How to design an open cuboid iron box with the capacity of10m 3 to save the most materials?

Let the bottom length of the box be l, the bottom width be w and the height be h.

L * w * h = 10。

For rectangular iron boxes, the material is the least when the bottom is square, that is, L = W.

The required material area is: y = l * w+2 * l * h+2 * w * h = l * l+4 * l * h formula (1).

Because l * w * h = 10, that is, l * l * h = 10, that is, h =10/l/l.

Substituting into the formula 1, there is y = l * l+40/l.

That is, the derivative of y is: y' = 2l-40/l/l.

When y' = 0, there is an extreme point. At this time, L = 20 is cubic, that is, l = w = 2.714417617m.

That is, h =10/l/w =1.5808m.

At this time, materials are the most economical.