Question

How do you find the dimensions of the box that minimize the total cost of materials used if a rectangular milk carton box of width w, length l, and height h holds 534 cubic cm of milk and the sides of the box cost 4 cents per square cm and the top and bottom cost 8 cents per square cm?

Answer 1

Note that varying the length and width to be other than equal reduces the volume for the same total (length + width); or, stated another way, `w = l` for any optimal configuration.

Using given information about the Volume, express the height (`h`) as a function of the width (`w`).

Write an expression for the Cost in terms of only the width (`w`).

Take the derivative of the Cost with respect to width and set it to zero to determine critical point(s).

Details:
Volume `= w xx l xx h = w^2 h = 534`
`rarr h = (534)/(w^2)`

Cost = (Cost of sides) + (Cost of top and bottom)
`C= (4 xx (4w xx h)) + ( 8 xx (2 w^2))`

`= (4 xx (4w xx (534)/(w^2)) + (8 xx (2 w^2))`

`= 8544 w^(-1) + 16 w^2`

`(d C)/(dw) = 0` for critical points
`- 85434 w^(-2) + 32 w = 0`
Assuming `w != 0` we can multiply by `w^2`
and with some simple numeric division:
`- 267 + w^3 = 0`
and
`w = (267)^(1/3)`
`= 6.44` (approx.)

`rarr l = 6.44`
and `h = 12.88` (approx.)