A cardboard box without a lid is to have a volume of 8,788 cm3. Find the dimensions that minimize the amount of cardboard used. (Let x, y, and z be the dimensions of the cardboard box.)

  x = y = 26 cm; z = 13 cm

Step-by-step explanation:

The generic solution for the minimum material in an open-top box is that the box is square and half as high as it is wide. It is half a cube of twice the volume.

The dimensions of the square base are ...

  ∛(2·8788 cm³) = 26 cm

Then the height is half that, or 13 cm.

  x = y = 26 cm; z = 13 cm

If you need to see the development, you can use the method of Lagrange multipliers to find the minimum area for the given volume;

  area = xy +2(xz +yz)

  volume = xyz = 8788

We require each of the partial derivatives of L with respect to x, y, z, and λ to be zero.

  L = xy +2(xz +yz) +λ(xyz -8788)

  partial with respect to x: 0 = y+2z +λyz

  partial with respect to y: 0 = x +2z +λxz

  partial with respect to z: 0 = 2x+2y +λxy

  partial with respect to λ: 0 = xyz -8788

From the first two equations, we have ...

  λ = (y +2z)/(yz) = 1/z +2/y

  λ = (x +2z)/(xz) = 1/z +2/x

Equating these expressions for λ, we find ...

  1/z +2/y = 1/z +2/x   ⇒   x = y

The third equation then tells us ...

  λ = (2x +2y)/(xy) = 2/y +2/x

Comparing this to either of the first two expressions for λ, we see ...

  1/z +2y = 2/x +2/y   ⇒   z = x/2

This is the result we started the answer with:

  x = y = 2z = ∛(2·8788 cm³)