Hello, janvdl!

I suggest you use more familiar variables . . .

The rectangular container should have a length twice the width.

It should have a vol of 1000 ml. .It should have a minimum exterior area.

Let L = length, W = width, H = height of the container.

Since the length is twice the width: .L = 2W

The box looks like this: Code:

* - - - - - - - *
/ / |
/ / | H
* - - - - - - - * |
| | *
H | | /W
| | /
* - - - - - - - *
2W

The volume of a "box" is: .V .= .L × W × H

So we have: .V .= .(2W)(W)(H) .= .1000 . → . H .= .500/W² .**[1]**

Now determine the exterior area of the box.

. . Front and back panels: .2 × (2W)(H) .= .4WH

. . Left and right panels: .2 × (W)(H) .= .2WH

. . Top and bottom panels: .2 × (2W)(W) .= .4W²

Hence, the total exterior area is: .A .= .4W² + 6WH .**[2]**

Substitute [1] into [2]: .A .= .4W² + 6W(500/W²)

. . and we have: .A .= .4W² + 3000W^{-1}

And **that** is the function we must minimize . . .