Question

What are the dimensions of a box that will use the minimum amount of materials, if the firm needs a closed box in which the bottom is in the shape of a rectangle, where the length being twice as long as the width and the box must hold 9000 cubic inches of material?

Answer 1

Let's begin by putting in some definitions.

If we call `h` the height of the box and `x` the smaller sides (so the larger sides are `2x`, we can say that volume

`V=2x*x*h=2x^2*h=9000` from which we extract `h`

`h=9000/(2x^2)=4500/x^2`

Now for the surfaces (=material)

Top&bottom: `2x*x` times `2->` Area=`4x^2`
Short sides: `x*h` times `2->` Area=`2xh`
Long sides: `2x*h` times `2->` Area=`4xh`

Total area:

`A=4x^2+6xh`

Substituting for `h`

`A=4x^2+6x*4500/x^2=4x^2+27000/x=4x^2+27000x^-1`

To find the minimum, we differentiate and set `A'` to `0`

`A'=8x-27000x^-2=8x-27000/x^2=0`

Which leads to `8x^3=27000->x^3=3375->x=15`

Answer :
Short side is `15` inches
Long side is `2*15=30` inches
Height is `4500/15^2=20` inches

Check your answer! `15*30*20=9000`