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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!

This is related to CAD design in Solidworks!

$$a=0.15m$$ enter image description here

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What have you tried? –  Sreekanth Karumanaghat Apr 7 '13 at 12:31
    
@SreekanthKarumanaghat: Cut that out; this doesn't look like homework. –  Henning Makholm Apr 7 '13 at 13:57

2 Answers 2

up vote 7 down vote accepted

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.

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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! –  Jerry 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 $h = \dfrac{a}{6\sqrt3}$.

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Oh woops, yours is definitely more elegant! –  Jerry Apr 7 '13 at 11:26
    
@Jerry Thanks. It's just that I like inequalities a lot. –  Macavity Apr 7 '13 at 11:32
    
Thank you Macavity. :) Cheers! –  user31113 Apr 7 '13 at 11:44

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