# How to make a box which has the largest possible volume?

I have sheet metal in form of an equilateral triangle and I want to fold it to make a container for the screws. How should I cut and fold to make the a box with largest volume?

Basically I cut the corners and then fold them. There is no "roof". Thank you!

$$a=0.15m$$ Let the side remaining after cutting off the vertices be A. The volume is given by:

$$V = \dfrac12 A^2 \sin(60) \times h = \dfrac{\sqrt3}{4} A^2h$$

The relation between $A$, $a$ and $h$ is:

$A = a - \dfrac{2h}{\tan(30)}$

(You get a small kite at the edge with two right angles, one angle of 60 degrees and one angle at 120 degrees)

You can substitute them now:

$$V = \dfrac{\sqrt3}{4} \left(a - \dfrac{2h}{\tan(30)}\right)^2h$$

$$V = \dfrac{\sqrt3}{4} (0.15 - 2\sqrt3h)^2h$$

$$V = 3 \sqrt3 h^3-0.45 h^2+0.00974279 h$$

Differentiate w.r.t. $h$ to get:

$$\dfrac{dV(h)}{dh} = 9 \sqrt3 h^2-0.9 h+0.00974279$$

Equate to zero and solve for $h$ to get $h\approx0.0433013$ and $h\approx0.0144338$.

The first one gives the minimum volume, so you don't want that. Take the second.

• Thank you Jerry. :) Cheers! You get the right answer, because you actually gave me two value and described them really nice! I could follow the process very easily!
– user31113
Apr 7 '13 at 11:46
• You're welcome! Apr 7 '13 at 11:47

If you cut the corner in the manner shown, by trigonometry at any corner, the new side length is smaller by $2 \sqrt3 h$. (Let me know if you have difficulty with this.)

Hence the box's volume is proportional to $\left(a-2 \sqrt3 h\right)^2 h$, which we try to maximise. Let $V(h) = \left(a-2 \sqrt3 h\right)^2 (4 \sqrt3 h)$. The value of $h$ which maximises $V(h)$ is exactly same as that maximising the volume we desire, due to proportionality. Now $V(h)$ can be looked at as the product of $3$ terms, $(a-2 \sqrt3 h), (a-2 \sqrt3 h)$ and $(4 \sqrt3 h)$, which sum to a constant $2a$. Hence the product is maximised when these three terms are equal. i.e.

$a-2 \sqrt3 h = 4 \sqrt3 h$ or when $h = \dfrac{a}{6\sqrt3}$.

• Oh woops, yours is definitely more elegant! Apr 7 '13 at 11:26
• @Jerry Thanks. It's just that I like inequalities a lot. Apr 7 '13 at 11:32
• Thank you Macavity. :) Cheers!
– user31113
Apr 7 '13 at 11:44
• @Macavity Your solution is certainly neat, but I get a slight butterfly feeling when you pull that factor out of the air. I seem to get the same numbers by a method at least I find simpler. Any comments? Dec 20 '14 at 16:24

Using exactly the same argument as in this question: Optimize volume of an open cardboard box made from flat square of cardboard... we consider an infinitesimal change $\delta$ in the position of the fold between walls and floor. At the maximum, this change must add/subtract the same volume from the change in height as from the moving in/out of the walls. Call the area of the base $B$, the perimeter of the base $P$, and the height $h$, then these changes are:

$\delta \times h \times P$ ...change over walls

$\delta \times B$ ...change over base

And the maximum is when the area of the base and the area of the walls are equal, so we have:

$h = B / P$

In this case, let $s$ be the side of the base. Then the area of the base (an equilateral triangle) and perimeter are given by:

$B = \frac{\sqrt 3}{4} s^2$

$P = 3s$

So

$h = \frac{\sqrt 3}{4} s^2 / 3s = \frac{1}{4\sqrt 3} s$

Now we get $s$ from the original triangle side $a$ and substitute:

$s = a - 2\sqrt 3 h$

$h = \frac{1}{4\sqrt 3} (a - 2\sqrt 3 h)$

$4\sqrt 3 h = a - 2\sqrt 3 h$

$h = \frac{a}{6\sqrt 3}$