Box Optimization Problem

[jetter2]

So I think I am approaching this correctly.

I have a piece of material that is 10x16. I want to create a closed top box that will maximize volume. I approached it as you can see in my paper. The top of the box really has no variance on the volume of the box..so I treated it like the classic "open top box" problem..shouldn't that approach work?

Can someone check my work? My process was to create a Volume Function, and then solve to the relative Max.

Does this look correct to you?

 

[soroban]

Hello, jetter2!

I can't see your work . . . but here's my approach.

I have a piece of material that is 10 x 16.
I want to create a closed-top box that will maximize volume.

The sheet of material looks like this:

      : - - - - - - - 16  - - - - - - :
    - *---*-----------*---*-----------* -
    : |///:           :///:///////////| z
    : * - * - - - - - * - * - - - - - * -
    : |   :           :   :           | :
   10 |   :           :   :           | y
    : |   :           :   :           | :
    : * - * - - - - - * - * - - - - - * -
    : |///:           :///:///////////| z
    - *---*-----------*---*-----------* -
      : z : - - x - - : z : - - x - - :

The shaded regions are cut away
and the remainder is folded into a closed-top box.

          *---------------*
         /               /|
        /               / *
       /               / /
      *---------------* / y
    z |               |/
      *---------------*
              x

The volume of the box is: .$\displaystyle V \:=\:xyz$ .[1]

In the first diagram, we see that the length of the sheet is:
. . $\displaystyle 2x + 2z \:=\:16 \quad\Rightarrow\quad x \:=\:8-z$ .[2]

And the width of the sheet is:
. . $\displaystyle y + 2z \:=\:10 \quad\Rightarrow\quad y \:=\:10-2z\;\;\text{ (Note that }z \,<\,5)$ .[3]


Substitute [2] and [3] into [1]: .$\displaystyle V \;=\;(8-x)(10-2z)z \;=\;2z^3 - 26z^2 + 80z$

Then: .$\displaystyle V' \;=\;6x^2 - 52z + 80 \:=\:0 \quad\Rightarrow\quad 2(z-2)(3x-20) \:=\:0$

. . $\displaystyle \text{Hence: }\:z \;=\;2,\:\tfrac{20}{3}$

$\displaystyle \text{Since }z < 5,\,\text{ the solution is: }z = 2$

Substitute into [2]: .$\displaystyle x \:=\:8-2 \:=\:6$

Substitute into [3]: .$\displaystyle y \:=\:10-2(2) \:=\:6$


$\displaystyle \text{Therefore: }\;\text{Length} \times \text{Width} \times \text{Height} \;=\;6\times6\times2$