Differentiation in a Max/min 'tin can' problem

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Asmayus

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Hello, I'm preparing for repeat exams in August (cringe!) and was trying to solve a problem that employs Differentiation, Maxima and Minima. I'm not entirely sure if this is the correct forum; I apologise if this should be located elsewhere.

The problem is posed as follows:

A tin can manufacturer wishes to manufacture a million 500ml cans so that he uses the least amount of surface area possible. The cans are cylindrical in
shape with a top and bottom. Determine the ratio of hight of cylinder to radius, as well as the area used for each can to achieve this minimum total area.

He gave us the solution in a workshop, which I am working through. I understand the following:

r = radius of cylinder, h = Height of Cylinder

S = Area of material
S = Area of top + Area of bottom + Curved surface area
S = Pi r^2 + Pi r^2 + 2Pi rh

Thus

S = 2Pi r^2 + 2Pi rh


Volume = V = Pi r^2h

h = V/Pi r^2

Thus if S = 2Pi r^2 + 2Pi rh,

then (substituting in the above value for h and cancelling)

S = 2Pi r^2 + 2V/r

The next lines in the notes throw me:

max/min dS/dt = 0

Therefore

2Pi(2r) + 2V(-1)/r^2 = 0

I understand that he's setting it to zero as it's the max or min we're looking for, but I can't see where he gets the above line from. If anyone can point me in the right direction I will give you a cookie :)

Thanks for reading,
Asmayus
 
\(\displaystyle 500={\pi}r^{2}h\).........[1]

\(\displaystyle S=2{\pi}rh+2{\pi}r^{2}\)........[2]

Solve [1] for h and sub into [2], gives:

\(\displaystyle S=\frac{1000}{r}+2{\pi}r^{2}\)

Differentiate:

\(\displaystyle \frac{dS}{dr}=4{\pi}r-\frac{1000}{r^{2}}\)

Set to 0 and solve for r:

\(\displaystyle r=\frac{5\cdot 2^{\frac{1}{3}}}{{\pi}^{\frac{1}{3}}}\approx 4.3\)

Plug this into the height equation obtained from solving [1] for h:

\(\displaystyle h=\frac{10\cdot 2^{\frac{1}{3}}}{{\pi}^{\frac{1}{3}}}\approx 8.6\)

See?. The height is twice the radius or equal to the diameter when the optimum is achieved.

Plug r and h into [2] to find the minimum surface area
 
Question: Does anyone know why most cans of produce, a can of corn, a can of beets, a can of spaghettios, etc. does not follow the dictum that diameter should equal height to minnimize the surface area, as most cans of the above mentioned food stuffs have a diameter of about 3 inches and a height of about 4+3/16 inches. Just curious, I am assuming, unless told otherwise, that the dimensions the food industries use are the most pleasing to the eye as advertising is everything. Any takers?
 
Yes, I have noticed that as well. That is probably why they do that. I have two cans before me. One is a 15 oz. can of sliced potatoes with volume of around 30 in^3

Its diameter is 2-15/16 and height 4-7/16

I have another 27 oz. can of vegetables with diameter 3-15/16 and height 4-5/8.

They must use some other criteria than calc to come up with their dimensions.

I thought maybe the Golden Ratio(diameter to height) if aestethics are the intent, but that does not jive either.

I tried calc on these cans considering their volumes, and the solution certainly does not match what they actually are.
 
Yes, galactus, I concur.

Note:

Normal can: V = ?r²h = ?(1.5)²(67/16) ? 29.6 cu in.

Min. Surface Area can: V =?(1.676)²(3.354) ? 29.6 cu in.

At least the food industry isn't cheating us out of our vittles, however:

Normal can: Surface Area = 53.6034 sq in.

Min. Surface Area can: Surface Area = 52.969 sq in.

This represents a difference of surface area of about .6344 sq inches.

Alhough small potatos for one can, when one considers millions (if not billions) of cans it becomes quite appreciative.

Who pays for the extra surface area, not the food industry as they pass it off on the consumer.

Hence, whenever possible, only buy consumer products where the diameter equals the height of the can. I've noticed that beef stew cans usually fall under this category (diameter = height of the can).
 
In general, the top and bottom (sometimes just the top) are made of different thickness (hence cost). That throws the 1:1 ratio out of whack.
 
Yes. Also, the top and bottom are probably stamped out of some kind of sheet. Which makes for a different calc problem. How to get the most circles from a sheet metal stamping, blah, blah, blah....
 
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