# Thread: A really hard Optimization problem

1. ## A really hard Optimization problem

1. An efficient packing of the discs is obtained by dividing the metal sheet into hexagons and cutting the circular lids and bases from the hexagons. Show that if this method is used, then the amount of metal used is minimized when h/r = (4 square-root of 3)/ pi= 2.21.

2. Without definition of h and r and a picture it is hard to see what you want.
Is the hexagon regular? for example.

3. ## A really hard Optimization problem

What do you mean by the definition of h and r. Can you give me an example? I know that h is height and r is the radius. I think this question is asking for a regular hexgon.

4. I mean the height is the height of what from which point of the hexagon or the sheet to which point.

5. ## A really hard Optimization problem

Well, I can describe how the picture looks like. Maybe this will help you understand the problem so that you can help me. Thanks though.
The picture: there are a total of 10 circles, 3 lies in front of 4 and another 3 lie below that 4 cirlces. These 10 circles are inside the hexagons. That means that there are 10 hexagons too, all sticks to one another. We are looking from the top of the picture not the sides.

6. ## A really hard Optimization problem

I think the equation is asking for the area that needs to be opitimize. But I dont know what formula to use for the area.

7. I'm very sorry but I can't see the optimization problem ...
but a regular hexagon is form with 6 equilateral triangles
therefore $\displaystyle r = \sqrt{\frac{h^2}{4}-\frac{h^2}{16}} =\frac{\sqrt{3}}{4} \cdot h \implies \frac{h}{r}=\frac{4}{\sqrt{3}}$
$\displaystyle A_{hex} = 6 \cdot \frac{h}{2}\frac{r}{2} = \frac{3hr}{2}$
$\displaystyle A_{cir} = \pi \cdot r^2$
$\displaystyle \frac{A_{hex}}{A_{circ}} = \frac{3hr}{2\pi \cdot r^2} = \frac{3h}{2\pi r} = \frac{3}{2\pi}\frac{4}{\sqrt{3}} =\frac{2\sqrt{3}}{\pi}$
So we miss a factor of two and I don't know if this is correct at all.

8. First you need to determine the area of the hexagon. It's fairly obvious that the sides of the hexagon will be tangent to the circle we cut in it since we want the circle to be as big as possible to waste as little metal as possible. Thus the area of the hexagon is given by

$\displaystyle A_{hex}=3rl$ where r is the radius of the circle and l is the length of each side of the hexagon. With a little geometry know-how we can find that $\displaystyle l=2rtan(30)$.

So the total area of the hexagon is $\displaystyle A_{hex}=6r^2tan(30)=2\sqrt{3} r^2$.

Now if I'm reading the problem right, h is the height of a can that we'd fashion by using the two discs as lids. So the total area that we're trying to minimize is

$\displaystyle A=4\sqrt{3}r^2+2\pi rh$

The second term comes from the height of the can (you can imagine that if we "unrolled" a cylinder we'd have a rectangle with length h and width 2*pi*r). The first term becomes 4 times the root of 3 because we have 2 lids (one on top and one on the bottom).

Now the next thing to realize, or assume, is that the volume of the can needs to be constant, otherwise the minimum area would be h=r=0. Also, the manufacturer of said can probably has a set volume of stuff to put in it. So, let's consider the volume of the can:

$\displaystyle V=\pi r^2h$

Derive wrt r:

$\displaystyle \frac{dV}{dr}=\pi (r^2\frac{dh}{dr}+2hr)$

Since V is constant, dV/dr=0, so we get:

$\displaystyle \frac{dh}{dr}=-\frac{2h}{r}$

So finally we're ready to derive A:

$\displaystyle \frac{dA}{dr}=8\sqrt{3}r+2\pi (r\frac{dh}{dr}+h)$

Set the derivative to zero (to optimize) and substitute the value of dh/dr that we found and we'll get

$\displaystyle 0=8\sqrt{3}r+2\pi (-2h+h)$
$\displaystyle 0=4\sqrt{3}r-\pi h$

$\displaystyle \frac{h}{r}=\frac{4\sqrt{3}}{\pi }$

Hope that helps.