Question

An open top box is to have a rectangular base for which the length is 5 times the width and a volume of 10 cubic feet. It's five sides are to have as small a total surface area as possible. What are the sides?

Answer 1

The box should have the following dimensions (2dp):

`{(1.34, "Width of box (feet)"), (6.69, "Length of box (feet)"), (1.13, "Height of box (feet)") :}`

I'll leave as an exercise conversion to inches (if required).

Explanation:

Let us set up the following variables:

`{(w, "Width of box (feet)"), (l, "Length of box (feet)"), (h, "Height of box (feet)"), (A, "Total Surface Area (sq feet)"), (V, "Total Volume (cubic feet)") :}`

Our aim is to get the total surface area `A` as a function of one variable (`l` or `w`) and hopefully be able to minimise `A` wrt that variable.

The width/length constraint gives us:

`l=5w`

The volume constraint gives us:

`\ \ \ \ \ \ V=wlh`
`:. 10 =w(5w)h`
`:. 10 = 5w^2h`
`:. \ h = 2/w^2`

And the surface area of the box is given by:

`A = ("area base") + 2 xx ("area ends") + 2 xx ("area sides")`
`\ \ \ = wl + 2(wh) + 2(lh)`
`\ \ \ = w(5w) + 2(w)(2/w^2) + 2(5w)(2/w^2)`
`\ \ \ = 5w^2 + 4/w + 20/w`
`\ \ \ = 5w^2 + 24/w`

Which we can differentiate to find the critical points

`(dA)/(dw) = 10w-24/w^2`
`(dA)/(dw) = 0 => 10w-24/w^2 = 0`
`:. 10w=24/w^2`
`:. w^3=2.4`
`:. w=root(3)(2.4) = 1.338865 ...`

With this dimension we have:

`l = 5w = 6.694329 ...`
`h = 2/w^2 = 1.115721 ...`
`A = 26.888428 ...`
`V = 10`

We should check that this value leads to a minimum (rather than a maximum) surface area by checking that `(d^2A)/(dw^2) > 0`.

`\ \ \ \ \ \ (dA)/(dw) = 10w-24/w^2`
`:. (d^2A)/(dw^2) = 10+8/w^3 > 0 " when "w=1.338865 ...`

Hence, The box should have the following dimensions (2dp):

`{(1.34, "Width of box (feet)"), (6.69, "Length of box (feet)"), (1.13, "Height of box (feet)") :}`

I'll leave as an exercise conversion to inches (if required).