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!
$$a=0.15m$$
On
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 $h = \dfrac{a}{6\sqrt3}$.
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.