Question

An engineer must design a rectangular box that has a volume of 9 m3 and that has a bottom whose length is twice its width. What are the dimensions of the box so that the total surface area (of all six sides) of the box is minimized

Answer 1

`Length =3`
`Height = 2` and
`Width = (3)/(2)`

Step-by-step explanation:

Given

`Volume = 9m^3`

Represent the height as h, the length as l and the width as w.

From the question:

`Length = 2 * Width`

`l = 2w`

Volume of a box is calculated as:

`V = l*w*h`

This gives:

`V = 2w *w*h`

`V = 2w^2h`

Substitute 9 for V

`9 = 2w^2h`

Make h the subject:

`h = (9)/(2w^2)`

The surface area is calculated as:

`A = 2(lw + lh + hw)`

Recall that:
`l = 2w`

`A = 2(2w*w + 2w*h + hw)`

`A = 2(2w^2 + 2wh + hw)`

`A = 2(2w^2 + 3wh)`

`A = 4w^2 + 6wh`

Recall that:
`h = (9)/(2w^2)`

So:

`A = 4w^2 + 6w * (9)/(2w^2)`

`A = 4w^2 + 6* (9)/(2w)`

`A = 4w^2 + (6* 9)/(2w)`

`A = 4w^2 + (3* 9)/(w)`

`A = 4w^2 + (27)/(w)`

To minimize the surface area, we have to differentiate with respect to w

`A' = 8w - 27w^(-2)`

Set A' to 0

`0 = 8w - 27w^(-2)`

Add
`27w^(-2)` to both sides

`27w^(-2) = 8w`

Multiply both sides by
`w^2`

`27w^(-2)*w^2 = 8w*w^2`

`27 = 8w^3`

Make
`w^3` the subject

`w^3 = (27)/(8)`

Solve for w

`w = \sqrt[3]{(27)/(8)}`

`w = (3)/(2)`

Recall that :
`h = (9)/(2w^2)` and
`l = 2w`

`h = (9)/(2 * (3)/(2)^2)`

`h = (9)/(2 * (9)/(4))`

`h = (9)/((9)/(2))`

`h = 9/(9)/(2)`

`h = 9*(2)/(9)`

`h= 2`

`l = 2w`

`l = 2 * (3)/(2)`

`l = 3`

Hence, the dimension that minimizes the surface area is:

`Length =3`
`Height = 2` and
`Width = (3)/(2)`