# 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?

Jan 14, 2017

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

$\left\{\begin{matrix}1.34 & \text{Width of box (feet)" \\ 6.69 & "Length of box (feet)" \\ 1.13 & "Height of box (feet)}\end{matrix}\right.$

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

#### Explanation:

Let us set up the following variables:

$\left\{\begin{matrix}w & \text{Width of box (feet)" \\ l & "Length of box (feet)" \\ h & "Height of box (feet)" \\ A & "Total Surface Area (sq feet)" \\ V & "Total Volume (cubic feet)}\end{matrix}\right.$

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:

$l = 5 w$

The volume constraint gives us:

$\setminus \setminus \setminus \setminus \setminus \setminus V = w l h$
$\therefore 10 = w \left(5 w\right) h$
$\therefore 10 = 5 {w}^{2} h$
$\therefore \setminus h = \frac{2}{w} ^ 2$

And the surface area of the box is given by:

$A = \left(\text{area base") + 2 xx ("area ends") + 2 xx ("area sides}\right)$
$\setminus \setminus \setminus = w l + 2 \left(w h\right) + 2 \left(l h\right)$
$\setminus \setminus \setminus = w \left(5 w\right) + 2 \left(w\right) \left(\frac{2}{w} ^ 2\right) + 2 \left(5 w\right) \left(\frac{2}{w} ^ 2\right)$
$\setminus \setminus \setminus = 5 {w}^{2} + \frac{4}{w} + \frac{20}{w}$
$\setminus \setminus \setminus = 5 {w}^{2} + \frac{24}{w}$

Which we can differentiate to find the critical points

$\frac{\mathrm{dA}}{\mathrm{dw}} = 10 w - \frac{24}{w} ^ 2$
$\frac{\mathrm{dA}}{\mathrm{dw}} = 0 \implies 10 w - \frac{24}{w} ^ 2 = 0$
$\therefore 10 w = \frac{24}{w} ^ 2$
$\therefore {w}^{3} = 2.4$
$\therefore w = \sqrt[3]{2.4} = 1.338865 \ldots$

With this dimension we have:

$l = 5 w = 6.694329 \ldots$
$h = \frac{2}{w} ^ 2 = 1.115721 \ldots$
$A = 26.888428 \ldots$
$V = 10$

We should check that this value leads to a minimum (rather than a maximum) surface area by checking that $\frac{{d}^{2} A}{{\mathrm{dw}}^{2}} > 0$.

$\setminus \setminus \setminus \setminus \setminus \setminus \frac{\mathrm{dA}}{\mathrm{dw}} = 10 w - \frac{24}{w} ^ 2$
$\therefore \frac{{d}^{2} A}{{\mathrm{dw}}^{2}} = 10 + \frac{8}{w} ^ 3 > 0 \text{ when } w = 1.338865 \ldots$

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

$\left\{\begin{matrix}1.34 & \text{Width of box (feet)" \\ 6.69 & "Length of box (feet)" \\ 1.13 & "Height of box (feet)}\end{matrix}\right.$

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