Minimization and Constraint (material for 163-mL juice can)

[marqade]

I'm trying to set up minimization and constraint equations. On the surface, the problem looks easy enough:

Minimize the amount of material to make a 163 ml juice can.

Okay...so I figured the area formula to be 2*Pi*r^2+2(163/r), which differentiates into 2*Pi*r+326/r^2. We'll solve it for 0, which results in 2.96 cm for the radius. Moving right along...this can is first cut from a sheet divided into squares with circles inscribed in them, which is easier but produces more waste material. Then this can is cut (more efficiently) from a sheet divided into hexagons with circles inscribed in them. How do I involve this information with the problem at the beginning of the post to minimize the amount of material in both cases? I would have assumed I simply needed to come up with a constraint equation, then provide area sums for both sheets, but there are blank spots for the new radius and height which implies that I'm missing something.

 

[stapel]

Minimize the amount of material to make a 163 ml juice can.

It is unfortunate that no other information was provided; we will have to make some assumptions. Go online to find the average thickness of metal for juice cans, and be sure to include this information within your solution. Also, we'll have to assume that this is an old-style can (needing a "church key" for accessing the juice), and we'll have to assume uniform thickness. We'll also have to absorb the rim and seals within what is over-estimated from the overlap of wall-thicknesses.

Okay...so I figured the area formula to be 2*Pi*r^2+2(163/r)...

I will guess that you actually mean "surface-area expression", with "r" being the radius of the right-cylindrical can, so that your formula actually started out as follows:

. . . . .\(\displaystyle SA(r,\,h)\, =\, 2 \pi r^2\, +\, 2 \pi h r\)

You plugged the given value into the volume formula (noting that "milliliters" is the same as "cubic centimeters"), and solved for the height h in terms of the radius r:

. . . . .\(\displaystyle 163\, =\, \pi r^2 h\)

. . . . .\(\displaystyle \frac{163}{\pi r^2}\, =\, h\)

Then you plugged this into your surface-area formula, to get your formula in terms only of the radius r:

. . . . .\(\displaystyle SA(r)\, =\, 2 \pi r^2\, +\, \frac{326}{r}\)

I don't get the same derivative (for instance, should the derivative of 2 pi r[sup:23krgzqb]2[/sup:23krgzqb] be 4 pi r?), and I'm afraid I don't follow what you're doing with the rest...? It sounds like there is a bunch of information, and possibly a graphic, for this exercise which remains missing...?

Please reply with clarification. Thank you!

Eliz.

 

[marqade]

Hmm...this is a Maple project exercise. I'll just paste the text of the instructions below...

You work for a company that produces aluminum cans. They have just received an order from a juice company for a can that will hold 5.5 fl. oz (163 mL) or their juice. You have been assigned to find the dimensions of such a can that will use the least amount of aluminum possible (and therefore, minimize the production cost).

You consider the problem. Sheets of aluminum will be used to cut out the material for each can. For the sides of the can this is a simple matter. It can be constructed from a rectangular piece of metal rolled up and joined at the ends. Once the dimensions are determined, rectangular cuts can be made from a large sheet of aluminum with no material wasted.

However, the top and bottom of each can are circular pieces. Cutting out circular pieces from a sheet of aluminum will produce waste material. Even though this waste material is not part of the final can, it is still part of producing a can, and so must be included as part of its cost (and part of its surface area).

You will look at this problem in two different ways. The first is simpler, but produces more waste material. The second is slightly more complicated and requires more cutting, but produces less waste material.

1) The simples way to cut circles out of a sheet of metal is to divide the sheet into squares and then cut out the inscribed circle within each square. In calculating the material used for top or bottom of the can, you must include the entire square, even though some of it ends up being waste material.

Minimization equation: (Here is where I put) V = Pi*r^2**h = 163; h = 163/(Pi*r^2); A = 2*Pi*r^2 + 2*Pi*r*h;
= A = 2*Pi*r(r + 163/(Pi*r^2)); = A = 2*Pi*r^2 + 2(163/r)

Constraint equation: (Here is where I put) S' = 2*Pi(r + 163/(Pi*r^2)) + 2*Pi*r(1 - 326/(Pi*r^3)); = S = 4*Pi*r - 326/r^2.

Radius: 2.96 cm
Height: Pi*r^2*h = 163; h = 163/(Pi*r^2) = 163/(Pi*2.96^2) = 5.92 cm

Amount of material used for each can: (for the ends of the can) 2r = length = 5.92; Area of square = 5.92^2 = 35.05 cm
(for the side of the can) height*circ = h*2*Pi*r = 5.92*2*Pi*5.92 = 220 cm
(total) 35.05 cm + 220 cm = 255.05 cm material for each can.


2) A slightly more complicated but more efficient design can be obtained by dividing the metal sheet into hexagons and cutting the circular tops and bottoms from the hexagons.

Before setting up the optimization problem, you need to find the area of a hexagon (in terms of the radius of the inscribed circle r). You can derive it by trying to fill the hexagon with right triangles and finding their area.

(So far I have...) Let h = r.
1/2 *b*h = A; (using 1 unit for the area); (b*r)/2 = 1; b*r = 2; b = 2/r; A(r) = 6*(1/2)*(2/r)*r

(and that's as far as I got before I became quite lost, because they now want to know the NEW radius of the more efficient can)

Minimization equation:

Constraint equation:

radius:

height:

Amount of material used for each can:

I appreciate your taking the time out to help me, thanks very much.

 

[galactus]

The area of the hexagon in terms of radius r would be: \(\displaystyle \frac{3\sqrt{3}r^{2}}{2}\)

If you break the hexagon up into 12 right triangles of height \(\displaystyle rcos(30)\) and base length \(\displaystyle rsin(30)\)

We have area of each\(\displaystyle \frac{r^{2}cos(30)sin(30)}{2}=\frac{\sqrt{3}r^{2}}{8}\)

But there are 12 of them, so we get \(\displaystyle \frac{3\sqrt{3}r^{2}}{2}\)

 

[marqade]

Okay...that makes sense. But how does that relate to the minimization equation? With the way I'm looking at things, it seems that the only information this changes is the amount of material used for each can - not the radius.

 

[galactus]

You know, I cut the hexagon out of the circle previously. We need to cut the circle out of the hexagon.

We have 12 right triangles as before. Each with angle 30 degrees and \(\displaystyle \text{Height} = r\) and \(\displaystyle \text{base}=rtan(30)=\frac{r}{\sqrt{3}}\)

So, the area of each is \(\displaystyle \frac{\sqrt{3}r^{2}}{6}\). Sinc there are 12, the area of the hexagon is \(\displaystyle 2\sqrt{3}r^{2}\)

You seem to have found the radius of the can OK. It was 2.96 cm and the height was 5.92

If you stamp this circle out of a square piece of metal, you'll get waste of \(\displaystyle (4-{\pi})r^{2}\)

If you stamp them out of a hexagonal piece of metal, you get waste of \(\displaystyle 2\sqrt{3}r^{2}-{\pi}r^{2}=(2\sqrt{3}-{\pi})r^{2}\)

So, using r=2.96, you get a circle of area 27.52. This is stamped from a hexagon of area 30.35. If it were stamped from a square piece it would have area

35.05. See how much more waste that is?.

This is less than with a square piece of metal. Of course, if you could conceivably use a 12-sided piece of metal it would be even better.

As the number of sides gets larger and larger the amount of waste will approach 0 because we are getting closer and closer to the circle area. This is how Archimedes estimated the area of a circle. By using n-gons