes < the given sum of cubes to see if they give the required A and B. With the given example A=48, B=94, and the initial triple (3, 4, 5), we get x^3+y^3+z^3 = 216 = 6^3. Hence we need only check triples (x, y, z) with 1 <= x, y, z < 6. A brute force computer search shows that indeed the six permutations of (3, 4, 5) are the only solution. I don't know how to make the computation tractable in general or how to show when a solution doesn't exist, but the question at least seems well-defined.

If x,y and z are the lenghts of the edges, then we have an integer programming problem:

Minimize v = x y z

Subject to

x + y + z = A/4

xy + xz + yz = B/2

x, y, z positive integers.

It's been some decades I haven't dealt with integer programming, but this is usually solved with branch and bound algorithms. I don't know if we can find a formula for x, y and z as functions of A and B. This is possible if we drop the constraint that the variables should be integer and use Lagrange Multipliers.

An approach sometimes acceptable is to solve the problem supposing the variables are real and then rounding them to the nearest integers (This usually done when the variables are money. Strictly speaking, money is discrete, not continuous. There's no such things as $ 3.765439 or $ √537.) But this may violate the constraints and, even if it doesn't, nothing ensures this will lead to the optimal solution.

This is a messy algebraic question.

I won't work it out totally, but here are some pointers towards the answer.

Let's call the edges a, b ,and c

Like you say, 4a + 4b + 4c = A and 2ab + 2ac + 2bc = B

The volume of the box is very simple V = abc

On the basis that you ant to consider integer values only I suspect that the answer will be

a = 1, b = 1, and that c = A - 4a - 4b = A - 8

Let's call the edges a, b ,and c

Like you say, 4a + 4b + 4c = A and 2ab + 2ac + 2bc = B

The volume of the box is very simple V = abc

(a+b+c)² ≥ 0 or

(a+b+c)² = a² + b² +c² + 2(ab+bc+ca) ≥ 0

From equality we have

(A²)/(16) = (a² + b² +c²) +B =

good luck man